Mean Value Theorem

Section 4.1 Mean Value Theorem

Let’s begin here with some weird questions. The questions aren’t weird because of what they’re asking. Instead, they’re weird because of the logic of how we interpret them compared to how we want to interpret them.

Why is the derivative of a constant function 0?
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Why do \(y=x^2+7\) and \(y=x^2-3\) have the same derivative?
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If a function is only increasing on the interval \((0,1)\text{,}\) what do we know about the derivative at any of these \(x\)-values in \((0,1)\text{?}\)
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These questions are ones we can think through and answer.

Here are some answers for these first three questions:

A constant function has all of the same \(y\)-values for any \(x\)-value in the domain: of course the slope everywhere is 0! There isn’t any change in the \(y\)-values!
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We can differentiate these functions term by term: we know that the \(x^2\) term has a derivative of \(2x\text{,}\) and then for each function’s constant, the derivative has to be 0. So it doesn’t matter what each constant was, the derivatives wouldn’t rely on that value!
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If a function is increasing on an interval, then it means that for any pair of \(x\)-values, \(x_1\lt x_2\text{,}\) we get \(y\)-values in the same order: \(f(x_1)\lt f(x_2)\text{.}\) If we find the limit of slopes as \(x_1\to x_2\text{,}\) we’ll always get a positive slope, so the derivative has to be positive.
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What’s tricky is that these don’t always say what we want to say. For instance, I might sometimes want to say the following:

If you know a function’s derivative is 0, then you know the function is constant. Another way of saying this is that the only functions whose derivative is 0 are constant functions.
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Similarly, we might want to say that if you know two functions that have the same derivative, then the only difference between the functions is a constant. There aren’t any other ways for functions to be different with the same derivative.
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We might want to say that if you know the derivative is positive on an interval, that means that the function has to be increasing.
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Do you see the (slight) difference?

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What we’ll secretly see is that all of these statements are actually correct, but require a result for us to say them. The Mean Value Theorem will be the result that we use to build important and helpful results throughout the rest of this course.

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Subsection Slopes

We have two different kinds of slopes that we think of right now: a slope between two points, and a slope at a single point.

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We can translate this into a rate of change! We think of two rates of change: an average rate of change on some interval and the instantaneous rate of change at some point.

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We will try to connect these two different slopes/rates of change for “well-behaved” functions. Let’s take a look at an example. In the graph below, we have a curve where we are considering some closed interval, as well as a point within that interval. Both slopes are visualized and calculated: the slope between the ending points of the interval is the average slope, while the slope of the line tangent to the curve at a point is the instantaneous slope. Move the point/interval around and get a feel for how these two different slopes relate (or don’t relate!) to each other.

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If you move the interval/point around enough, you’ll see that sometimes these two slopes are really similar and sometimes they’re very different. But what if the point in the middle of the interval wasn’t so “set” at being stuck in the exact middle of the interval? What if we stuck the interval in place, and then had the freedom to move the point anywhere in the interval?

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Do you think we will always be able to do this? What kinds of functions will have these points where the slope at the point matches the average slope on the interval?

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Subsection The Mean Value Theorem

Theorem 4.1.1. Mean Value Theorem.

For a function \(f(x)\) that is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\text{,}\) then there is some value \(x=c\) with \(a\lt c\lt b\) where:

\begin{equation*} f'(c) = \frac{f(b)-f(a)}{b-a}\text{.} \end{equation*}

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This theorem is really just guaranteeing the existence of at least one of the points we found above: a point where the slope of the line tangent to the curve is the same as the slope between the endpoints of the interval. We can (and, very briefly, we will) use this theorem to find the point that is guaranteed to exist, but we will, more generally, use this theorem as a tool for connection.

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We will try to use it as a way of connecting derivatives to the behavior of a function. The Mean Value Theorem gives us this equation where on one side we have a derivative, and on the other side we have function outputs. This is really similar to the definition of the Derivative at a Point, except that in this case we have no limit: we just get to use the behavior of the function on an interval around the point.

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Secretly, the Mean Value Theorem is the driving force behind most of the results we will look at from here on out, at least when the requirements include continuity and differentiability on an interval. You can almost guarantee that if a theorem or result has these two requirements, then the Mean Value Theorem is likely at work.

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Let’s look at one example, at least, before we move on.

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Example 4.1.2.

Let’s say that a person is planning on biking to their college campus, 8 miles away. At 2:39 p.m., they send a text to their friend with a selfie of them on their bike about to start their trip, captioned “Beautiful day for a ride!” At 3:27 p.m., they post a picture on their social media of them in front of the bike rack on campus.

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(a)

What was the average velocity of the student between 2:39 p.m. and 3:27 p.m.?

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Solution.

The total trip took 48 minutes (or 0.8 hours), and the student traveled a total distance of 8 miles.

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The student’s average velocity is \(\dfrac{8}{0.8} = 10\) miles per hour.

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(b)

The Mean Value Theorem connects average slopes with slopes of tangent lines. What does that mean, in general, in this context?

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Solution.

In this case, the average rate of change of the function on the interval is the average velocity of the biker. So then the instantaneous rate of change must correspond with an instantaneous velocity, of their velocity at some specific point in time.

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(c)

Make a claim about the instantaneous velocity of the student on their bike at some point in time.

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Solution.

We know that at some point between 2:39 p.m. and 3:27 p.m., the cyclist had to be traveling at exactly 10 miles per hour.

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Example 4.1.3.

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For a function \(f(x)=\sqrt{x}+1\) on the interval \([1,4]\text{,}\) find the point that the Mean Value Theorem guarantees the existence of, and explain what it is.

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Solution.

Let’s calculate the average slope on the interval:

\begin{align*} \frac{f(4)-f(1)}{4-1} \amp = \frac{(\sqrt{4}+1)-(\sqrt{1}+1)}{3} \\ \amp = \frac{1}{3} \end{align*}

So we know that there is some \(x\)-value between 1 and 4 where \(f'(x)=\dfrac{1}{3}\text{.}\)

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The derivative is \(f'(x)=\dfrac{1}{2\sqrt{x}}\text{,}\) so we can solve for \(x\text{:}\)

\begin{align*} f'(x) \amp = \frac{1}{3}\\ \frac{1}{2\sqrt{x}} \amp = \frac{1}{3}\\ 2\sqrt{x} \amp= 3\\ \sqrt{x} \amp = \frac{3}{2}\\ x \amp = \frac{9}{4} \end{align*}

So the point guaranteed to exist by the Mean Value Theorem is located at \(\left(\frac{9}{4}, \frac{5}{2}\right)\text{,}\) where the slope of the tangent line is \(f'\left(\frac{9}{4}\right) = \frac{1}{3}\text{.}\)

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These examples are fine, but the real power of the Mean Value Theorem comes in how we can use it to generate more interesting results. Let’s check those out!

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Subsection More Results due to the Mean Value Theorem

First, let’s remind ourselves of what it means for a function to be increasing or decreasing on an interval.

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Definition 4.1.4. Increasing and Decreasing on an Interval.

A function \(f(x)\) is increasing on the interval \((a,b)\) if, for every pair of \(x\)-values \(x_1\) and \(x_2\) (with \(x_1 \lt x_2\)), \(f(x_1)\ \lt f(x_2)\text{.}\)

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If \(f(x_1)\gt f(x_2)\text{,}\) then we say that \(f\) is decreasing on the interval \((a,b)\text{.}\)

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Note that we classify a function as increasing or decreasing based on comparing two function outputs. This is a perfect time to use the Mean Value Theorem, since it can connect these pairs of function outputs with a derivative!

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Theorem 4.1.5. Sign of the Derivative and Direction of a Function.

If \(f\) is a function that is differentiable on the interval \((a,b)\) and \(f'(x)\gt 0\) for all \(x\) in the interval \((a,b)\text{,}\) then \(f\) must be increasing on \((a,b)\text{.}\)

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Similarly, if \(f'(x)\lt 0\) for all \(x\) in the interval \((a,b)\text{,}\) then \(f\) must be decreasing on \((a,b)\text{.}\)

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Proof.

Before we begin, let’s just agree to think about only the case where \(f'(x)\gt 0\) on the interval \((a,b)\text{.}\) The other case (where \(f'(x)\lt 0\)) ends up being the exact same argument, with some changes in signs and directions of inequalities. So we’ll say \(f'(x)\gt 0\) for all \(x\)-values in the interval \((a,b)\text{.}\)

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Ok, let’s begin!

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Let’s pick two arbitrary \(x\)-values from the interval \((a,b)\text{.}\) Call them \(x_1\) and \(x_2\text{,}\) and we’ll make sure that we name them in a way where \(x_1\lt x_2\text{.}\) Now, \(f\) must be continuous on \([x_1,x_2]\) and differentiable on \((x_1, x_2)\text{.}\) We also know that \(f'(x)\gt 0\) for every \(x\) in the interval (\(x_1\lt x\lt x_2\)).

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The Mean Value Theorem says that there is some \(x=c\) in \((x_1,x_2)\) with

\begin{equation*} f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}\text{.} \end{equation*}

Equivalently, this means that

\begin{equation*} f'(c)(x_2-x_1) = f(x_2)-f(x_1)\text{.} \end{equation*}

Notice that \(f'(c)\gt 0\) (since the derivative is positive everywhere in the interval) and \((x_2-x_1)\gt 0\) (by the way we named these \(x\)-values). This means that \(f'(c)(x_2-x_1)\gt 0\text{,}\) and so \((f(x_2)-f(x_1))\gt 0\) as well.

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So notice what just happened: we arbitrarily chose \(x\)-values \(x_1\) and \(x_2\) and noticed that for any of these pairs where \(x_1\lt x_2\text{,}\) we found that \(f(x_1)\lt f(x_2)\text{.}\) This is exactly what it means for \(f\) to be increasing on the interval \((a,b)\) (based on Definition 4.1.4).

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We’ll use this result pretty extensively in the next couple of topics. It is helpful for us to use this to connect the signs of a derivative with our intuition about what that must mean for the direction of a function.

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Theorem 4.1.6. If a Function’s Derivative is 0, it’s Constant.

If \(f(x)\) is a function defined on \((a,b)\) with \(f'(x)=0\) for all \(x\)-values in the interval \((a,b)\text{,}\) then \(f(x)\) is a constant function.

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Proof.

We can almost exactly replicate the proof from Theorem 4.1.5 here!

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Let’s pick two arbitrary \(x\)-values from the interval \((a,b)\text{.}\) Call them \(x_1\) and \(x_2\text{,}\) and we’ll again make sure that we name them in a way where \(x_1\lt x_2\text{.}\) Now, \(f\) must be continuous on \([x_1,x_2]\) and differentiable on \((x_1, x_2)\text{.}\) This time, we know that \(f'(x)= 0\) for every \(x\) in the interval (\(x_1\lt x\lt x_2\)).

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The Mean Value Theorem says that there is some \(x=c\) in \((x_1,x_2)\) with

\begin{equation*} f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}\text{.} \end{equation*}

Equivalently, this means that

\begin{equation*} f'(c)(x_2-x_1) = f(x_2)-f(x_1)\text{.} \end{equation*}

Notice that \(f'(c)= 0\) (since the derivative is zero everywhere in the interval). This means that \(f'(c)(x_2-x_1)= 0\text{,}\) and so \((f(x_2)-f(x_1))= 0\) as well.

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This means that every \(y\)-value from the function is the same as every other one, since we picked these points arbitrarily. So \(f\) must be a constant function!

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Theorem 4.1.7. Equal Derivatives Correspond with Related Functions.

For two functions \(f\) and \(g\text{,}\) both differentiable on \((a,b)\text{,}\) if \(f'(x)=g'(x)\) for all \(x\)-values on \((a,b)\text{,}\) then we know that \(f(x)=g(x)+C\) for some real number constant \(C\text{.}\) That is, the only differences in \(f\) and \(g\) are due to a difference in the constant term.

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Proof.

This one is pretty easy to prove: we’re going to use a fun little trick where we connect this theorem to Theorem 4.1.6.

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Let’s think about these two functions \(f\) and \(g\) with \(f'(x)=g'(x)\text{,}\) and let’s think about a function \(h(x)=f(x)-g(x)\text{.}\) Now we can notice that \(h'(x)=f'(x)-g'(x)\) which has to be 0.

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Oh wow, \(h(x)\) is a function with \(h'(x)=0\) on the interval \((a,b)\text{.}\) It must be a constant function (based on Theorem 4.1.6)! Let’s call it \(h(x)=C\text{,}\) where \(C\) is some real number constant.

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This means that \(f(x)-g(x)=C\text{,}\) and we can see that the only difference between these two functions is a constant.

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We’ll use this result a bit later on, but it’s a great example of how we can use the Mean Value Theorem to connect information about the derivative to information about how a function might work.

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Let me interject my own opinion here: I think the Mean Value Theorem is extremely important. But I also don’t think that it is that important for you to use.

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I think you should know the general idea of connecting the slopes of the line tangent to the curve and the average slope on an interval. I think you should remember the picture we looked at. I think you should be content to see some results later on in the course and smile fondly about how the Mean Value Theorem is under the surface, churning away and getting us cool results to think about. I think we should be happy that the Mean Value Theorem is here and we can recognize it as, maybe, the most important result in this textbook.

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But we don’t need to pretend that we need to actually use it directly...we aren’t going to need to compute a lot or anything like that. We’ll just know it.

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