Lecture Outlines
You will cover a wide variety of materials during classtime, so your constant attendance is important. To help you in organizing your study materials, the list below gives an overview of the basic concepts covered during a given lecture period.
Exam 1 content
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We spent the first part of class today getting to know each other and talking about the basic motivation for this course (both why you might be taking the course, and what the course itself tries to work towards).
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Class Notes: One copy of the notes and another copy of the notes Text sections: N/A Lecture Summary: At the start of the class we talked a bit about logistical details related to the syllabus for the course, then looked at the online homework system we'll use for the course this semester.
Next we dove into some real mathematics. We observed that mathematics is one of the key lenses through which we understand the world, and that fluency in the language of mathematics is therefore essential for us to navigate our lives. In classes before calculus, one typically focuses on developing an understanding of how some basic functions behave. Recall that a function is just a way to associate some set of inputs with corresponding outputs. For example, the squaring function has the property that if you give it some input $x$, then it will return an associated output $y$. There is a lot to say about understanding how functions behave, but one of the key insights is that one can depict a function graphically. For example, the graph of the squaring function is a parabola.
Whereas math before calculus is typically focused on how one chosen input is associated to its output, in calculus we'll start to study certain relationships outputs associated to a pairs of ("nearby") inputs. In this setting, one wants to know how the "closeness" of two inputs $x_1$ and $x_2$ is related to the "closeness" of their associated outputs $y_1$ and $y_2$. Since we want to understand the relationship between these two quantities, it makes the most sense to think about them proportionally: we want to understand the change outputs relative to the change in inputs. This is the motivation for our first key concept of the semester: the average rate of change of a function.
If we have some function $f$ and some inputs $a$ and $b$, then we define the average rate of change for $f$ from $x=a$ to $x=b$ as the fraction $$\frac{\text{change in outputs}}{\text{change in inputs}} = \frac{f(b)-f(a)}{b-a}.$$ (Some people represent this quantity as $\frac{\Delta y}{\Delta x}$.) We spent some time computing average rates of change for some particular functions.
Facts to know: $$\text{avg rate of change of $f$ from $a$ to $b$ is }\frac{\Delta y}{\Delta x} = \frac{f(b)-f(a)}{b-a}$$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Consider the function $g(x) = x^3-x$. Rank the following from smallest to largest: the average rate of change from $x=-2$ to $x=2$, the average rate of change from $x=-2$ to $x=0$, and the average rate of change from $x=0$ to $x=2$. (Solution)
- I'm going on a long car trip. When I'm 100 miles into my journey it's noon, and when I'm 500 miles into my journey its 8pm. What is my average speed from noon to 8pm? (Solution)
- My friend is going on a long car trip. They tell me that over the first 200 miles of their journey, their average speed is 55mph. If their whole journey is 600 miles and their average speed over the entire 600 miles is 65mpg, what does this tell me about their average speed need to be on the last 400 miles of the journey? (Solution)
- You are hired as an economic consultant for a upstart artisanal butter manufacturing company. You have analyzed their production facilities and sales forecasts, and using this you have created a function $f$ so that if $x$ is the amount of butter they sell, then $f(x)$ is the net profit their make by selling $x$ units of butter. You are asked to present your finding to the CEO of the company.
- Explain to the CEO what it means when $f(x)$ is positive, and what it means if $f(x)$ is negative.
- You calculate the average rate of change from $x=100$ to $x=200$ to be $3.50. The CEO isn't sure what to do with this information. What do you advise?
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.1, 2.2 Lecture Summary: Last class period we studied the notion of average rate of change of a function $f$ from $x=a$ to $x=b$, and we got the formula $$\frac{\Delta y}{\Delta x} = \frac{f(b)-f(a)}{b-a}.$$ Today we saw that this quantity is the same as the slope of the secant line that connects the points $(a,f(a))$ and $(b,f(b))$. We also did some warmup problems in class that showed us that we can use this average rate of change as a way to predict what our function might be doing as it moves beyond $x=b$.
On the other hand, one problem with using average rates of change is that they tell us how outputs for a function are changing over some interval of inputs, and those intervals can be quite large! This means that while it can be quite good at capturing "global" information about how the function behaves on that interval, it's not necessarily the case that this "global" information gives us a good sense of what the function might be doing at a particular point. Instead, we hoped that we could get a deeper sense of how the function behaves at a point $x=a$ by trying to understand the instantaneous rate of change for $f$ at $x=a$. But how do we compute this? A natural first attempt is to just let $b=a$ in our formula for average rate of change, though unfortunately this leaves us with an undefined quantity $\frac{f(a)-f(a)}{a-a}=\frac{0}{0}$. Instead, our idea was to study the situations when $b$ is a number very close to $a$; this means that we want to study the case where $b =a+h$, where $h$ is some "tiny" number. If we do this, then we find that the instantaneous rate of change is approximated by $$\frac{f(a+h)-f(a)}{(a+h)-a}=\frac{f(a+h)-f(a)}{h}$$ for tiny values of $h$. In order to get a good approximation for this, we want $h$ to be really really tiny, but without ever actually plugging in $h=0$.
So how to do study this bizarre situation, where we want to know what a function does for small values, but where we aren't allowed to know what it does at $0$? To answer this, we needed to invent a new idea: the concept of a limit. (Limits are the theoretically tie that binds all of the various mathematical ideas that fall under the umbrella of calculus.) If $f$ is a function, then we say the limit of $f$ at $a$ --- denoted $\displaystyle \lim_{x \to a}f(x)$ --- is the values that outputs of $f$ are approaching as inputs approach $a$. We saw several examples of computations of limits, both when we are told about the function graphically and when we are told about the function algebraically.
Facts to know: $$\text{instantaneous rate of change of $f$ from $a$ to $b$ }\approx \frac{f(a+h)-f(a)}{h} \text{ for tiny $h$}$$
$\displaystyle \lim_{x \to a}f(x)$ is the values that outputs of $f$ are approaching as inputs approach $a$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Using a calculator, compute the average rate of change for $f(x)=x^3$ from $a=2$ to $a+h$ when
- $h=0.1$
- $h=0.01$
- $h=0.001$
- $h=0.000001$
(Solution)
- Here is a picture of two functions; the function $f(x)$ is in red, and the function $g(x)$ is in black. Compute each of
- $\displaystyle \lim_{x \to -0.5}f(x)$
- $\displaystyle \lim_{x \to -0.5}g(x)$
- $\displaystyle \lim_{x \to 1}f(x)$
- $\displaystyle \lim_{x \to 1}g(x)$
- (Challenge!) Consider the function $g(x)=x\sin\left(\frac{1}{x}\right)$. (This is a scary function!) You can find a graph of this function here so you can see what the function looks like, zoom in at particular points, etc.
- Explain why this function is undefined at $x=0$. (This just means that you should explain why you can't plug $x=0$ into this function.)
- Based on the graph, what do you expect $\lim_{x \to 0}g(x)$ is? Explain your answer.
- Using a calculator, compute the average rate of change for $f(x)=x^3$ from $a=2$ to $a+h$ when
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.2, 2.3 Lecture Summary: Last class we saw that in order to define the notion of instantaneous rate of change we will need to have some notion of limit, since in that case the instantaneous rate of change for $f$ at $a$ "wants to be" the limit of the average rate of change of $f$ from $a$ to $b$ as $b$ gets closer and closer to $a$. (Or, if you're thinking geometrically, the slope of the tangent line should be the limit of the slopes of the secant lines we get as those secant lines approach the tangent.) But to compute these limits, we need to understand better how they work. With this in mind, we spent today's class becoming more familiar with limits and how we can compute them. In our warmup exercise we thought about this question from two perspectives: an algebraic one as well as a graphical one. The former was particularly useful when we were evaluating the limit of a function which was described to us algebraically (i.e., according to some explicit formula), whereas the latter was useful when our function is instead defined graphically. We then continued by discussing some of the properties of limits, and in particular the way they interact with arithmetic operations. We wrote down a theorem that told us that if we know $$\displaystyle \lim_{x \to a} f(x) \qquad \text{ and } \qquad \lim_{x \to a}g(x),$$ then we can use this information to compute limits of related functions like $f(x)+g(x)$, $f(x)g(x)$, and so on.
We saw examples of functions which fail to have a limit at some point $a$ because inputs approach different numbers when they are getting close to $a$ "from the left" as compared to when inputs get close to $a$ "from the right." We used this as motivation for defining the direction limits $$\displaystyle \lim_{x \to a^-}f(x) \qquad \text{ and } \qquad \lim_{x \to a^+} f(x),$$ and we saw that the "two-sided limi" $\displaystyle \lim_{x \to a} f(x)$ exists if and only if the two one-sided limits $\displaystyle \lim_{x \to a^-}f(x)$ and $\displaystyle \lim_{x \to a^+}f(x)$ are equal to each other.
Facts to know: The theorem we called "arithmetic of limits"
How to determine the limit of a function described algebraically by plugging inputs that get closer and closer to the limiting point and studying their corresponding outputs
How to determine the limit of a function described geometrically by analyzing its graph
How to compute directional limits algebraically and graphically
Practice problems: No practice problems this time! -
Class Notes: One copy of the notes and another copy of the notes Text sections: 2.2, 2.3, 2.5 Lecture Summary: We spent a large part of today's class working through this handout. This gave us practice with computing limits of functions described graphically. Towards the end of the class we defined two ways in which limits can "interact with infinities" by defining what it means to say that $\displaystyle \lim_{x \to a}f(x)=\infty$, as well as defining what $\displaystyle \lim_{x\to\infty} f(x)$. (We also discussed some variants of this idea that involved $-\infty$ and/or directional limits.)
Facts to know: $\displaystyle \lim_{x \to a}f(x)=\infty$ means that as inputs get closer and closer to $a$, outputs grow without bound
(Analogous definition for $\lim_{x \to a}f(x) =-\infty$, as well as directional limit analogs)
$\displaystyle \lim_{x \to \infty} f(x)$ asks where outputs are headed as inputs get arbitrarily large
(Analogous definition for $\displaystyle \lim_{x \to -\infty} f(x)$)
Practice problems: To test your knowledge of this material, you might consider re-working the problems on the handout we worked on in class today! - Do problems (1)--(8) of today's handout. Show enough work that a friend in class can follow your line of thought. (Solution)
- Do problems (9)--(16) of today's handout. Show enough work that a friend in class can follow your line of thought. (Solution)
- Do problems (17)--(23) of today's handout. Show enough work that a friend in class can follow your line of thought. (Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.6 Lecture Summary: We started today with some practice problems that built on the ideas we have been working on in the last week: $$\lim_{x \to 1}\frac{x^2+2x+3}{2x-4} \qquad \text{ and }\lim_{x \to 1^+} \frac{x^2+1}{x-1}.$$ In the first problem, we could evaluate the limit by simply "plugging in" the limiting point. This was because our function was a fraction of two "well-behaved" (i.e., continuous) functions, and we didn't run into any scary computational red lights (namely, division by 0). This meant we could use our "arithmetic of limits" theorem to compute the overall limit in this problem as the quotient of the limits in the numerator and denomiator: $$\lim_{x \to 1}\frac{x^2+2x+3}{2x-4}=\frac{\lim_{x \to 1} x^2+2x+3}{\lim_{x \to 1}2x-4}=\frac{6}{-2}=-3.$$ For problem 2, our function was again a fraction of two well-behaved functions, but this time the limit in the denominator was $0$, and so we could no longer make use of our "arithmetic of functions" theorem. Instead, we had to more carefully analyze what each function was doing as inputs approached the limiting value from the right. We concluded that values in the numberator were getting close to $2$, whereas values in the denominator were very tiny and positive. Hence the quotient of a number close to $2$ by a number that was tiny and positive yielded a whopping large number. This allowed us to conclude $$\lim_{x \to 1^+} \frac{x^2+1}{x-1}=\infty.$$
With those problems completed, we turned our attention to a problem that we first thought about in Lecture 3: computing the instantaneous rate of change of a function $f$ at a point $a$. In lecture 3 we said that we thought this instantaneous rate of change should be the result of analyzing what happens to the average rate of change of $f$ from $x=a$ to $x=b$ as values for $b$ get closer and closer to $a$. Now that we have the language of limit in our back pocket, this means we can define the instantaneous rate of change of $f$ at $a$ to be $$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ (assuming this limit exists; if it doesn't exist, then we say that $f$ is not differentiable at $a$.) We said that a synonym for "instantaneous rate of change of $f$ at $a$" is "the derivative of $f$ at $a$," and we denoted this quantity by $f'(a)$ or $\left.\frac{dy}{dx}\right|_{x=a}.$
We finished class by doing some example problems involving derivatives as limits. In the case of $f(x)=x^2$ at $a=3$, we set up the relevant limit as $$\lim_{x \to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x \to 3}\frac{x^2-9}{x-3}.$$ Note that both the numerator and denominator of this fraction are continuous, so the limit of each of the numerator and denominator can be evaluated by just "plugging in" $x=3$. Unfortunately, if we do this, we see that the denominator has limit $0$, as does the numerator; sadly this "$\frac{0}{0}$" expression is one that is inconclusive as it stands, so to compute our desired limit we needed to do something else. Our big idea was to begin by simplifying our function $\frac{x^2-9}{x-3}$ in the hopes that we might resolve this issue. Fortunately we noticed that $x^2-9$ factors as $(x-3)(x+3)$, and so we found that $$\lim_{x \to 3} \frac{x^2-9}{x-3}=\lim_{x \to 3} x+3.$$ This latter expression is easy to evaluate since the function $x+3$ is "well behaved" everywhere; we concluded that $$f'(3)=\lim_{x\to3}\frac{x^2-9}{x-3}=\lim_{x\to 3}x+3=6.$$
At the very end of class, I asked everyone to write out $f'(-1)$ as a limit in the case that $f(x)=x^2+x-3$. We didn't quite get to complete this problem, so we will return to it on Wednesday.
Facts to know: Instantaneous rate of change for $f$ at $a$ is $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$
"Derivative of $f$ at $a$" is the same thing as "instantaneous rate of change of $f$ at $a$"
We use $f'(a)$ and $\left.\frac{dy}{dx}\right|_{x=a}$ as notations for "derivative of $f$ at $a$"
If $f$ is continuous at $a$, then $\displaystyle \lim_{x \to a}f(x)=f(a)$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Suppose that $f(x)=7$. (I.e., $f$ is the constant function with value $7$ at every input.) Write out $f'(5)$ as a limit, and then evaluate the limit carefully. (Solution)
- Suppose that $f(x)=\frac{1}{x}$. Write out $f'(2)$ as a limit, and then evaluate the limit carefully. (Solution)
- Suppose that $f(x)=|x|$. Write out $f'(0)$ as a limit, and then evaluate the limit carefully. [Hint: the absolute value function behaves one way to the left of zero and a different way to the right of zero. For this reason, you might find it easier to analyze the given limit by considering first the limit from the left and the limit from the right.] (Solution)
- (Challenge!) Suppose that $f(x)=x^3$. Write out $f'(-2)$ as a limit, and then evaluate the limit carefully. (Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.6, 2.7 Lecture Summary: At the start of class we picked up where we left off on Monday: computer $f'(-1)$ for the function $f(x)=x^2+x-3$. We had already seen last time that the definition of instantaneous rate of change gives us $$f'(-1)=\lim_{x\to -1} \frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to -1}\frac{(x^2+x-3)-(-3)}{x+1}=\lim_{x \to -1}\frac{x^2+x}{x+1}.$$ Unfortunately if we try the naive method for evaluating this limit by just plugging $-1$ into the top and the bottom, both give a limit of $0$, and so we're stuck with the uninterpretable "$\frac{0}{0}$" expression. To gain real information, we therefore need to rethink our approach to this limit. Fortunately we saw that the numerator can be factored, and this makes the computation possible: $$\lim_{x\to -1}\frac{x^2+x}{x+1}=\lim{x\to -1}\frac{x(x+1)}{x+1}=\lim_{x \to -1} x=-1.$$ Hence we found that $f'(-1)=-1$.
Next we asked what would happen to our computation if we kept the same function $f$, but this time asked for $f'(3)$. We set this up according to the definition of the derivative as a limit, and in computing this limit we followed the same basic line of thought. In this case our computation required a more subtle factorization than the one we saw in the previous problem. Since factorization can be so hard to perform, this motivated an alternative approach to computing derivatives; specifically, we saw that if you want the derivative of a function $f$ at a point $x=a$, one can use the formula $$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}.$$ We did an example to see precisely what this looked like when computing $f'(3)$ for $f(x)=x^2+x-3$.
To wrap up the class today, we noted that it can be tiresome to compute the derivative for a function at many different points: we're stuck evaluating lots and lots of limits, all of which have the same basic idea. Why not try to create something which computes $f'(a)$ for all $a$ at the same time? We can achieve this with a function $f'(x)$ which gives us the derivative of $f$ at any point $x$. Fortunately, although this function is computing the derivative of $f$ at all points simultaneously, it doesn't require any new ideas for setup or execution. We had the formula $$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h},$$ and in computing these kinds of limits we just need to use the same kinds of ideas we've already seen when computing derivatives at specific points. To showcase this, we ended class by computing $f'(x)$ for $f(x)=x^2$.
Facts to know: Alternative definition for $f'(a)$: $$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}.$$
The derivative function $f'(x)$: $$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.$$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Suppose that $f(x)=7$. (I.e., $f$ is the constant function with value $7$ at every input.) Write out $f'(x)$ as a limit, and then evaluate the limit carefully. (Solution)
- Suppose that $f(x)=\frac{1}{x}$. Write out $f'(x)$ as a limit, and then evaluate the limit carefully. (Solution)
- Suppose that $f(x)=|x|$. Write out $f'(x)$ as a limit, and then evaluate the limit carefully. [Hint: one gets different answers depending on whether $x< 0$ or $x=0$ or $x> 0$. So you might consider breaking up your analysis based on these cases.] (For solution, see notes from Lecture 9.)
- (Challenge!) Suppose that $f(x)=x^3$. Write out $f'(x)$ as a limit, and then evaluate the limit carefully. (Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.7 Lecture Summary: Today we spent the whole class period computing derivatives of functions using the definition (i.e., using limits). We computed derivatives of $f(x)=\sqrt{x}$ and $f(x)=\frac{1}{x^2}$.
Practice problems: No practice problems today! -
Class Notes: One copy of the notes and another copy of the notes Text sections: 2.8 Lecture Summary: We have thought very carefully about how one can compute the instantaneous rate of change for a function $f$ at an arbitrary point (i.e., how one can compute $f'(x)$) using limits. Our formula is $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h},$$ and we've done lots of problems that ask us to use our algebra skills to compute limits of this type for various functions. But we also know the quantity $f'(a)$ captures something geometric: we have seen before that it is the slope of the tangent line to the graph $y=f(x)$ at the point $(a,f(a))$. In today's class we worked to build on this fact so that we can start to make connections between the geometry of the graph of $f(x)$ and qualitative properties of $f'(x)$.
We started by computing $f'(x)$ for the absolute value function $f(x)=|x|$, with the approach we used again relying on algebraic computations. We found that $$f'(x) = \left\{\begin{array}{ll}1,&\text{ if }x> 0\\-1,&\text{ if }x< 0 \end{array}\right.$$ But what does that mean when we look at the graph of $y=|x|$ itself? We noticed that for any positive $a$, the tangent line to $y=|x|$ at $(a,|a|)$ is just the line of slope $1$ passing through $(a,|a|)$, simply because the graph of $y=|x|$ is itself a line of slope $1$ at this point! Likewise if $a< 0$, then the tangent line to $y=|x|$ at $(a,|a|)$ is a line of slope $-1$ passing through $(a,|a|)$. The takeaway was this: if the graph of $y=f(x)$ at a point $(a,f(a))$ looks like a line segment passing through the point $(a,|a|)$, then the slope of that line segment is the value of $f'(a)$.
In that same example we saw that the point $(0,|0|)$ on the graph of $y=|x|$ is "pointy," since its the place where two line segments of different slope are intersecting. Because we have a "pointy point" at $(0,|0|)$, there's no way to draw a single tangent line at this point. For this reason, we saw that if the graph of a function $y=f(x)$ has a "pointy point" at $(a,f(a))$, then $f'(a)$ does not exist.
Building on ideas like this, we explored more ideas in this ilk. We saw, for example, that if the graph of $y=f(x)$ "flattens out" at some point $(a,f(a))$, then we know $f'(a)=0$. On the other hand, if the tangent line at the point $(a,f(a))$ is a line of positive slope (i.e., if it's a line that moves from bottom-left to top-right), then $f'(a)>0$; if instead the tangent line at $(a,f(a))$ is a line of negative slope (i.e., a line that moves from top-left to bottom-right), then $f'(a)< 0$. We can even say more: if a function $f(x)$ is increasing along an interval $[a,b]$, then $f'(x)$ will be positive for values of $x$ in $[a,b]$; a similar statement tells us that if $f(x)$ is decreasing along an interval $[a,b]$, then $f'(x)$ is negative along that interval.
Facts to know: If the graph of $y=f(x)$ at $(a,f(a))$ is a line segment of slope $m$, then $f'(a)=m$
If the graph of $y=f(x)$ at $(a,f(a))$ is a "pointy point," then $f'(a)$ does not exist
If the graph of $y=f(x)$ "flattens out" at $(a,f(a))$, then $f'(a)=0$
If the graph of $y=f(x)$ is increasing along an interval $[a,b]$, then $f'(x)$ is positive along that interval
If the graph of $y=f(x)$ is decreasing along an interval $[a,b]$, then $f'(x)$ is negative along that interval
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Consider the graph of the function $f(x)$ shown in this image. Plot $f'(x)$. (Solution)
- Consider the graph of the function $f(x)$ shown in this image. For what values of $a$ is $f'(a)=0$? For what values of $a$ is $f'(a)$ positive? For what values of $a$ is $f'(a)$ negative?(Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 2.8 Lecture Summary: We continued to build on our discussion connecting derivative information and the geometry of the graph of a function $f(x)$. Today we explored how the concavity of the graph of $f$ is influenced by the positivity or negativity of the second derivative of $f$. Specifically we noted that if $f''(x)$ is positive on an interval, then the graph of the function $y=f(x)$ will be concave up along this interval. (Similarly if $f''(x)$ is negative along an interval, then the graph of $f$ will be concave down.) We saw how we could use this information, together with the information from last class period, to give sketches of graphs based on derivative information.
As a final thought on the geometry of derivatives, we noted that since $f'(a)$ measures the slope of the tangent line to $y=f(x)$ at $(a,f(a))$, we can use this information (and previous know-how related to the "point-slope" form of a line) to write out an equation that expresses the tangent line at a given point. Specifically, the equation of the line tangent to the graph of $f(x)$ at a point $(a,f(a))$ is given by $$y=f(a)+f'(a)(x-a).$$ We put this information to use by writing out some equations of tangent lines.
Facts to know: If $f'(x)$ is increasing along an interval $[a,b]$, then the graph of $y=f(x)$ is concave up along that interval
If $f'(x)$ is decreasing along an interval $[a,b]$, then the graph of $y=f(x)$ is concave down along that interval
$f''(x)$ means the derivative of $f'(x)$
If $f''(x)$ is positive along an interval $[a,b]$, then the graph of $y=f(x)$ is concave up along that interval
If $f''(x)$ is negative along an interval $[a,b]$, then the graph of $y=f(x)$ is concave down along that interval
The tangent line to $y=f(x)$ at $(a,f(a))$ has an equation given by $$y=f(a)+f'(a)(x-a)$$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Consider the graph of the function $f(x)$ shown in this image. Along what intervals is $f''(x)$ positive? along what intervals is $f''(x)$ negative? When is $f''(x)=0$? (Solution)
- Suppose you are told that a given function $f(x)$ has the property that $f(3)=2$, and also that
- $f'(5)=f'(9)=0$
- $f'(x)$ is positive between $x=3$ and $x=5$, and again between $x=9$ and $x=12$
- $f'(x)$ is negative between $x=5$ and $x=9$
- $f''(x)$ is negative between $x=3$ and $x=7$
- $f''(x)$ is positive between $x=7$ and $x=12$
- During lecture 7 we computed the derivative of $f(x)=x^2+x-3$. Use this information to compute the equation of the tangent line to $f(x)$ at the point $(1,-1)$. (Solution)
- Compute $\frac{d}{dx}\left[x^3\right]$, then use this to give the equation for the tangent line to $y=f(x)$ when $x=2$. (Solution)
Exam 2 content
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.1, 3.3 Lecture Summary: Our main point of focus for the last many class periods has been to understand the derivative of a function. We already know this derivative is defined in terms of limits; specifically, we have $$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}.$$ And while we have seen how to compute these limits in a lot of different situations, there is no denying that each computation requires its own set of ideas to unravel. This is fine as we're getting started with examining derivatives of relatively simple functions, but what happens when our functions get more complicated? Are we expected to break out increasingly complicated techniques every time we're met with a new function whose derivative we want to study? It would be much better if we had a set of rules we could follow that would let us decode derivatives easily (ie, without wresting with limits). Ideally these rules would give us a library of derivatives of "basic functions," as well as the ways that derivatives interact with the basic operations we use to combine functions into new, more complicated functions.
With this as our motivation, today we started to uncover some of these rules. Our first such rule as a theorem we called the linearity of the derivative, which tells us that for any functions $f(x)$ and $g(x)$, and for any constant $c$, we have $$\frac{d}{dx}\left[f(x)+g(x)\right] = f'(x)+g'(x) \hspace{.5in} \text{ and } \hspace{.5in} \frac{d}{dx}\left[cf(x)\right]=cf'(x).$$ We also gave a rule which lets us compute derivatives for power functions (i.e., those of the form $f(x)=x^n$ for some numeric value $n$); this "power rule" says that $$\frac{d}{dx}\left[x^n\right]=nx^{n-1}.$$ Finally, we discussed some derivatives of familiar trigonometric functions: $$ \begin{align*} \frac{d}{dx}\left[\sin(x)\right]&=\cos(x)\\ \frac{d}{dx}\left[\cos(x)\right]&=-\sin(x)\\ \frac{d}{dx}\left[\tan(x)\right]&=\sec^2(x)\\ \frac{d}{dx}\left[\sec(x)\right]&=\sec(x)\tan(x)\\ \end{align*}$$
We put a number of these principles into practice with some sample calculations.
Facts to know: The theorem we called "linearity of the derivative"
The theorem we called "the power rule"
The theorem which gives us the derivatives of 4 of the trigonometric functions
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Compute $\frac{d}{dx}\left[6x^7-4x^5+2x^2-\frac{1}{x^8}\right]$. (Solution pending)
- What is the equation of the line tangent to $y=\root{3}\of{x^2}+x-\frac{1}{x}$ at the point corresponding to $x=1$? (Solution)
- At what points $x$ does the function $f(x)=3x^4-4x^2-12x^2+7$ have a tangent line that is horizontal?
- At what points $x$ does the function $f(x)=\frac{1}{3}x^3-\frac{1}{2}x^2-4x$ have a tangent line that is parallel to the line $y=2x+4$? [Note: Two lines are parallel if they have the same slope.] (Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.1, 3.2 Lecture Summary: Last time we began the work of finding some "quick rules" that let us compute derivatives without having to go through the hassle of evaluating the corresponding limit. In today's class we continued this work, uncovering rules for computing derivatives of exponential functions, products of functions, and quotients of functions.
An exponential function is a function of the form $f(x)=a^x$, where $a$ is a positive number. (Notice that this format is similar to --- but meaningfully different from --- what we see in power functions. A power function features a variable (like $x$) raised to a constant, whereas an exponential function features a constant raised to a variable.) We saw that $$\frac{d}{dx}\left[a^x\right] = (\ln(a)) a^x,$$ where here $\ln(a)$ means the natural logarithm of $a$ (i.e., the logarithm of $a$ with $e$, where $e \approx 2.71\cdots$).
We then went on to give rules for derivatives of products and quotients of functions. These were the so-called "product rule" and "quotient rule." We saw $$ \begin{align*} \frac{d}{dx}\left[f(x)g(x))\right]&=f(x)g'(x)+f'(x)g(x)\\ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]&=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}.\\ \end{align*}$$
We put a number of these principles into practice with some sample calculations.
Facts to know: The difference between an exponential function and a power function
The formula for computing the derivative of an exponential
The theorem we called "the product rule"
The theorem we called "the quotient rule"
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Compute $\frac{d}{dx}\left[x^{10}-10^x\right]$
- Compute $\frac{d}{dx}\left[2\cos(x)\sqrt{x}\right]$ (Solution)
- A manufacturer produces quantity $q$ of a given good, but the price they can charge depends on $q$ (e.g., if they produce very little of the product, then demand will far outweigh supply, and so they can charge a premium. So we write the price as $p(q)$ to record the fact that price depends on quantity. Likewise the cost of production depends on $q$, so we write this cost as $c(q)$. The overall revenue from $q$ sales is therefore $qp(q)$, and the net profit is $n(q) = qp(q)-c(q)$.
- Give a formula for $n'(q)$.
- Explain what $n'(q)$ measures. Based on your explanation, what business decision should the manufacturing company make if $n'(1000)=10$? What if $n'(1000)=-5$?
[You may assume that the manufacturer is primarily interested in maximizing their net profit.]
- Compute $\displaystyle \frac{d}{dx}\left[\frac{\root{3}\of{x}\cos(x)}{x^2e^x}\right]$.
[Hint: You might want to do a little algebra before diving into calculus.]
(Solution)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.4 Lecture Summary: Today we introduced our last big result we have for computing derivatives: the chain rule. The chain rule gives us a way to compute derivatives of functions which are expressed as a composition of other functions. Recall that a composite function is one where we have plugged the outputs of one function into another function. The chain rule tells us that for functions $f(x)$ and $g(x)$, the derivative of the composition $(f\circ g)(x)$ is $$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x))g'(x).$$
We saw this rule being implemented in a number of different situations, including some where we had a triple composition.
Facts to know: How to recognize when a function can be written as a composition (and identifying the "outer" and "inner" parts of that composition)
The theorem we called "the chain rule"
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Compute $\frac{d}{dx}\left[e^x \sin((x+3)^4) \right]$ (Solution covered in class on Lecture 14)
- Compute $\frac{d}{dx}\left[\frac{\tan(x)-\sqrt{\sin(x)}}{x^2-x+5}\right]$ (Solution)
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Class Notes: One copy of the notes Text sections: 3.1--3.4 Lecture Summary: At the start of class we did the first of the two example problems from last class period. We then spent time working individually and with friends on this worksheet.
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Problem 1 from the "Warmup" section of the in-class handout. (Solution pending)
- Problem 2 from the "Warmup" section of the in-class handout.
- Problem 3 from the "Warmup" section of the in-class handout.
- Problem 4 from the "Warmup" section of the in-class handout. (Solution)
- Problem 5 from the "Warmup" section of the in-class handout.
- Problem 6 from the "Warmup" section of the in-class handout. (Solution)
- Problem 1 from the "Grab Bag" section of the in-class handout.
- Problem 2 from the "Grab Bag" section of the in-class handout.
- Problem 3 from the "Grab Bag" section of the in-class handout.
- Problem 4 from the "Grab Bag" section of the in-class handout. (Solution)
- Problem 5 from the "Grab Bag" section of the in-class handout.
- Problem 6 from the "Grab Bag" section of the in-class handout.
- Problem 7 from the "Grab Bag" section of the in-class handout. (Solution)
- Problem 8 from the "Grab Bag" section of the in-class handout.
- Problem 9 from the "Grab Bag" section of the in-class handout.
- Problem 10 from the "Grab Bag" section of the in-class handout.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.1--3.4, 3.5, 3.7 Lecture Summary: Today we spent the first have of class practicing derivative computations using the worksheet from last class period. Next, we introduced the notion of implicit differentiation. Whereas we typically think about how to compute slopes of tangent lines to graphs of functions $y=f(x)$, the idea behind implicit differentiation is that we could equally well be interested in the slope of the line tangent to some curve that is not the graph of a function. In such a situation, the curve (or, more specifically, the equation that gives rise to the curve) gives us some relationship between the variables $x$ and $y$. In these situations, $y$ implicitly depends on the variable $x$ by virtue of that defining equation (whereas when $y$ is a function of $x$, we have some explicit way to describe $y$ in terms of $x$ via the equation $y=f(x)$).
For example, the unit circle consists of all those points which satisfy the equation $x^2+y^2=1$. In this equation, we see that for a given value of $x$, there are only so many values of $y$ which can make the equation true; it is in this sense that $y$ depends implicitly on $x$, even if we don't know exactly what that dependence is! Nevertheless, we ought to be able to compute the slope of the tangent line to this curve if we work with the knowledge that $y$ depends on $x$ in some way. Specifically, in the equation, we can think of each occurrence of $y$ according to this unknown (but implicit) relationship, so we often think of $y$ as $y(x)$ to emphasize this connection. If we then differentiate both sides of $x^2+y^2=1$, then the chain rule gives $$2x+2y\frac{dy}{dx} = \frac{d}{dx}\left[x^2+y^2\right]=\frac{d}{dx}[1]=0,$$ and so we can solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{-2x}{2y}=-\frac{x}{y}.$$ Hence if we have a point $(x_0,y_0)$ on the unit circle, this tells us that the slope of the line tangent to the circle at $(x_0,y_0)$ is $-\frac{x_0}{y_0}.$
At the very end of class we said we can use this process of implicit differentiation to compute other derivatives. For instance, we saw the rule that $$\frac{d}{dx}\left[\ln(x)\right] = \frac{1}{x}.$$ We will see how implicit differentiation gives us this result next class period.
Facts to know: $\frac{dy}{dx}$ gives the slope of the tangent to any curve (not just graphs of functions)
$\displaystyle \frac{d}{dx}\left[\ln(x)\right]=\frac{1}{x}$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - An ellipse centered at $(0,0)$ has an equation of the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$ where $a$ is half the "width" of the ellipse and $b$ is half the "height" of the ellipse. Use implicit differentiation to compute the slope of the line tangent to the ellipse at some point $(x_0,y_0)$. What is the equation of the tangent line going through $(x_0,y_0)$?
- We have all seen the parabola $y=x^2$ before; it's a parabola that opens "upward." A cousin of this parabola is $x=y^2$, which has the same basic shape but instead opens "rightward." (Here's a picture of what this parabola looks like.) Use implicit differentiation to compute the equation of the line tangent to the curve at the point $(9,-3)$.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.5 Lecture Summary: Last class period we discussed the notion of implicit differentiation, and we said it could be used to compute $\frac{d}{dx}\left[\ln(x)\right]$. At the start of today's class, we made good on this promise. The key idea was to transform the equation $y=\ln(x)$ (which involves the function $\ln(x)$ that we don't know how to differentiate) into an equivalent equation $x=e^y$ (which involves functions we do know how to differentiate). If we compute $\frac{d}{dx}$ of both sides and use the chain rule and implicit differentiation, we get an equation $$1 = e^y \frac{dy}{dx}.$$ Solving for $\frac{dy}{dx}$ --- and remembering that $e^y=x$ --- then gives us $$\frac{d}{dx}\left[\ln(x)\right] = \frac{dy}{dx} = \frac{1}{e^y}=\frac{1}{x}.$$
We next used this same idea to compute $\frac{d}{dx}\left[\arctan(x)\right]$. Before jumping into the calculus side of this problem, we recalled that $\arctan(x)$ is the function inverse of $\tan(x)$. This means that if we want to differentiate $y=\arctan(x)$, we might instead consider differentiating the equivalent equation $x=\tan(y)$. We carried this process out and were able to establish $$\frac{d}{dx}\left[\arctan(x)\right]=\frac{1}{1+x^2}.$$
Facts to know: $\frac{d}{dx}\left[\ln(x)\right]=\frac{1}{x}$
$\frac{d}{dx}\left[\arctan(x)\right]=\frac{1}{1+x^2}$
Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - We already have a way to compute the derivative of $y=\sqrt{x}$ using the power rule. If we instead rewrite the equation $y=\sqrt{x}$ to be the equation $x=y^2$, what does implicit differentiation tells us is the value of $\frac{dy}{dx}$? [Hint: it should be the same answer!]
- (Challenge!) Find a formula for $\frac{d}{dx}\left[\arcsin(x)\right]$.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 3.9 Lecture Summary: Now that we have a number of derivative rules under our belt, we are going to move into a new unit of material. The focus on this new unit will be on applications of differentiation. (Said another way: we've focused for a while on the "how?" of computing derivatives. Now we'll give lots of new answers to the question "why derivatives?") Today we covered two basic applications of differentiation, each of which is rooted in concepts related to derivatives that we've already considered.
Our first application is called linearization. The basic problem it addresses is the fact that most functions are hard to compute. Linear functions, on the other hand, are always very easy to compute. Because the tangent line to $y=f(x)$ at a point $(a,f(a))$ is a close to the graph of $y=f(x)$ (at least for values of $x$ near $a$), this means we can approximation values for $f(x)$ by looking at the corresponding value on the tangent line. Symbolically, this just means that $$f(x) \approx f(a)+f'(a)(x-a)$$ for values of $x$ "close to" $a$. We used this idea to approximate $\sqrt{9.1}$; our strategy was to compute the tangent line at the nearby point $(9,\sqrt{9})$ and then plug $x=9.1$ into that tangent line equation. With $a=9$, $f(a)=\sqrt{9}=3$ and $f'(a)=\frac{1}{2\sqrt{9}}=\frac{1}{6}$, we found $$\sqrt{9.1} \approx 3+\frac{1}{6}(9.1-9) = 3+\frac{1}{60}.$$
The second application we discussed is a related concept called "differentials." The key idea here is one that we've seen before: for a point $(a,f(a))$ on the curve and a "nearby" point $(a+h,f(a+h))$, the slope of the secant line between these points (namely, $\frac{f(a+h)-f(a)}{h}$) is pretry approximately equal to the slope of the tangent line at $(a,f(a))$ (namely $f'(a)$). Symbolically, this just means $$f'(a) \approx \frac{f(a+h)-f(a)}{h} = \frac{\Delta y}{\Delta x}.$$ This gives the related expression $$\Delta y \approx f'(a) \Delta x.$$ In other words, we can approximate how much outputs change from $x=a+h$ to $x=a$ by multiplying $f'(a)$ by $h$. We used this in a specific example to compute the possible error in a glass manufacturing process.
Facts to know: Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Approximate $\arctan(1.2)$ using linearization.
- It is a fact that $e \approx 2.718$. Use this, together with linearization, to approximate the value of $\sqrt[11]{e^{10}}$.
- The edge of a cube was found to be $50$cm with a possible error in measurement of $0.1$cm. Estimate the error in computing the volume of the cube.
- The edge of a cube was found to be $50$cm with a possible error in measurement of $0.1$cm. Estimate the error in computing the surface area of the cube.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.1 Lecture Summary: Often it happens that two quantities are related to each other (which, in mathematics-speak, means there's an equation that ties them together). In a situation like this, if we know how one of the variables is changing (with respect to some third --- often implicit --- variable), we can use this equation to determine how the other variable is changing. These kinds of scenarios are called related rates problems.
Today we introduced the notion of related rates by studying the following problem: suppose that we spill paint onto a canvas, so that over time it makes an ever-growing circle. If the area of the circle grows at a rate of 1 square centimeter per second, then what is the rate of change of the radius of the circle at the moment that the area is $14\pi$?
After working through this problem, we outlined some basic steps one can follow to solve related rates problems.
Facts to know: Practice problems: To test your knowledge of this material, you might consider working on these problems. Solutions will be crowd-sourced and posted when available. - Section 4.1: Problem 3
- Section 4.1: Problem 15
- Section 4.1: Problem 17
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.1 Lecture Summary: After reviewing the setup of related rates problems, we worked through two related rates problems in teams.
Facts to know: Practice problems: No practice problems today! Check out the optional problems from the homework tab if you want more practice. -
Class Notes: One copy of the notes and another copy of the notes Text sections: 4.1, 4.2 Lecture Summary: We finished our discussion of related rates problems by working through Wednesday's problems in detail at the board.
Afterwards we introduced the next section of material by defining and discussing local maxs and local mins (which we collectively refer to as extreme values). We thought about what these mean graphically, and then gave a theorem which tells us that extreme values have to occur at so-called critical points --- places where the derivative of the function is either $0$ or undefined.
Facts to know: What is a local extreme?
What does Fermat's extreme value theorem tell us?
Practice problems: TBP -
Class Notes: One copy of the notes Text sections: 4.2 Lecture Summary: Today we dug deeper into understanding why Fermat's Extreme Value Theorem tells us that local extreme values for a function $f(x)$ occur at places where the derivative $f'(x)$ is either zero or not defined. We said that a point $x=x_0$ is a critical point (or critical value) of a function if $f'(x_0)$ is either zero or not defined. We made the important observation that although all local extremes are critical points, we don't know that all critical points are local extremes! We will need some new technology to determine whether a given critical point is a bona fide local extreme.
To get comfortable with the idea of critical points, we computed the critical points for two functions together in class.
Facts to know: How to compute critical points for a function.
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Class Notes: One copy of the notes Text sections: 4.2, 4.3 Lecture Summary: Last class we devised a method for whittling down the location of possible location extremes for a function $f(x)$ by using Fermat's extreme value theorem. It tells us that every local extreme for $f$ must be a critical point for $f$. Unfortunately, it's not the case that every critical point has to be a local extreme, so we're stuck with the question of distinguishing legitimate local extremes from those critical points that are not local extremes. How do we tell these apart?
As a first answer to that question we devised the 2nd derivative test. The idea is that if we have a sense for the concavity of the function at a given critical point, then this might be able to tell us if the critical point is a local max or a local min. For instance, if concavity at a critical point $x_0$ is positive, this tells us that the function is "bowed up" at the critical point, which means that the critical point must be a local minimum. Likewise if $f''(x_0)< 0$ for a critical point $x_0$, then this tells us that the function has a local max at $x_0$. The fact that these two scenarios give us definitive conclusions about the nature of the extreme at these critical points is the heart of the 2nd derivative test.
The downside to the second derivative test is that it's not fool-proof: if $x_0$ is a critical point so that either $f''(x_0)=0$ or $f''(x)$ does not exist, then the 2nd derivative test can't tell us anything about whether $x_0$ is a critical point for $f$ or not. In these case, we say that the 2nd derivative test fails, and we have to go look for some other way to determine if $x_0$ is a critical point.
We saw several examples of how to apply the 2nd derivative test to discern whether a critical point is a local extreme, including examples where the test works and examples where the test fails.
Facts to know: How to run the 2nd derivative test to check for local extremes
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.2, 4.3 Lecture Summary: We saw last time the that 2nd derivative test can be a good way to determine whether a given critical point is a bona fide local extreme, but that sometimes the test is inconclusive. What do we do in these situations? We gave an answer to this question today when we discussed the first derivative test for local extremes. The nice feature of this test is that it always works, and it only requires information that we would have already computed in the course of finding critical points. It tells us that if $x_0$ is a critical point, then
- if $f'(x)$ is positive just to the left of $x_0$ and $f'(x)$ is negative just to the right of $x_0$, then $x_0$ is a local max
- if $f'(x)$ is negative just to the left of $x_0$ and $f'(x)$ is positive just to the right of $x_0$, then $x_0$ is a local min
- if $f'(x)$ is positive just to the left of $x_0$ and $f'(x)$ is also positive just to the right of $x_0$, then $x_0$ is not a local extreme
- if $f'(x)$ is negatve just to the left of $x_0$ and $f'(x)$ is also negative just to the right of $x_0$, then $x_0$ is not a local extreme.
We put this test to practice in several situations to determine the local extremes for a few functions.
Facts to know: How to run the 1st derivative test to check for local extremes
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Class Notes: None today Text sections: 4.2, 4.3 Lecture Summary: Today we practiced using the skills from last week to detect and classify local extremes for functions off of this worksheet. You can find solutions here.
Exam 3 content
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.2 Lecture Summary: We have been thinking for some time about local extremes and how to find them. Today we asked a parallel question: how do we find global extremes for a function? We started by defining what global extremes are, and we lots of examples of examples of functions that either did or did not have global maxs and/or global mins. We studied conditions under which global extremes are guaranteed to occur. One case where this occurs is when we have a function which has exactly one critical point on its domain, and the derivative changes sign at that critical point (i.e., if you look to the left and to the right of that critical point, your derivative will return positive values on one side and negative values on the other). In this case we saw that one can use the 1st derivative test to detect and classify the global exteme for your function.
The other case was when the function we're studying is continuous on a domain which is a closed, finite interval. In that case, the extreme value theorem tells us not only that the function is guaranteed to have at least one global max and at least one global min on its domain, but additionally that any global extremes must occur either at critical points or at endpoints. We worked through an example that showed how to put this idea into practice.
Facts to know: What a global max (or a global min) is
How to detect global extremes graphically
How to find global extremes for continuous functions on a closed, finite interval
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.6 Lecture Summary: Over the past several lectures we have developed a lot of skills for determining whether (and where) a function takes on maximum or minimum values, both in the case of "local" extremes and "global" extremes. In this class period we put those skills to use by working on so-called optimization problems. These are essentially word problems that can be boiled down into some corresponding problem about finding extreme values for a particular function. The challenge in these problems is twofold: first we need to determine how to convert the given word problem into some explicit extreme value problem, and second we need to use our calculus know-how to solve the extreme value problem. The former requires us to identify the relevant variables involved with the problem, any relationships they may have to each other, and some "target function" that we're trying to optimize. Any relationships we find between the variables of interest are typically used to convert the target function into a function of just one variable; we also need to determine the appropriate domain for the target function. Once we've done all this, phase 2 can kick in, and we can use the tools we have for finding extreme values.
Facts to know: The process for converting an optimization problem into a calculus problem
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.6 Lecture Summary: We continued our discussion of optimization today (but we also went on a few tangents...sorry...)
Facts to know: Why the Pythagoreans didn't eat beans
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.5 Lecture Summary: At the start of class we finished the optimization problem we were working on in the previous class. There was also a request to see a couple more optimization problems, so here's a link to a PDF of another couple of optimization problems being worked out.
Today we discussed our last application of derivatives: l'Hopital's rule. The idea is that we can use derivatives as a tool for evaluating limits that are "indeterminate." Specifically, if we are interested in computing a limit of the form $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ and the naive approach yields a limit of the form "0/0" or "$\infty/\infty$," then l'Hopital's rule says that if $$\lim_{x \to a} \frac{f'(x)}{g'(x)}=L,$$ then our original limit is $L$ as well
Facts to know: When you can apply l'Hopital's rule to evaluate a limit
How to apply l'Hopital's rule to evaluate a limit
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Class Notes: One copy of the notes and another copy of the notes Text sections: 5.1 Lecture Summary: We have spent all our time this semester discussing and applying derivatives. The motivation for derivatives was to understand the notion of instantaneous rate of change. Today we switch gears and think about an entirely new problem, one which replaces our search for instantaneous rates of change for functions with instead some measurement for cumulative change of a function over an interval. We are, in essence, replacing our "local" study of functions with a new kind of "global" analysis of their behavior.
To motivate this problem we worked on a problem related to a given road trip. On this trip, I marked my speed at 30 minute intervals. The question was whether we could take that information and use it to estimate the total distance I traveled on my trip. One issue we faced right at the beginning is that we didn't have as much data as we would have liked to answer this problem. In particular, we had to make (hopefully) reasonable guesses about how fast we were moving between those 30 minutes, and from there extrapolate the distance traveled during each 30 minute window of time. To get the total distance traveled, we then added up each of these approximate distances.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 5.2 Lecture Summary: Last time we finished with an introduction to the area problem, which attempts to compute the "net area under the curve" for a function $f(x)$ along some interval $[a,b]$. Today we discussed this notion more completely. We did some examples to compute this "net area" in cases where the graph of our function was made up of "simple" geometric shapes (like triangles, rectangles, and parts of circles). We also saw that the notion of "net area" meant that regions below the line $y=0$ count as "negative area" in our calculations. We introduced the notation $\int_a^b f(x)~dx$ to represent the "net area" we are interested in for the area problem, and called this quantity the definite integral for $f$ along $[a,b]$. Although we had some success in computing $\int_a^b f(x)~dx$ for some functions, we didn't have to look to hard to find an example of a function whose graph was too complicated to be approached using elementary geometry. Specifically, we saw that $\int_1^4 x^2~dx$ was too swoopy to be computing using geometric formulas we had on hand. Eventually we will work to build some theory that lets us compute definite integrals for functions like these, but before doing that we took advantange of the geometry we do have to uncover some properties the definite integral enjoys. We saw that
- if $c$ is a constant, then $\displaystyle \int_a^b c~dx = (b-a)c$;
- if $a$ is any number and $f(x)$ is any function, then $\displaystyle \int_a^a f(x)=0$; and
- if $a \leq k \leq b$, then for any function $f(x)$ we have $\displaystyle \int_a^b f(x)~dx = \int_a^k f(x)~dx+ \int_k^b f(x)~dx$.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 5.2 Lecture Summary: Last class we thought about the basic ideas surrounding the area problem, and we also invented the notation $\int_a^b f(x)~dx$ as a shorthand for the quantity more explicitly described as "the net area under the curve $y=f(x)$ along the interval $[a,b]$." We saw that if $f(x)$ has a graph whose net area is made up of "easy" geometric shapes like triangles, rectangles, and nice parts of circles, then one can compute $\int_a^b f(x)~dx$ just by using elementary geometry and some scissors. But we also saw that lots of functions have graphs whose areas are not made up of such simple shapes. What do we do in that setting?
For today, we decided to forego getting a "perfect" calculation for this area by instead crafting an idea that would let us approximate this area; the idea we created was built on the foundation of our "road trip" problem from a few days ago. If we want to approximate $\int_a^b f(x)~dx$, the process worked liked this:
- First we would cut the interval $[a,b]$ into $n$ subintervals, each of width $\Delta x = \frac{b-a}{n}$; we let the left endpoint of the $i$th subinterval be represented by $x_{i-1}$, and the right endpoint be represented by $x_i$.
- Then we would pretend like the function behaves in some really nice way along each subinterval. For example, we could assume that the function is constant along each subinterval, with the value of the function along the $i$th subinterval $[x_{i-1},x_i]$ being given by $f(x_{i-1})$. (This is called the "left hand approximation.")
- Then we could use this "pretend" version of the function to approximate the net area along each subinterval $[x_{i-1},x_i]$. In the case where we assume that the function is constant along $[x_{i-1},x_i]$ with value $f(x_{i-1})$, the area of interest becomes a rectangle of width $\Delta x$ and height $f(x_{i-1})$, and so the area $\int_{x_{i-1}}^{x_i} f(x)~dx \approx f(x_{i-1})\Delta x$.
- Finally, now that we have approximations for the net area along each subinterval, we can approximate the total area by just adding these values up. In the case of our left-hand approximation, this gives $$\int_a^b f(x)~dx = f(x_0)\Delta x + f(x_1)\Delta x+ \cdots + f(x_{n-1})\Delta x = \sum_{i=1}^n f(x_{i-1})\Delta x.$$(We often denote the quantity on the right side of this expression by $L_n$, since it's giving us the left-hand approximation for $\int_a^b f(x)~dx$ using $n$ subintervals.)
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Class Notes: One copy of the notes and another copy of the notes Text sections: 5.3 Lecture Summary: We have seen in previous classes that computing $\int_a^b f(x)~dx$ is possible when the resultant area is made up of "nice" geometric figures, but what do we do when the graph isn't so nice? Last class we saw that we can use Riemann sums to approximate the value of $\int_a^b f(x)~dx$. We had four different methods for creating estimates (the left- and right-hand approximations, the midpoint approximation, and the trapezoidal approximation). We also commented that --- generally speaking --- for a fixed number of subintervals, the trapezoidal approximation is typically the most accurate, followed by the midpoint rule, and then followed by the left- and right-hand approximations. But what if we want to give an even better approximation, the key is that increasing the value of $n$ (i.e., increasing the number of subintervals) should result in better approximations. Indeed, today we said that for any reasonably nice function $f$, we have $$\int_a^b f(x)~dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\Delta x.$$ This is an incredibly deep observation with a lot of geometric intuition built into it, and it finally allows us to bring our calculus skill set to solve the area problem!
The downside to this observation is that it's incredibly laborious to implement this idea from a computational standpoint. It's not impossible, but it takes a great deal of effort, and (even worse) the ideas we need to bring to each such problem change dramatically depending on the function we're thinking about. What we'd prefer is some other way to compute definite integrals. Fortunately, just such a theorem exists in the fundamental theorem of calculus (Part II), also known as the Evaluation Theorem. This theorem says that if $F$ is a function so that $\frac{d}{dx}\left[F(x)\right] = f(x)$, then $$\int_a^b f(x)~dx = F(b)-F(a).$$ We used this result to compute $\int_1^4 x^2~dx$ quite easily; this is impressive, since it's an area calculation that has stumped us for many class periods, and the fundamental theorem lets us solve the problem with very little effort!
We will see in future class periods how to use this important theorem to evaluate lots of areas of interest.
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Class Notes: One copy of the notes Text sections: 4.8, 5.3 Lecture Summary: Last class period we saw that if we want to evaluate a net area captured by the graph $y=f(x)$ for values $a \leq b$ (i.e., compute a definite integral $\int_a^b f(x)~dx$), then we're able to shortcut evaluating this in the "geometric way" via limits of Riemann sums: the evaluation theorem (aka, the fundamental theorem of calculus part II) says if we can find a function $F$ so that $\frac{d}{dx}\left[F(x)\right]=f(x)$, then we get $$\int_a^b f(x)~dx = F(b)-F(a).$$ In class we said that a function $F$ is an antiderivative of $f$ if $\frac{d}{dx}[F(x)]=f(x)$, so that the evaluation theorem allows us to reframe our pursuit of the area problem: instead of being a geometric problem, it's really now a search for antiderivatives. For the rest of the term, we'll spend our time looking for antiderivative rules so that we can use them to compute some basic areas. Our guiding principle will be that every derivative rule can be rephrased into a corresponding antiderivative rule.
For example, we saw that the linearity properties for the derivative give us linearity properties for antiderivatives: $$\begin{align*}\int f(x)+g(x)~dx &= \int f(x)~dx + \int g(x)~dx \\ \int cf(x)~dx &= c\int f(x)~dx.\end{align*}$$ We also saw the antipower rule, as well as an antiderivative rule involving $\ln(x)$:$$\begin{align*}\int x^n~dx &= \frac{1}{n+1} x^{n+1}+C \qquad \text{(for n \neq -1)} \\ \int \frac{1}{x}~dx &= \ln(x)+C.\end{align*}$$ We used these rules to compute some specific definite integrals.
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Class Notes: One copy of the notes and another copy of the notes Text sections: 4.8, 5.3 Lecture Summary: At the start of class we recalled that our pursuit for an answer to the area problem (i.e., computing definite integrals $\int_a^b f(x)~dx$) has been replaced with a pursuit for antiderivatives. This is because the Evaluation theorem (aka, the Fundamental Theorem of calculus Part II) tells us that if $F$ is an antiderivative for $f$ along $[a,b]$, then $$\int_a^b f(x)~dx = F(b)-F(a).$$ We also saw last class period that every derivative rule can be turned into an antiderivative rule. We already saw a number of antiderivative rules last time, including linearity of antiderivatives as well as rules for antiderivatives of power functions (almost all of which are covered by the antipower rule. Today we recorded a number of other such rules, namely $$\begin{align*} \int \cos(x)~dx &= \sin(x)+C\\ \int \sin(x)~dx &= -\cos(x)+C\\ \int \sec^2(x)~dx &= \tan(x)+C\\ \int \sec(x)\tan(x)~dx &= \sec(x)+C\\ \int \frac{1}{1+x^2}~dx &= \arctan(x)+C\\ \int e^x~dx &= e^x+C\\ \int a^x~dx &= \frac{1}{\ln(a)}a^x+C. \end{align*}$$ We then practiced antiderivatives using this worksheet. (Here are the solutions.)
Additional content
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Class Notes: One copy of the notes Text sections: 5.5 Lecture Summary: Our recent work has focused on implementing the philosophy behind the Evaluation Theorem. This theorem tells us that if we want to compute a definite integral $\int_a^b f(x)~dx$, then we can do this easily if we know an antiderivative for $f$ (i.e., a function $F$ so that $F'(x)=f(x)$). Specifically, if we have such a function, then $$\int_a^b f(x)~dx = F(b)-F(a).$$ One way to view this theorem is that it lets us replace our search for answers to the area problem with a new search instead: a search to come up with good rules for finding antiderivatives. Based on the definition of antiderivative, we know that every derivative rule we learn can be reinterpreted as an antiderivative rule. We've already used this to great effect, discovering that antiderivatives are "linear" (in the sense that they play quite nice with respect to scaling and adding), creating an "antipower rule," and many other results.
In today's class, we looked for the antiderivative counterpart to the chain rule. In some sense, this is easy enough: since the chain rule says that $$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x))\cdot g'(x),$$ we can "read this equation backwards" to get the antiderivative rule $$\int f'(g(x))g'(x)~dx = f(g(x))+C.$$ In some sense, this is all there is to say. The only improvement we could make to this is trying to come up with better ways to "see" when some integral we're given fits this very particular form. For that, we introduced the notion of revisioning the "form" of this integral with a convenient shorthand. We write $u=g(x)$, which tells us that $\frac{du}{dx} = g'(x)$. (Or, equivalently, that $du = g'(x)~dx$.) With this notation shift in mind, we find that our antiderivative formula becomes $\int f'(g(x))\cdot g'(x)~dx$ becomes $$\int f'(u)~du = f(u)+C.$$ This latter expression is, of course, way easier to work with than our original expression, which is an advertisement this this kind of "substitution" setup is useful. Making a substitution of this form is what we call $u$-substitution.
We spent time in class going over some examples of how this procedures works with some specific antiderivatives.
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Class Notes: One copy of the notes Text sections: 5.5 Lecture Summary: We finished our semester together by doing some more work with $u$-substitution. We reviewed the ideas we discussed in the previous class, and even recorded some useful observations about how one implements $u$-substitution:
- When we look at an antiderivative problem, the first thing we do is look to see if it's some familiar derivative (in which case, we can compute the integral easily). Only if the function isn't something we recognize as a derivative would we consider using $u$-substitution
- From our setup of the chain rule, we know that the "$u$'' function appears as the interior function in a composition (namely $u=g(x)$, where $f'(g(x))$ is one of the factors in our integral). So when trying to find a good candidate for $u$, consider looking for a composition, and think about picking the "inside" function of that composition,
- If you have a guess for what $u$ ought to be, it's not a bad idea to see if the derivative of your $u$ function appears as a factor in the integrand (at least up to some constant multiple).
- $\displaystyle \int \frac{\sin(x)}{\left(\cos(x)\right)^2}~dx$
- $\displaystyle \int x^2e^{-x^3}~dx$
- $\displaystyle \int \frac{1}{x(\ln(x))^1}~dx$
- $\displaystyle \int \frac{1}{x^2}\sec\left(\frac{1}{x}\right)\tan\left(\frac{1}{x}\right)~dx$
- $\displaystyle \int \sin(5\theta)~d\theta$
- $\displaystyle \int \frac{\arctan(x)}{1+x^2}~dx$
Here are the solutions for those problems!