Extreme Value Theorem, Rolle’s Theorem, and Mean Value Theorem

Extreme Value Theorem	Mean Value Theorem
Rolle’s Theorem	More Practice

Curve Sketching isn’t my favorite subject in Calculus, since it’s so abstract, but it’s useful to be able to look at functions and their characteristics by simply taking derivatives and thinking about the functions. Before we get into curve sketching, let’s talk about two theorems that seems sort of obvious, but we need to go over them nonetheless.

To sketch curves in Calculus, we’ll be looking at minimums and maximums of functions in certain intervals, so we have to talk about a few theorems first. We’ll need these theorems to know that if a function is differentiable and the derivative at a certain point is 0, then that point is either a minimum or maximum. Thus, before you to get to actual curve sketching, you’ll probably see some problems as in this section.

Extreme Value Theorem

The Extreme Value Theorem states that a function on a closed interval must have both a minimum and maximum in that interval. (A closed interval is an interval that includes its endpoints, which are the points at the very beginning and end of the interval). If we didn’t include the endpoints, we’d have an open interval, and we may never have a minimum or maximum, since the function could get closer and closer to a point but never touch it (like we saw with Limits). When we do include these endpoints, there will definitely be a minimum or maximum; for example, if the function is increasing for the whole interval, the minimum and maximum would be at these endpoints.

Rolle’s Theorem

Rolle’s Theorem states that if the function in an interval comes up and back down (or down and back up) and ends up exactly where it started, you’ll have at least one maximum or minimum (where the derivative is 0). Here is the formal form of Rolle’s Theorem:

Rolle’s Theorem:

If a function is continuous on a closed interval $ [a,b]$ and differentiable on the open interval $ (a,b)$, and $ f\left( a \right)=f\left( b \right)$ (the $ y$’s on the endpoints are the same), then there is at least one number $ c$ in $ (a,b)$, where $ {f}’\left( c \right)=0$.

Note that if this function was not differentiable, you would still have either a maximum or minimum, but you may not be able to take the derivative at that point; for example, you may have a sharp turn instead of a nice curve at that point.

Here are pictures of a differentiable and non-differentiable functions. For the differentiable graph, do you see how if the graph goes up and comes back down, we have to have at least one point where the derivative is 0 (at the maximum)?

Mean Value Theorem

Note that the Mean Value Theorem for Integrals can be found here in the Definite Integration section.

The Mean Value Theorem for Derivatives is a little bit more important and is proven using the Rolle’s theorem. It says that somewhere inside a closed interval $ [a,b]$ there exists a point $ c$ where the derivative at this point is the same as the slope between points $ a$ and $ b$. Think of the Mean Value Theorem as Rolle’s Theorem, but possibly “tilted”.

Here’s the formal form of the Mean Value Theorem and a picture; in this example, the slope of the secant line is 1, and also the derivative at the point $ \boldsymbol {(2,3)}$ (tangent line) is also 1. We’ll see more examples below.

Mean Value Theorem

If a function is continuous on a closed interval $ \left[ {a,b} \right]$ and differentiable on the open interval $ \left( {a,b} \right)$, then there is at least one number $ c$ in $ \left( {a,b} \right)$, where $ \displaystyle {f}’\left( c \right)=\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}$.

Here are some problems that you might see with these theorems:

Rolle’s Theorem Problem	Solution
For which of the following graphs does Rolle’s Theorem apply for the indicated interval? a) b) c) d) e) $ \displaystyle f\left( x \right)=\sin \left( {\frac{x}{2}} \right),\,\,\left[ {0,4\pi } \right]\,\,$ f) $ \displaystyle f\left( x \right)=\tan \left( {\frac{x}{2}} \right),\,\,\left[ {0,4\pi } \right]\,\,$ g) $ f\left( x \right)=\sqrt{{1-{{x}^{{\frac{2}{3}}}}}},\,\,\left[ {-1,1} \right]\,\,$	a) Does not apply, since the function is not differentiable everywhere on $ (a,b)$. b) Does apply, since the function is continuous and differentiable on $ [a,b]$. c) Does not apply since the function is not continuous on $ [a,b]$. d) Does not apply since $ f\left( a \right)\ne f\left( b \right)$. e) Does apply, since the function is continuous and differentiable on $ [a,b]$ and $ \displaystyle f\left( a \right)=f\left( b \right):\,\,\sin \left( {\frac{0}{2}} \right)=\sin \left( {\frac{{4\pi }}{2}} \right)=0$. To find a $ c$-value where $ {f}’\left( c \right)=0$: $ \displaystyle \begin{align}{f}’\left( c \right)&=\cos \left( {\frac{c}{2}} \right)\cdot \frac{1}{2}=0;\,\,\,\,\cos \left( {\frac{c}{2}} \right)=0\\\,\frac{c}{2}&={{\cos }^{{-1}}}\left( 0 \right);\,\,\,c=\frac{\pi }{2}\cdot 2=\pi \end{align}$ and this lies in the interval $ \left[ {0,4\pi } \right]$. f) Even though $ f\left( a \right)=f\left( b \right)$, does not apply since the function has asymptotes and thus is not continuous on $ [a,b]$. g) Even though $ f\left( a \right)=f\left( b \right)$, does not apply since the function is not differentiable at $ x=0$ (which is inside the interval).
For which of the following functions does the Mean Value Theorem apply for $ [-1,1]$? a) $ y=\sqrt[3]{x}$ b) $ y=\left\| {x+2} \right\|$ c) $ \displaystyle y=\frac{1}{x}$	a) Does not apply, since the function is not differentiable (has a vertical tangent) at $ x=0$. b) Does apply, since the function is continuous and differentiable on the interval. Note that the function is not differentiable at $ x=-2$, but this is not in the interval $ [-1,1]$. c) Does not apply since the function is not continuous on $ [-1,1]$ (vertical asymptote at $ x=0$).

Rolle’s Theorem Problem

Solution

For which of the following graphs does Rolle’s Theorem apply for the indicated interval?

a) b)

c) d)

e) $ \displaystyle f\left( x \right)=\sin \left( {\frac{x}{2}} \right),\,\,\left[ {0,4\pi } \right]\,\,$ f) $ \displaystyle f\left( x \right)=\tan \left( {\frac{x}{2}} \right),\,\,\left[ {0,4\pi } \right]\,\,$

g) $ f\left( x \right)=\sqrt{{1-{{x}^{{\frac{2}{3}}}}}},\,\,\left[ {-1,1} \right]\,\,$

a) Does not apply, since the function is not differentiable everywhere on $ (a,b)$.

b) Does apply, since the function is continuous and differentiable on $ [a,b]$.

c) Does not apply since the function is not continuous on $ [a,b]$.

d) Does not apply since $ f\left( a \right)\ne f\left( b \right)$.

e) Does apply, since the function is continuous and differentiable on $ [a,b]$ and $ \displaystyle f\left( a \right)=f\left( b \right):\,\,\sin \left( {\frac{0}{2}} \right)=\sin \left( {\frac{{4\pi }}{2}} \right)=0$. To find a $ c$-value where $ {f}’\left( c \right)=0$:

$ \displaystyle \begin{align}{f}’\left( c \right)&=\cos \left( {\frac{c}{2}} \right)\cdot \frac{1}{2}=0;\,\,\,\,\cos \left( {\frac{c}{2}} \right)=0\\\,\frac{c}{2}&={{\cos }^{{-1}}}\left( 0 \right);\,\,\,c=\frac{\pi }{2}\cdot 2=\pi \end{align}$

and this lies in the interval $ \left[ {0,4\pi } \right]$.

f) Even though $ f\left( a \right)=f\left( b \right)$, does not apply since the function has asymptotes and thus is not continuous on $ [a,b]$.

g) Even though $ f\left( a \right)=f\left( b \right)$, does not apply since the function is not differentiable at $ x=0$ (which is inside the interval).

For which of the following functions does the Mean Value Theorem apply for $ [-1,1]$?

a) $ y=\sqrt[3]{x}$ b) $ y=\left| {x+2} \right|$ c) $ \displaystyle y=\frac{1}{x}$

a) Does not apply, since the function is not differentiable (has a vertical tangent) at $ x=0$.

b) Does apply, since the function is continuous and differentiable on the interval. Note that the function is not differentiable at $ x=-2$, but this is not in the interval $ [-1,1]$.

c) Does not apply since the function is not continuous on $ [-1,1]$ (vertical asymptote at $ x=0$).

Here are a few more typical Mean Value Theorem (MVT) problems. Note that when we get our value of $ \boldsymbol {c}$, we have to make sure it lies in the interval we’re given. Note also that these problems may be worded something like this: For what value of $ c$ on a certain open interval would the tangent to the graph of a certain function be parallel to the secant line in that closed interval?

Mean Value Theorem Problem	Solution
Assume that the function $ f\left( x \right)=4x-\sqrt{x}$ satisfies the Mean Value Theorem on the interval $ [0,4]$. Find the value of “$ c$” that is guaranteed by the theorem.	First make sure the MVT applies. The function is continuous on closed interval $ [0,4]$ and differentiable on the open interval $ (0,4)$; therefore, the MVT applies. Find the slope: $ \displaystyle \frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}=\frac{{f\left( 4 \right)-f\left( 0 \right)}}{{4-0}}=\frac{{\left[ {4\left( 4 \right)-\sqrt{4}} \right]-\left[ {4\left( 0 \right)-\sqrt{0}} \right]}}{{4-0}}=\frac{{14}}{4}=\frac{7}{2}$ Find the derivative: since $ f\left( x \right)=4x-{{x}^{{\frac{1}{2}}}}$, $ \displaystyle {f}’\left( x \right)=4-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$. By the MVT, $ \displaystyle {f}’\left( x \right)=4-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$, so we need to find $ c$ where $ \displaystyle 4-\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=\frac{7}{2}$: $ \displaystyle 4-\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=\frac{7}{2};\,\,\,\,\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=4-\frac{7}{2};\,\,\,\,\,{{\left( {{{c}^{{-\frac{1}{2}}}}} \right)}^{{-2}}}={{\left( {\frac{1}{2}\cdot 2} \right)}^{{-2}}};\,\,\,c={{1}^{{-2}}}=1$. This value lies in the interval $ [0,4]$, so the value of $ c$ that is guaranteed by the MVT is $ c=1$.
Does the function $ f\left( x \right)={{x}^{3}}-4{{x}^{2}}+2x$ satisfy the Mean Value Theorem on the interval $ [0,2]$? Justify answer and then find the value of “$ c$” guaranteed by the theorem.	First make sure the MVT applies. The function is continuous on closed interval $ [0,2]$ and differentiable on the open interval $ (0,2)$; therefore, the MVT applies. Find the slope: $ \displaystyle \begin{align}\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}&=\frac{{f\left( 2 \right)-f\left( 0 \right)}}{{2-\left( 0 \right)}}=\frac{{\left[ {{{{\left( 2 \right)}}^{3}}-4{{{\left( 2 \right)}}^{2}}+2\left( 2 \right)} \right]-\left[ {{{{\left( 0 \right)}}^{3}}-4{{{\left( 0 \right)}}^{2}}+2\left( 0 \right)} \right]}}{2}\\&=\frac{{-4-\left( 0 \right)}}{2}=-2\end{align}$ Find the derivative: since $ f\left( x \right)={{x}^{3}}-4{{x}^{2}}+2x$, $ {f}’\left( x \right)=3{{x}^{2}}-8x+2$. By the MVT, $ {f}’\left( c \right)=-2$, so we need to find $ c$ where $ 3{{c}^{2}}-8c+2=-2$. $ \displaystyle 3{{c}^{2}}-8c+2=-2;\,\,\,\,3{{c}^{2}}-8c+4=0;\,\,\,\,\,\left( {3c-2} \right)\left( {c-2} \right)=0;\,\,\,c=\frac{2}{3},\,\,2$ Check to make sure the value(s) we got lie in the interval $ [0,2]$. Since only $ \displaystyle c=\frac{2}{3}$ lies in the interval, the only value of $ c$ that is guaranteed by the MVT is $ \displaystyle c=\frac{2}{3}$.
Does the function $ \displaystyle f\left( x \right)=\frac{1}{{x-2}}$ satisfy the Mean Value Theorem on the interval $ [1,4]$? Justify answer and then find the value of “$ c$” guaranteed by the theorem.	Since the function is not continuous (nor totally differentiable) in the interval $ [1,4]$, the MVT does not apply.

Mean Value Theorem Problem

Solution

Assume that the function $ f\left( x \right)=4x-\sqrt{x}$ satisfies the Mean Value Theorem on the interval $ [0,4]$.

Find the value of “$ c$” that is guaranteed by the theorem.

First make sure the MVT applies. The function is continuous on closed interval $ [0,4]$ and differentiable on the open interval $ (0,4)$; therefore, the MVT applies.

Find the slope:

$ \displaystyle \frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}=\frac{{f\left( 4 \right)-f\left( 0 \right)}}{{4-0}}=\frac{{\left[ {4\left( 4 \right)-\sqrt{4}} \right]-\left[ {4\left( 0 \right)-\sqrt{0}} \right]}}{{4-0}}=\frac{{14}}{4}=\frac{7}{2}$

Find the derivative: since $ f\left( x \right)=4x-{{x}^{{\frac{1}{2}}}}$, $ \displaystyle {f}’\left( x \right)=4-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$.

By the MVT, $ \displaystyle {f}’\left( x \right)=4-\frac{1}{2}{{x}^{{-\frac{1}{2}}}}$, so we need to find $ c$ where $ \displaystyle 4-\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=\frac{7}{2}$:

$ \displaystyle 4-\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=\frac{7}{2};\,\,\,\,\frac{1}{2}{{c}^{{-\frac{1}{2}}}}=4-\frac{7}{2};\,\,\,\,\,{{\left( {{{c}^{{-\frac{1}{2}}}}} \right)}^{{-2}}}={{\left( {\frac{1}{2}\cdot 2} \right)}^{{-2}}};\,\,\,c={{1}^{{-2}}}=1$.

This value lies in the interval $ [0,4]$, so the value of $ c$ that is guaranteed by the MVT is $ c=1$.

Does the function $ f\left( x \right)={{x}^{3}}-4{{x}^{2}}+2x$ satisfy the Mean Value Theorem on the interval $ [0,2]$?

Justify answer and then find the value of “$ c$” guaranteed by the theorem.

First make sure the MVT applies. The function is continuous on closed interval $ [0,2]$ and differentiable on the open interval $ (0,2)$; therefore, the MVT applies.

Find the slope:

$ \displaystyle \begin{align}\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}&=\frac{{f\left( 2 \right)-f\left( 0 \right)}}{{2-\left( 0 \right)}}=\frac{{\left[ {{{{\left( 2 \right)}}^{3}}-4{{{\left( 2 \right)}}^{2}}+2\left( 2 \right)} \right]-\left[ {{{{\left( 0 \right)}}^{3}}-4{{{\left( 0 \right)}}^{2}}+2\left( 0 \right)} \right]}}{2}\\&=\frac{{-4-\left( 0 \right)}}{2}=-2\end{align}$

Find the derivative: since $ f\left( x \right)={{x}^{3}}-4{{x}^{2}}+2x$, $ {f}’\left( x \right)=3{{x}^{2}}-8x+2$.

By the MVT, $ {f}’\left( c \right)=-2$, so we need to find $ c$ where $ 3{{c}^{2}}-8c+2=-2$.

$ \displaystyle 3{{c}^{2}}-8c+2=-2;\,\,\,\,3{{c}^{2}}-8c+4=0;\,\,\,\,\,\left( {3c-2} \right)\left( {c-2} \right)=0;\,\,\,c=\frac{2}{3},\,\,2$

Check to make sure the value(s) we got lie in the interval $ [0,2]$. Since only $ \displaystyle c=\frac{2}{3}$ lies in the interval, the only value of $ c$ that is guaranteed by the MVT is $ \displaystyle c=\frac{2}{3}$.

Does the function $ \displaystyle f\left( x \right)=\frac{1}{{x-2}}$ satisfy the Mean Value Theorem on the interval $ [1,4]$?

Justify answer and then find the value of “$ c$” guaranteed by the theorem.

Since the function is not continuous (nor totally differentiable) in the interval $ [1,4]$, the MVT does not apply.

Here’s one more where we need to understand visually the Mean Value Theorem:

Mean Value Theorem Problem	Solution
A differentiable function $ f$ is graphed below on the closed interval $ \left[ {-8,4.5} \right]$. How many values of $ x$ in the open interval $ \left( {-8,4.5} \right)$ satisfy the Mean Value Theorem for $ f$ on $ \left[ {-8,4.5} \right]$?	First draw the secant line from $ x=a=-8$ and $ x=b=4.5$; this is the bottom line. Notice that there are three places in the open interval $ \left( {-8,4.5} \right)$ where the function touches either this secant line or a line parallel to the secant line (top line). At these three places, the slope of the tangent line through the points (“$ c$”) is equal to the slope of the secant line. Thus, there are three values of $ x$ where $ \displaystyle {f}’\left( c \right)=\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}$.

Mean Value Theorem Problem

Solution

A differentiable function $ f$ is graphed below on the closed interval $ \left[ {-8,4.5} \right]$.

How many values of $ x$ in the open interval $ \left( {-8,4.5} \right)$ satisfy the Mean Value Theorem for $ f$ on $ \left[ {-8,4.5} \right]$?

First draw the secant line from $ x=a=-8$ and $ x=b=4.5$; this is the bottom line.

Notice that there are three places in the open interval $ \left( {-8,4.5} \right)$ where the function touches either this secant line or a line parallel to the secant line (top line). At these three places, the slope of the tangent line through the points (“$ c$”) is equal to the slope of the secant line.

Thus, there are three values of $ x$ where $ \displaystyle {f}’\left( c \right)=\frac{{f\left( a \right)-f\left( b \right)}}{{b-a}}$.

Learn these rules, and practice, practice, practice!

Click on Submit (the blue arrow to the right of the problem) and click on Find Where the Mean Value Theorem is Satisfied to see the answer. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.

If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.

You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!

On to Curve Sketching – you are ready!