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## Section2.10The Mean Value Theorem

###### Motivating Questions
• Can we use global information about a function (for instance, the average rate of change) to say anything about the local behavior of the function (such as the instantaneous rate of change)?

When you know how a function behaves at every point, you know quite a lot about that function. This is the whole idea behind plotting points on a coordinate plane in order to determine what the graph of a function looks like. Indeed, given enough local information about a function, we can describe it very well globally. However, the opposite is not generally true. Suppose, for instance, that you know the graph of $y=f(x)$ is a parabola. Knowing that, can you determine the value of $f(3)\text{?}$ The answer, of course, is that you do not have enough information.

In general, global information is not enough to determine local information. In other words, knowing how a function behaves overall is not usually enough to say anything about how the function behaves at a specific point. However, there are some exceptions to this generalization. One such special case is the mean value theorem.

### SubsectionThe Mean Value Theorem

###### Example2.112

Suppose it took you exactly three quarters of an hour to drive the $59$ miles from Lincoln, Nebraska to Omaha, Nebraska. Is it possible that you drove less than $75$ miles per hour the entire trip?

Since $59$ miles were driven in $\frac34$ hours, the average speed (i.e. average rate of change) on the trip was

\begin{equation*} \frac{59\text{ miles}}{\frac34\text{ hours}}=\frac{59\cdot4}3\approx78.67\text{ mph}\text{.} \end{equation*}

Because this is an average, our intuition tells us that at some point on the trip the car must have been going at least as fast as $78.67$ miles per hour. This is precisely what the mean value theorem addresses.

###### The Mean Value Theorem

Let $f$ be a function that is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)\text{.}$ Then there exists a number $c\text{,}$ with $a\lt c \lt b\text{,}$ such that

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

In other words, $f(b)-f(a)=f'(c)(b-a)\text{.}$

While a proof of the mean value theorem is beyond the scope of this textbook, it is worth rephrasing the theorem in a way that is perhaps more clear. The mean value theorem essentially states that given any two points on a continuous curve, there is a point somewhere in the middle at which the line tangent to the curve is parallel to the secant line that connects the two points. A geometric representation of this is illustrated below in Figure2.113.

###### Example2.114

Consider the function $f(x)=2x^3-x$ on the interval $[1,3]\text{.}$ Find a value of $c$ on the interval $(1,3)$ such that

\begin{equation*} f'(c)=\frac{f(3)-f(1)}{3-1}. \end{equation*}
Hint

Use differentiation rules to find the derivative, then evaluate at $x=c$ and solve for $c\text{.}$

Answer

$c=\sqrt{\frac{13}{3}}\text{.}$

Solution

We start by evaluating the function at the endpoints of the interval, finding that

\begin{equation*} f(3)=2(27)-3=51, \quad \text{and} \quad f(1)=2(1)-1=1\text{.} \end{equation*}

So we are trying to find $c$ in the interval $(1,3)$ such that $f'(c)=\frac{51-1}{3-1}=25.$

Using the power rule and the sum rule, we have that $f'(x)=6x^2-1\text{,}$ so $f'(c)=6c^2-1\text{.}$ Solving the equation $6c^2-1=25\text{,}$ we find that $c=\pm\sqrt{\frac{13}{3}}\text{.}$ Since we are looking only on the interval $(1,3)\text{,}$ the value we want is $c=+\sqrt{\frac{13}{3}}\approx2.082\text{.}$

### SubsectionConsequences of the Mean Value Theorem

The mean value theorem has several important consequences that confirm some intuitive ideas that you may already hold. First, the increasing function theorem establishes that continuous functions with a positive derivative on an interval are increasing on that interval. More formally,

###### The Increasing Function Theorem

Suppose that $f$ is continuous on $a \leq x \leq b$ and differentiable on $a \lt x \lt b\text{.}$

• If $f'(x)>0$ on $a \lt x \lt b\text{,}$ then $f$ is increasing on $a \leq x \leq b\text{.}$
• If $f'(x) \geq 0$ on $a \lt x \lt b\text{,}$ then $f$ is non-decreasing on $a \leq x \leq b\text{.}$

This theorem is demonstrated in Figure2.115. When the slope of the tangent line is positive for all $x$ values on a given interval, we see that the function is increasing.

We can also state that a function with a derivative equal to zero on an interval is constant on that interval.

###### Constant Function Theorem

Suppose that $f$ is continuous on $a \leq x \leq b$ and differentiable on $a \lt x \lt b\text{.}$ If $f '(x) = 0$ on $a \lt x \lt b\text{,}$ then $f$ is constant on $a \leq x \leq b\text{.}$ Equivalently, if $f$ is not constant everywhere on $[a,b]\text{,}$ then there is some point $c$ on the interval $(a,b)$ at which $f'(c)\neq0\text{.}$

The following example will provide some intuition for the constant function theorem.

###### Example2.116

Suppose a car is being driven with cruise control set to $55$ miles per hour, and that the drive continues for $30$ minutes. Thinking of the car's velocity (in miles per hour) as a function of time (in minutes), the instantaneous rate of change of velocity (i.e., the instantaneous acceleration) at each minute of the drive is $0$ miles per hour per minute. In terms of function notation, we have $v=f(t)$ denoting the velocity as a function of time, and $a=f'(t)$ denoting the acceleration as the derivative of velocity, with $f'(t)=0$ for each $0\lt t\lt30\text{.}$

Suppose now that the car approaches a red light, leading the driver to turn off cruise control in order to slow down and eventually stop. However, cruise control is only off for $15$ seconds before the light turns green again, at which point the driver resets the cruise control and regains a speed of $55$ mph within a further $15$ seconds. When cruise control was turned off, the car slowed down a little bit before returning to cruising speed. Consequently, the instantaneous acceleration was nonzero during that $30$ second period. In other words, when the (velocity) function is not constant over an interval, its derivative (acceleration) is nonzero on that same interval.

We have one final result that can be used to formalize another intuitive idea. The racetrack principle argues that if two objects begin at the same place and the same time and one is traveling faster than the other, then the faster object will be ahead of the slower object. Formally, we write the principle in terms of functions.

###### The Racetrack Principle

Suppose that $g$ and $h$ are continuous on $a \leq x \leq b$ and differentiable on $a\lt x\lt b\text{,}$ and that $g'(x) \leq h'(x)$ for $a\lt x \lt b\text{.}$

• If $g(a)=h(a)\text{,}$ then $g(x) \leq h(x)$ for $a \leq x \leq b\text{.}$
• If $g(b)=h(b)\text{,}$ then $g(x) \geq h(x)$ for $a \leq x \leq b\text{.}$

###### Example2.117

Erica and Mason want to race each other down the street. After choosing start and finish lines, they get set to race. Starting together, Erica runs faster than Mason the entire time. Because of this, Erica is ahead of Mason for the whole race and reaches the finish line first.

Now suppose that Mason wants a rematch, but he isn't willing to play fair this time. Mason and Erica line up at the starting line, but Mason takes an early start. Erica again runs faster than Mason the entire time, and they reach the finish line together. This means Erica spent the race catching up to Mason, who was ahead of her for the whole race.

###### Example2.118

Use the racetrack principle to show that $\ln(x)\leq x-1$ for all $x\geq 1.$ Is the statement also true for all $0\lt x \leq 1\text{?}$

Hint

Compare function values at $x=1\text{.}$ How do the derivatives compare on the interval in question?

Answer

$\ln(x)\le x-1$ for every positive value of $x\text{.}$

Solution

Set $g(x)=\ln(x)$ and $h(x)=x-1\text{.}$ Both functions are continuous on $[1,\infty)$ and differentiable on $(1,\infty)\text{.}$

We take the derivative of each function:

\begin{equation*} g'(x)=\frac{1}{x}, \end{equation*}
\begin{equation*} h'(x)=1. \end{equation*}

Since $x\ge1$ and $\frac1x\le1$ are equivalent statements, we thus have $g'(x)\leq h'(x)$ for all $x\geq 1\text{.}$

Comparing function values when $x=1\text{,}$ we see that $g(1)=\ln(1)=0$ and $h(1)=1-1=0\text{.}$ Therefore the racetrack principle says $g(x)\leq h(x)$ (i.e. $\ln(x)\leq x-1$) for all $x\geq 1\text{.}$

Similarly, if we consider values $0\lt x \leq 1\text{,}$ we have that $g'(x)\geq h'(x)\text{.}$ Since $g(1)=h(1)\text{,}$ we have that $h(x)\geq g(x)$ or $x-1\geq \ln(x)$ for all $0\lt x\leq 1\text{.}$ Note that we have to be careful in applying the racetrack principle here, since $\ln(x)$ is not continuous (because it's undefined) at $x=0\text{.}$ But we can choose the lower bound of the interval to be any positive number less than $1\text{,}$ and then take the limit as that number approaches $0$ to obtain the result.