Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

The Mean Value Theorem

The Mean Value Theorem relates a function and its derivative. Several important results in calculus are proved using this theorem, among them a very important theorem we will study in the next section: the Fundamental Theorem of Calculus.

The geometric interpretation of the Mean Value Theorem is intuitively simple. This theorem says that if $f$ is a continuous function on a closed interval ${\displaystyle \left[a,b\right]}$ and its graph is smooth, then the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ is parallel to at least one tangent line to the curve. See Figure 6.2, below.

Figure 6.2: The Mean Value Theorem

So we know that the slopes of the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ and the indicated tangent line are equal.

The slope of the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ is ${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\text{.}}$

The slope of the tangent line is ${\displaystyle f^{\prime }\left(c\right)}$ for some ${\displaystyle a<c<b\text{.}}$

Therefore,

${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}=f^{\prime }\left(c\right)\text{.}}$

Theorem 6.3. The Mean Value Theorem.

Let $f$ be a function satisfying the two conditions identified below.

1. It is continuous on a closed interval ${\displaystyle \left[a,b\right]\text{.}}$
2. It is differentiable on the open interval ${\displaystyle \left(a,b\right)\text{.}}$

Then

${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}=f^{\prime }\left(c\right)\text{ for some }a<c<b}$

or, equivalently

${\displaystyle f\left(b\right)-f\left(a\right)=f^{\prime }\left(c\right)\left(b-a\right)\text{.}}$

The second condition in this theorem guarantees that the graph of the function is smooth.

In terms of rate of change, this theorem is saying that for at least one ${\displaystyle a<c<b\text{,}}$ the instantaneous rate of change, ${\displaystyle f^{\prime }\left(c\right)\text{,}}$ is equal to the average rate of change,

${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}},}$

on the interval ${\displaystyle \left[a,b\right]\text{.}}$

Example 6.23. If a car travels 250 km in 2 hours and 15 minutes, then its average velocity is

$\dfrac{{250}}{{2.25}} = 111.11\;{\text{km/hr}}$,

and by the Mean Value Theorem, at some time the car’s velocity is exactly equal to its average velocity. Hence, the car exceeded the legal speed limit of 100 km/hr at least once during the trip.

Example 6.24. In 2005, Asafa Powell’s world record in the $100\;\mbox{m}$ race was ${\displaystyle 9.77}$ s, and in 1996, Michael Johnson’s world record for the $200\;\mbox{m}$ race was ${\displaystyle 19.32}$ s. The Mean Value Theorem says that at some time during his race, Powell’s velocity was equal to his record’s average velocity of

${\displaystyle {\displaystyle \frac{100}{9.77}}=10.235\text{m/s;}}$

and also at some time during his race, Johnson’s velocity was equal to his record’s average velocity of

${\displaystyle {\displaystyle \frac{200}{19.32}}=10.352\text{m/s.}}$

Therefore, over the $200\;\mbox{m}$ distance, Johnson was running faster than Powell did over the $100\;\mbox{m}$ distance.

You may want to check previous records and see, based on the Mean Value Theorem, how often it is the case that a $200\;\mbox{m}$ runner is faster than a $100\;\mbox{m}$ runner.^[1]

For a $100\;\mbox{m}$ runner to be as fast as Johnson was in 1996 when he established his record, his $100\;\mbox{m}$ record would have to be

${\displaystyle {\displaystyle \frac{100}{10.352}}=9.66\text{s}\text{.}}$

The second-best speed in the $200\;\mbox{m}$ race is $19.89\;\mbox{s}$, established by Tyson Gay in 2007; his average velocity was

${\displaystyle {\displaystyle \frac{200}{19.62}}=10.19\text{m/s.}}$

Example 6.25. In 1988, Florence Griffith-Joyner established the world’s records for the $100\;\mbox{m}$ and $200\;\mbox{m}$ races at $10.49\;\mbox{s}$, and $21.34\;\mbox{s}$, respectively. By the Mean Value Theorem, at some time during each of these races, Griffith-Joyner was running at a velocity equal to her average velocity for that race. That is, in the $100\;\mbox{m}$ race, she ran at

${\displaystyle {\displaystyle \frac{100}{10.49}}=9.53\text{m/s;}}$

and in the $200\;\mbox{m}$ race, she ran at

${\displaystyle {\displaystyle \frac{200}{21.34}}=9.37\text{m/s.}}$

At some point, she was faster in the $100\;\mbox{m}$ race than in the $200\;\mbox{m}$ race.

For a $200\;\mbox{m}$ runner to run as fast as Griffith-Joyner ran in the $100\;\mbox{m}$ race, her record would have to be

${\displaystyle {\displaystyle \frac{200}{9.53}}=20.98\text{s.}}$

The second-best speed for the $100\;\mbox{m}$ race is $10.84\;\mbox{s}$, established by Chandra Sturrup in 2005, with an average velocity of

${\displaystyle {\displaystyle \frac{100}{10.84}}=9.22\text{m/s,}}$

and the second best world record of the $200\;\mbox{m}$ race is $22.13\;\mbox{s}$, established by Allyson Felix in 2006 with an average velocity of

${\displaystyle {\displaystyle \frac{200}{22.11}}=9.04\text{m/s.}}$

By the Mean Value Theorem, if we are told that ${\displaystyle f\left(a\right)=f\left(b\right)\text{,}}$ then the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ is horizontal, with slope $0$.

Then, ${\displaystyle f^{\prime }\left(c\right)=0}$ for some ${\displaystyle a<c<b\text{.}}$ This particular case is known as Rolle’s Theorem. See Figure 6.3.

Figure 6.3: Rolle’s Theorem

Theorem 6.4. Rolle’s Theorem

Let $f$ be a function satisfying the conditions identified below.

1. It is continuous on a closed interval ${\displaystyle \left[a,b\right]\text{.}}$
2. It is differentiable on the open interval ${\displaystyle \left(a,b\right)\text{.}}$
3. ${\displaystyle f\left(a\right)=f\left(b\right)\text{.}}$

Then

${\displaystyle f^{\prime }\left(c\right)=0\text{ for some }a<c<b\text{.}}$

Let us apply the Mean Value Theorem to prove several results we have already learned.

Theorem 6.5. If ${\displaystyle f^{\prime }\left(x\right)=0}$ for all $x$ on an interval ${\displaystyle \left(a,b\right)\text{,}}$ then ${\displaystyle f\left(x\right)=C}$ for some constant ${\displaystyle C\text{.}}$

Proof. We need to show that ${\displaystyle f\left(x_1\right)=f\left(x_2\right)}$ for any two numbers ${\displaystyle x_1}$ and ${\displaystyle x_2}$ in the interval ${\displaystyle \left(a,b\right)\text{.}}$

We take ${\displaystyle a<x_1<x_2<b\text{,}}$ and we consider the closed interval ${\displaystyle \left[x_1,x_2\right]\text{.}}$ We must check the conditions of the Mean Value Theorem for this interval. Since $f$ is differentiable on ${\displaystyle \left(a,b\right)\text{,}}$ it is continuous on ${\displaystyle \left[x_1,x_2\right]}$ and differentiable on ${\displaystyle \left(x_1,x_2\right)\text{.}}$ So, we conclude that

${\displaystyle {\displaystyle \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}}=f^{\prime }\left(c\right)\text{ for some }{x}_1<c<x_2\text{.}}$

For this ${\displaystyle c\text{,}}$ we know that ${\displaystyle f^{\prime }\left(c\right)=0\text{.}}$ [Why?]

So, ${\displaystyle {\displaystyle \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}}=0,}$ and ${\displaystyle f\left(x_2\right)-f\left(x_1\right)=0\text{.}}$

This result implies that $f(x)$ is constant on the interval ${\displaystyle \left(a,b\right)\text{.}}$

Q.E.D

This theorem says that the antiderivative of the zero function is a constant function, as we indicated before.

Theorem 6.6. [See Theorem 5.5.]

1. If ${\displaystyle f^{\prime }\left(x\right)>0}$ for any $x$ on the interval ${\displaystyle J\text{,}}$ then the function $f$ is increasing on the interval ${\displaystyle J\text{.}}$
2. If ${\displaystyle f^{\prime }\left(x\right)<0}$ for any $x$ on the interval ${\displaystyle J\text{,}}$ then the function $f$ is decreasing on the interval ${\displaystyle J\text{.}}$

Proof. To prove statement (a), by Definition 5.4, we must show that ${\displaystyle f\left(a\right)<f\left(b\right)}$ for any ${\displaystyle a<b}$ in the interval ${\displaystyle J\text{.}}$

We consider the subinterval ${\displaystyle \left[a,b\right]}$ of ${\displaystyle J\text{.}}$ Since we are assuming that ${\displaystyle f^{\prime }\left(x\right)>0}$ for all $x$ in ${\displaystyle J\text{,}}$ the function is differentiable in the interval ${\displaystyle J\text{,}}$ and so it is continuous on ${\displaystyle \left[a,b\right]}$ and differentiable on ${\displaystyle \left(a,b\right)\text{.}}$ By the Mean Value Theorem

${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}=f^{\prime }\left(c\right)\text{ for some }a<c<b\text{.}}$

For this ${\displaystyle c\text{,}}$ we know that ${\displaystyle f^{\prime }\left(c\right)>0}$; hence,

${\displaystyle {\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}>0\text{.}}$

We also know that ${\displaystyle b-a>0\text{.}}$ [Why?].

Therefore, the quotient is positive only if

${\displaystyle f\left(b\right)-f\left(a\right)>0,}$

which implies that ${\displaystyle f\left(a\right)<f\left(b\right)\text{.}}$

The proof of statement (b) is similar.

Q.E.D

The second derivative test is also a consequence of the Mean Value Theorem. To see why, we must formalize the definition of concavity and prove a lemma.

In Definition 5.11, we indicated that a function is concave up on an interval $J$ if the graph of the function is below the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ for any ${\displaystyle a<b}$ in ${\displaystyle J\text{.}}$ This is the same as saying that $f(x)$ is below the line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ for any ${\displaystyle a<x<b\text{.}}$

The line segment from ${\displaystyle \left(a,f\left(a\right)\right)}$ to ${\displaystyle \left(b,f\left(b\right)\right)}$ is given by the following equation for ${\displaystyle a\le x\le b\text{.}}$ [Why?]

${\displaystyle y={\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\left(x-a\right)+f\left(a\right)\text{.}}$

Thus, a function is concave up if

${\displaystyle f\left(x\right)<{\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\left(x-a\right)+f\left(a\right)}$

for any ${\displaystyle a<x<b\text{.}}$

This statement is equivalent to saying that

${\displaystyle {\displaystyle \frac{f\left(x\right)-f\left(a\right)}{x-a}}<{\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}}$

The following definition is equivalent to Definition 5.11.

Lemma 6.8. Let $f$ be a differentiable function with ${\displaystyle f^{\prime }}$ increasing. If ${\displaystyle a<b}$ and ${\displaystyle f\left(a\right)=f\left(b\right)\text{,}}$ then ${\displaystyle f\left(x\right)<f\left(a\right)=f\left(b\right)}$ for any ${\displaystyle a<x<b\text{.}}$

Proof. Let us argue by contradiction. We suppose that ${\displaystyle f\left(x\right)\ge f\left(a\right)}$ for some $x$ in ${\displaystyle \left(a,b\right)\text{.}}$ This assumption will yield a contradiction; therefore, ${\displaystyle f\left(x\right)<f\left(a\right)}$ for any ${\displaystyle a<x<b\text{.}}$

Figure 6.4: Lemma 6.8, proof diagram ${\displaystyle a\text{,}}$ ${\displaystyle f\left(x\right)>f\left(a\right)=f\left(b\right)}$

If ${\displaystyle f\left(x\right)>f\left(a\right)=f\left(b\right)}$ for some ${\displaystyle a<x<b\text{,}}$ then the maximum of $f$ occurs at some point ${\displaystyle x_0}$ in ${\displaystyle \left(a,b\right)\text{.}}$ See Figure 6.4, above.

Hence, ${\displaystyle f\left(x_0\right)>f\left(a\right)}$ and ${\displaystyle f^{\prime }\left(x_0\right)=0}$ (local maxima occur at critical numbers). Applying the Mean Value Theorem in the interval ${\displaystyle \left[a,x_0\right]\text{,}}$ we find that there is an ${\displaystyle a<x_1<x_0}$ such that

${\displaystyle f^{\prime }\left(x_1\right)={\displaystyle \frac{f\left(x_0\right)-f\left(a\right)}{x_0-a}}>0=f^{\prime }\left(x_0\right)\text{.}}$

This equation shows that ${\displaystyle f^{\prime }\left(x_1\right)>f^{\prime }\left(x_0\right)\text{,}}$ but we know that ${\displaystyle x_1<x_0}$ and ${\displaystyle f^{\prime }}$ is increasing. By Definition 5.4, this cannot be. Therefore, ${\displaystyle f\left(x\right)\le f\left(a\right)}$ for any ${\displaystyle a<x<b\text{.}}$

Figure 6.5: Lemma 6.8, proof diagram ${\displaystyle b\text{,}}$ ${\displaystyle f\left(x\right)=f\left(a\right)}$

Suppose that ${\displaystyle f\left(x\right)=f\left(a\right)}$ for some ${\displaystyle a<x<b\text{.}}$ The function must then be as in Figure 6.5, above, and $f$ has a local maximum at ${\displaystyle x\text{,}}$ hence ${\displaystyle f^{\prime }\left(x\right)=0\text{.}}$

Since ${\displaystyle f^{\prime }}$ is increasing, it is not a constant on the interval ${\displaystyle \left[a,x\right]\text{.}}$ Therefore, there is ${\displaystyle a<x_1<x}$ such that ${\displaystyle f\left(x_1\right)<f\left(a\right)=f\left(x\right)\text{.}}$ Applying the Mean Value Theorem on the interval ${\displaystyle \left[x_1,x\right]\text{,}}$ we find that there is an ${\displaystyle x_1<x_2<x}$ such that

${\displaystyle f^{\prime }\left(x_2\right)={\displaystyle \frac{f\left(x\right)-f\left(x_1\right)}{x-x_1}}>0=f^{\prime }\left(x\right)\text{.}}$

This equation says that ${\displaystyle f^{\prime }\left(x_2\right)>f^{\prime }\left(x\right)}$ for ${\displaystyle x_2<x\text{,}}$ again contradicting the fact that ${\displaystyle f^{\prime }}$ is increasing.

Therefore it must be the case that ${\displaystyle f\left(x\right)<f\left(a\right)}$ for any ${\displaystyle a<x<b\text{.}}$

Q.E.D

Theorem 6.9. The Concavity Test

1. If ${\displaystyle f^{″}\left(x\right)>0}$ for all $x$ on an interval ${\displaystyle J\text{,}}$ then the function is concave up on the interval ${\displaystyle J\text{.}}$
2. If ${\displaystyle f^{″}\left(x\right)<0}$ for all $x$ on an interval ${\displaystyle J\text{,}}$ then the function is concave down on the interval ${\displaystyle J\text{.}}$

Proof. To prove the first statement, we must recognize that, if ${\displaystyle f^{″}\left(x\right)>0}$ for all $x$ in ${\displaystyle J\text{,}}$ then ${\displaystyle f^{\prime }\left(x\right)}$ is increasing on the interval ${\displaystyle J\text{.}}$ Let ${\displaystyle a<b}$ in $J$ and let ${\displaystyle g\left(x\right)}$ be the function defined for ${\displaystyle a\le x\le b}$ as follows:

${\displaystyle g\left(x\right)=f\left(x\right)-{\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\left(x-a\right)\text{.}}$

Thus, ${\displaystyle g^{″}\left(x\right)=f^{″}\left(x\right)>0}$ (check it), and ${\displaystyle g^{\prime }}$ is increasing. Moreover, ${\displaystyle g\left(a\right)=g\left(b\right)=f\left(a\right)\text{.}}$

Applying Lemma 6.8 to ${\displaystyle g\text{,}}$ we find that for any ${\displaystyle a<x<b\text{,}}$

${\displaystyle g\left(x\right)<g\left(a\right)=f\left(a\right)\text{.}}$

This equation states that, if ${\displaystyle a<x<b\text{,}}$ then

${\displaystyle f\left(x\right)-{\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\left(x-a\right)<f\left(a\right)}$

or

${\displaystyle {\displaystyle \frac{f\left(a\right)-f\left(x\right)}{x-a}}<{\displaystyle \frac{f\left(b\right)-f\left(a\right)}{b-a}}\text{.}}$

The proof for the second statement is left as an exercise (see below).

Q.E.D

Footnotes

^[1] Note that the same athlete can usually run a 200 m race faster than double their time in a $100\;\mbox{m}$ race, largely because they enter the second hundred metres already running at full speed.