5.1 Important theorems

Achievable AP Calculus AB

5. Analytical uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Important theorems

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What you’ll learn:

3 more big theorems in calculus

Mean value theorem (MVT): instantaneous rate = average rate somewhere on the interval
Rolle’s theorem: special case of the MVT
Extreme value theorem: finding absolute max and min on closed interval

In addition to the Intermediate value theorem (from section 1.6 on Continuity), there are two other main existence theorems in calculus: the Mean value theorem (with Rolle’s theorem as a special case) and the Extreme value theorem.

Mean value theorem

Mean value theorem:

If a function $f (x)$ satisfies these 2 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$

Here’s the idea in plain language: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then somewhere between $a$ and $b$ the instantaneous rate of change (the derivative) matches the average rate of change over the whole interval.

Examples

Determine if the Mean value theorem can be applied to $f (x)$ on the interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2}, [1, 4]$
b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$
c) $f (x) = 1 - x^{2}, [- 1, 0]$

Solutions

a) $f (x) = x^{2}, [1, 4]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[1, 4]$ is:

$\frac{f ( 4 ) - f ( 1 )}{4 - 1}$

$= \frac{16 - 1}{3}$

$= 5$

The derivative of $f (x)$ is

$f^{'} (x) = 2 x$

Set the derivative equal to the average rate of change:

$2 c = 5$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ is the point where the instantaneous rate of change equals the average rate of change.

b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$

(spoiler)

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[0, 3]$ is:

$\frac{f ( 3 ) - f ( 0 )}{3 - 0}$

$= \frac{6 - 0}{3}$

$= 2$

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2} - 6 x + 2$

Set the derivative equal to the average rate of change:

$3 c^{2} - 6 c + 2 = 2$

$3 c^{2} - 6 c = 0$

$3 c (c - 2) = 0$

$c = 0, 2$

Because $c$ must be in the open interval $(0, 3)$ , we eliminate $c = 0$ . So $c = 2$ is the only value that satisfies the MVT.

c) $f (x) = 1 - x^{2}, [- 1, 0]$

(spoiler)

$f (x)$ is the top half of the unit circle. It is continuous on the interval $[- 1, 0]$ even though at the endpoint $x = - 1$ , it doesn’t have a two-sided limit.

This is still fine, because continuity on a closed interval $[a, b]$ only requires:

continuity from the right at the left endpoint: $x \to a^{+} lim f (x) = f (a)$
continuity from the left at the right endpoint: $x \to b^{-} lim f (x) = f (b)$

Also, while the derivative does not exist at the endpoint $x = - 1$ , the MVT only requires differentiability on the open interval $(- 1, 0)$ . Since $f (x)$ is differentiable on $(- 1, 0)$ , the MVT applies.

The average rate of change over $[- 1, 0]$ is

$\frac{f ( 0 ) - f ( - 1 )}{0 - ( - 1 )}$

$= \frac{1 - 0}{1}$

$= 1$

The derivative of $f (x)$ is

$f^{'} (x) = - \frac{x}{1 - x ^{2}}$

Set the derivative equal to the average rate of change:

$- \frac{c}{1 - c ^{2}} = 1$

$- c = 1 - c^{2}$

$c^{2} = 1 - c^{2}$

$2 c^{2} = 1$

$c = \pm \frac{2}{2}$

Because we squared both sides, we may have introduced an extraneous solution. Checking both values in the original equation shows that $c = - \frac{2}{2}$ is the only solution that satisfies the MVT.

A car drives 100 miles between 9 AM and 11 AM on a long stretch of road with a posted speed limit of 35 mph. Has the driver broken the speed limit laws at any point?

Solution

(spoiler)

The average speed is distance divided by time:

$\frac{100 miles}{2 hours} = 50 mph$

If the car’s position is a continuous and differentiable function of time (a reasonable model for real driving), then the MVT guarantees that at some point the car’s instantaneous speed was 50 mph. Since $50 > 35$ , the driver must have exceeded the speed limit at some point.

Rolle’s theorem

Rolle’s theorem is a special case of the Mean value theorem where the function starts and ends at the same value, meaning $f (a) = f (b)$ . In that situation, Rolle’s theorem guarantees at least one point in the interval where the tangent line is horizontal (so the derivative is $0$ ).

Rolle’s theorem:

If a function $f (x)$ satisfies these 3 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$
$f (a) = f (b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = 0$

Examples

Determine if Rolle’s theorem can be applied to $f (x)$ on the closed interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$
b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$
c) $f (x) = ∣ x ∣, [- 2, 2]$

Solutions

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere.

Also,

$f (0) = 1 and f (5) = 25 - 25 + 1 = 1,$

so $f (0) = f (5)$ . Rolle’s theorem applies.

The derivative is

$f^{'} (x) = 2 x - 5.$

Set $f^{'} (c) = 0$ :

$2 c - 5 = 0$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ . This matches the graph: there is a horizontal tangent at $x = \frac{5}{2}$ .

b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$

(spoiler)

$f (x)$ is a rational function and is discontinuous at $x = 3$ , which lies in the interval $[- 1, 6]$ . Since the continuity condition fails, Rolle’s theorem cannot be applied. A horizontal tangent might still exist, but the theorem does not guarantee one.

c) $f (x) = ∣ x ∣, [- 2, 2]$

(spoiler)

$f (x)$ is continuous on $[- 2, 2]$ , but it is not differentiable at $x = 0$ , which is inside the interval. Since differentiability on $(a, b)$ fails, Rolle’s theorem does not apply. In fact, there is no point where the tangent line is horizontal.

Extreme value theorem

The extreme value theorem guarantees that a continuous function has both a highest value and a lowest value on a closed interval. These absolute extrema can occur at interior points or at the endpoints.

Extreme value theorem

If $f (x)$ is continuous on a closed interval $[a, b]$ , then it must attain both an absolute maximum and an absolute minimum value on that interval.

These extrema occur at either critical points (where $f^{'} (x) = 0$ or is undefined) or at the endpoints ( $a$ and $b$ ) of the interval.

To use the extreme value theorem to find the absolute maximum and minimum on an interval:

Verify continuity on the interval. If there are any breaks, jumps, or asymptotes, the EVT is not applicable.
Find critical points: Solve $f^{'} (x) = 0$ or determine where it is undefined within $[a, b]$ .
Evaluate function values: Compute $f (x)$ at the critical points and at the endpoints - $f (a)$ and $f (b)$ .
Compare function values: the largest function value is the absolute maximum and the smallest is the absolute minimum on that interval.

Examples

Find the absolute extrema of $f (x) = x^{2} - 4 x + 3$ on $[0, 5]$ .

Solution

(spoiler)

1. Verify continuity

$f (x)$ is a polynomial, so it’s continuous.

2. Find critical points

$f^{'} (x) = 2 x - 4$

Solve $f^{'} (x) = 0$ :

$2 x - 4 = 0$

$x = 2$

3. Evaluate function values

At the critical point $x = 2$ ,

$f (2) = 2^{2} - 4 (2) + 3 = - 1$

At the endpoints,

$f (0) = 0^{2} - 4 (0) + 3 = 3$

$f (5) = 5^{2} - 4 (5) + 3 = 8$

4. Compare function values

The absolute maximum is $8$ , which occurs at $x = 5$ .

The absolute minimum is $- 1$ , which occurs at $x = 2$ .

Find the absolute extrema of

$f (x) = \frac{x}{x + 1}$

on $[0, 3]$ .

Solution

(spoiler)

1. Verify continuity

This rational function has a vertical asymptote at $x = - 1$ , but $- 1$ is not in $[0, 3]$ . So $f$ is continuous on the given interval.

2. Find critical points

$f^{'} (x) = \frac{( x + 1 ) ( 1 ) - ( x ) ( 1 )}{( x + 1 ) ^{2}}$

$= \frac{1}{( x + 1 ) ^{2}}$

Since $f^{'} (x) > 0$ for all $x$ in $[0, 3]$ , there are no critical points in the interval.

3. Evaluate function values

With no critical points, only evaluate the endpoints:

$f (0) = \frac{0}{0 + 1}$

$= 0$

$f (3) = \frac{3}{3 + 1}$

$= \frac{3}{4}$

4. Compare function values.

The absolute minimum is $0$ at $x = 0$ .

The absolute maximum is $\frac{3}{4}$ at $x = 3$ .

Mean value theorem (MVT)

Applies if $f (x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$
Guarantees $\exists c \in (a, b)$ such that $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$
Instantaneous rate equals average rate at some point

MVT examples

Check continuity and differentiability before applying MVT
Solve $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for $c$ in $(a, b)$
Real-world: If average speed exceeds speed limit, MVT guarantees speed limit was broken at some instant

Rolle’s theorem

Special case of MVT: $f (a) = f (b)$
Requires continuity on $[a, b]$ , differentiability on $(a, b)$ , and $f (a) = f (b)$
Guarantees $\exists c \in (a, b)$ such that $f^{'} (c) = 0$
- At least one horizontal tangent in $(a, b)$

Rolle’s theorem examples

Must check all three conditions (continuity, differentiability, $f (a) = f (b)$ )
If any condition fails, theorem does not apply
Find $c$ by solving $f^{'} (c) = 0$ in $(a, b)$

Extreme value theorem (EVT)

If $f (x)$ is continuous on $[a, b]$ , $f$ attains absolute maximum and minimum on $[a, b]$
Extrema occur at:
- Critical points ( $f^{'} (x) = 0$ or undefined in $[a, b]$ )
- Endpoints $a$ and $b$
Procedure:
- Verify continuity on $[a, b]$
- Find all critical points in $[a, b]$
- Evaluate $f (x)$ at critical points and endpoints
- Largest value = absolute max; smallest = absolute min

EVT examples

For polynomials: always continuous, so EVT applies
For rational functions: check for discontinuities in $[a, b]$
If no critical points, compare only endpoint values

Important theorems

What you’ll learn:

3 more big theorems in calculus

Mean value theorem (MVT): instantaneous rate = average rate somewhere on the interval
Rolle’s theorem: special case of the MVT
Extreme value theorem: finding absolute max and min on closed interval

Mean value theorem

Mean value theorem:

If a function $f (x)$ satisfies these 2 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$

Examples

Determine if the Mean value theorem can be applied to $f (x)$ on the interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2}, [1, 4]$
b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$
c) $f (x) = 1 - x^{2}, [- 1, 0]$

Solutions

a) $f (x) = x^{2}, [1, 4]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[1, 4]$ is:

$\frac{f ( 4 ) - f ( 1 )}{4 - 1}$

$= \frac{16 - 1}{3}$

$= 5$

The derivative of $f (x)$ is

$f^{'} (x) = 2 x$

Set the derivative equal to the average rate of change:

$2 c = 5$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ is the point where the instantaneous rate of change equals the average rate of change.

b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$

(spoiler)

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[0, 3]$ is:

$\frac{f ( 3 ) - f ( 0 )}{3 - 0}$

$= \frac{6 - 0}{3}$

$= 2$

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2} - 6 x + 2$

Set the derivative equal to the average rate of change:

$3 c^{2} - 6 c + 2 = 2$

$3 c^{2} - 6 c = 0$

$3 c (c - 2) = 0$

$c = 0, 2$

Because $c$ must be in the open interval $(0, 3)$ , we eliminate $c = 0$ . So $c = 2$ is the only value that satisfies the MVT.

c) $f (x) = 1 - x^{2}, [- 1, 0]$

(spoiler)

$f (x)$ is the top half of the unit circle. It is continuous on the interval $[- 1, 0]$ even though at the endpoint $x = - 1$ , it doesn’t have a two-sided limit.

This is still fine, because continuity on a closed interval $[a, b]$ only requires:

continuity from the right at the left endpoint: $x \to a^{+} lim f (x) = f (a)$
continuity from the left at the right endpoint: $x \to b^{-} lim f (x) = f (b)$

The average rate of change over $[- 1, 0]$ is

$\frac{f ( 0 ) - f ( - 1 )}{0 - ( - 1 )}$

$= \frac{1 - 0}{1}$

$= 1$

The derivative of $f (x)$ is

$f^{'} (x) = - \frac{x}{1 - x ^{2}}$

Set the derivative equal to the average rate of change:

$- \frac{c}{1 - c ^{2}} = 1$

$- c = 1 - c^{2}$

$c^{2} = 1 - c^{2}$

$2 c^{2} = 1$

$c = \pm \frac{2}{2}$

Because we squared both sides, we may have introduced an extraneous solution. Checking both values in the original equation shows that $c = - \frac{2}{2}$ is the only solution that satisfies the MVT.

A car drives 100 miles between 9 AM and 11 AM on a long stretch of road with a posted speed limit of 35 mph. Has the driver broken the speed limit laws at any point?

Solution

(spoiler)

The average speed is distance divided by time:

$\frac{100 miles}{2 hours} = 50 mph$

Rolle’s theorem

Rolle’s theorem:

If a function $f (x)$ satisfies these 3 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$
$f (a) = f (b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = 0$

Examples

Determine if Rolle’s theorem can be applied to $f (x)$ on the closed interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$
b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$
c) $f (x) = ∣ x ∣, [- 2, 2]$

Solutions

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere.

Also,

$f (0) = 1 and f (5) = 25 - 25 + 1 = 1,$

so $f (0) = f (5)$ . Rolle’s theorem applies.

The derivative is

$f^{'} (x) = 2 x - 5.$

Set $f^{'} (c) = 0$ :

$2 c - 5 = 0$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ . This matches the graph: there is a horizontal tangent at $x = \frac{5}{2}$ .

b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$

(spoiler)

c) $f (x) = ∣ x ∣, [- 2, 2]$

(spoiler)

Extreme value theorem

Extreme value theorem

If $f (x)$ is continuous on a closed interval $[a, b]$ , then it must attain both an absolute maximum and an absolute minimum value on that interval.

These extrema occur at either critical points (where $f^{'} (x) = 0$ or is undefined) or at the endpoints ( $a$ and $b$ ) of the interval.

To use the extreme value theorem to find the absolute maximum and minimum on an interval:

Verify continuity on the interval. If there are any breaks, jumps, or asymptotes, the EVT is not applicable.
Find critical points: Solve $f^{'} (x) = 0$ or determine where it is undefined within $[a, b]$ .
Evaluate function values: Compute $f (x)$ at the critical points and at the endpoints - $f (a)$ and $f (b)$ .
Compare function values: the largest function value is the absolute maximum and the smallest is the absolute minimum on that interval.

Examples

Find the absolute extrema of $f (x) = x^{2} - 4 x + 3$ on $[0, 5]$ .

Solution

(spoiler)

1. Verify continuity

$f (x)$ is a polynomial, so it’s continuous.

2. Find critical points

$f^{'} (x) = 2 x - 4$

Solve $f^{'} (x) = 0$ :

$2 x - 4 = 0$

$x = 2$

3. Evaluate function values

At the critical point $x = 2$ ,

$f (2) = 2^{2} - 4 (2) + 3 = - 1$

At the endpoints,

$f (0) = 0^{2} - 4 (0) + 3 = 3$

$f (5) = 5^{2} - 4 (5) + 3 = 8$

4. Compare function values

The absolute maximum is $8$ , which occurs at $x = 5$ .

The absolute minimum is $- 1$ , which occurs at $x = 2$ .

Find the absolute extrema of

$f (x) = \frac{x}{x + 1}$

on $[0, 3]$ .

Solution

(spoiler)

1. Verify continuity

This rational function has a vertical asymptote at $x = - 1$ , but $- 1$ is not in $[0, 3]$ . So $f$ is continuous on the given interval.

2. Find critical points

$f^{'} (x) = \frac{( x + 1 ) ( 1 ) - ( x ) ( 1 )}{( x + 1 ) ^{2}}$

$= \frac{1}{( x + 1 ) ^{2}}$

Since $f^{'} (x) > 0$ for all $x$ in $[0, 3]$ , there are no critical points in the interval.

3. Evaluate function values

With no critical points, only evaluate the endpoints:

$f (0) = \frac{0}{0 + 1}$

$= 0$

$f (3) = \frac{3}{3 + 1}$

$= \frac{3}{4}$

4. Compare function values.

The absolute minimum is $0$ at $x = 0$ .

The absolute maximum is $\frac{3}{4}$ at $x = 3$ .

Achievable AP Calculus AB

5. Analytical uses

Our AP Calculus AB course is currently in development and is a work-in-progress.

Important theorems

8 min read

Font

Discuss

Feedback

What you’ll learn:

3 more big theorems in calculus

Mean value theorem (MVT): instantaneous rate = average rate somewhere on the interval
Rolle’s theorem: special case of the MVT
Extreme value theorem: finding absolute max and min on closed interval

Mean value theorem

Mean value theorem:

If a function $f (x)$ satisfies these 2 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$

Examples

Determine if the Mean value theorem can be applied to $f (x)$ on the interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2}, [1, 4]$
b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$
c) $f (x) = 1 - x^{2}, [- 1, 0]$

Solutions

a) $f (x) = x^{2}, [1, 4]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[1, 4]$ is:

$\frac{f ( 4 ) - f ( 1 )}{4 - 1}$

$= \frac{16 - 1}{3}$

$= 5$

The derivative of $f (x)$ is

$f^{'} (x) = 2 x$

Set the derivative equal to the average rate of change:

$2 c = 5$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ is the point where the instantaneous rate of change equals the average rate of change.

b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$

(spoiler)

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[0, 3]$ is:

$\frac{f ( 3 ) - f ( 0 )}{3 - 0}$

$= \frac{6 - 0}{3}$

$= 2$

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2} - 6 x + 2$

Set the derivative equal to the average rate of change:

$3 c^{2} - 6 c + 2 = 2$

$3 c^{2} - 6 c = 0$

$3 c (c - 2) = 0$

$c = 0, 2$

Because $c$ must be in the open interval $(0, 3)$ , we eliminate $c = 0$ . So $c = 2$ is the only value that satisfies the MVT.

c) $f (x) = 1 - x^{2}, [- 1, 0]$

(spoiler)

$f (x)$ is the top half of the unit circle. It is continuous on the interval $[- 1, 0]$ even though at the endpoint $x = - 1$ , it doesn’t have a two-sided limit.

This is still fine, because continuity on a closed interval $[a, b]$ only requires:

continuity from the right at the left endpoint: $x \to a^{+} lim f (x) = f (a)$
continuity from the left at the right endpoint: $x \to b^{-} lim f (x) = f (b)$

The average rate of change over $[- 1, 0]$ is

$\frac{f ( 0 ) - f ( - 1 )}{0 - ( - 1 )}$

$= \frac{1 - 0}{1}$

$= 1$

The derivative of $f (x)$ is

$f^{'} (x) = - \frac{x}{1 - x ^{2}}$

Set the derivative equal to the average rate of change:

$- \frac{c}{1 - c ^{2}} = 1$

$- c = 1 - c^{2}$

$c^{2} = 1 - c^{2}$

$2 c^{2} = 1$

$c = \pm \frac{2}{2}$

Because we squared both sides, we may have introduced an extraneous solution. Checking both values in the original equation shows that $c = - \frac{2}{2}$ is the only solution that satisfies the MVT.

A car drives 100 miles between 9 AM and 11 AM on a long stretch of road with a posted speed limit of 35 mph. Has the driver broken the speed limit laws at any point?

Solution

(spoiler)

The average speed is distance divided by time:

$\frac{100 miles}{2 hours} = 50 mph$

Rolle’s theorem

Rolle’s theorem:

If a function $f (x)$ satisfies these 3 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$
$f (a) = f (b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = 0$

Examples

Determine if Rolle’s theorem can be applied to $f (x)$ on the closed interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$
b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$
c) $f (x) = ∣ x ∣, [- 2, 2]$

Solutions

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere.

Also,

$f (0) = 1 and f (5) = 25 - 25 + 1 = 1,$

so $f (0) = f (5)$ . Rolle’s theorem applies.

The derivative is

$f^{'} (x) = 2 x - 5.$

Set $f^{'} (c) = 0$ :

$2 c - 5 = 0$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ . This matches the graph: there is a horizontal tangent at $x = \frac{5}{2}$ .

b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$

(spoiler)

c) $f (x) = ∣ x ∣, [- 2, 2]$

(spoiler)

Extreme value theorem

Extreme value theorem

If $f (x)$ is continuous on a closed interval $[a, b]$ , then it must attain both an absolute maximum and an absolute minimum value on that interval.

These extrema occur at either critical points (where $f^{'} (x) = 0$ or is undefined) or at the endpoints ( $a$ and $b$ ) of the interval.

To use the extreme value theorem to find the absolute maximum and minimum on an interval:

Verify continuity on the interval. If there are any breaks, jumps, or asymptotes, the EVT is not applicable.
Find critical points: Solve $f^{'} (x) = 0$ or determine where it is undefined within $[a, b]$ .
Evaluate function values: Compute $f (x)$ at the critical points and at the endpoints - $f (a)$ and $f (b)$ .
Compare function values: the largest function value is the absolute maximum and the smallest is the absolute minimum on that interval.

Examples

Find the absolute extrema of $f (x) = x^{2} - 4 x + 3$ on $[0, 5]$ .

Solution

(spoiler)

1. Verify continuity

$f (x)$ is a polynomial, so it’s continuous.

2. Find critical points

$f^{'} (x) = 2 x - 4$

Solve $f^{'} (x) = 0$ :

$2 x - 4 = 0$

$x = 2$

3. Evaluate function values

At the critical point $x = 2$ ,

$f (2) = 2^{2} - 4 (2) + 3 = - 1$

At the endpoints,

$f (0) = 0^{2} - 4 (0) + 3 = 3$

$f (5) = 5^{2} - 4 (5) + 3 = 8$

4. Compare function values

The absolute maximum is $8$ , which occurs at $x = 5$ .

The absolute minimum is $- 1$ , which occurs at $x = 2$ .

Find the absolute extrema of

$f (x) = \frac{x}{x + 1}$

on $[0, 3]$ .

Solution

(spoiler)

1. Verify continuity

This rational function has a vertical asymptote at $x = - 1$ , but $- 1$ is not in $[0, 3]$ . So $f$ is continuous on the given interval.

2. Find critical points

$f^{'} (x) = \frac{( x + 1 ) ( 1 ) - ( x ) ( 1 )}{( x + 1 ) ^{2}}$

$= \frac{1}{( x + 1 ) ^{2}}$

Since $f^{'} (x) > 0$ for all $x$ in $[0, 3]$ , there are no critical points in the interval.

3. Evaluate function values

With no critical points, only evaluate the endpoints:

$f (0) = \frac{0}{0 + 1}$

$= 0$

$f (3) = \frac{3}{3 + 1}$

$= \frac{3}{4}$

4. Compare function values.

The absolute minimum is $0$ at $x = 0$ .

The absolute maximum is $\frac{3}{4}$ at $x = 3$ .

Mean value theorem (MVT)

Applies if $f (x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$
Guarantees $\exists c \in (a, b)$ such that $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$
Instantaneous rate equals average rate at some point

MVT examples

Check continuity and differentiability before applying MVT
Solve $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for $c$ in $(a, b)$
Real-world: If average speed exceeds speed limit, MVT guarantees speed limit was broken at some instant

Rolle’s theorem

Special case of MVT: $f (a) = f (b)$
Requires continuity on $[a, b]$ , differentiability on $(a, b)$ , and $f (a) = f (b)$
Guarantees $\exists c \in (a, b)$ such that $f^{'} (c) = 0$
- At least one horizontal tangent in $(a, b)$

Rolle’s theorem examples

Must check all three conditions (continuity, differentiability, $f (a) = f (b)$ )
If any condition fails, theorem does not apply
Find $c$ by solving $f^{'} (c) = 0$ in $(a, b)$

Extreme value theorem (EVT)

If $f (x)$ is continuous on $[a, b]$ , $f$ attains absolute maximum and minimum on $[a, b]$
Extrema occur at:
- Critical points ( $f^{'} (x) = 0$ or undefined in $[a, b]$ )
- Endpoints $a$ and $b$
Procedure:
- Verify continuity on $[a, b]$
- Find all critical points in $[a, b]$
- Evaluate $f (x)$ at critical points and endpoints
- Largest value = absolute max; smallest = absolute min

EVT examples

For polynomials: always continuous, so EVT applies
For rational functions: check for discontinuities in $[a, b]$
If no critical points, compare only endpoint values

Important theorems

What you’ll learn:

3 more big theorems in calculus

Mean value theorem (MVT): instantaneous rate = average rate somewhere on the interval
Rolle’s theorem: special case of the MVT
Extreme value theorem: finding absolute max and min on closed interval

Mean value theorem

Mean value theorem:

If a function $f (x)$ satisfies these 2 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$

Examples

Determine if the Mean value theorem can be applied to $f (x)$ on the interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2}, [1, 4]$
b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$
c) $f (x) = 1 - x^{2}, [- 1, 0]$

Solutions

a) $f (x) = x^{2}, [1, 4]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[1, 4]$ is:

$\frac{f ( 4 ) - f ( 1 )}{4 - 1}$

$= \frac{16 - 1}{3}$

$= 5$

The derivative of $f (x)$ is

$f^{'} (x) = 2 x$

Set the derivative equal to the average rate of change:

$2 c = 5$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ is the point where the instantaneous rate of change equals the average rate of change.

b) $f (x) = x^{3} - 3 x^{2} + 2 x, [0, 3]$

(spoiler)

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere. The MVT applies.

The average rate of change over $[0, 3]$ is:

$\frac{f ( 3 ) - f ( 0 )}{3 - 0}$

$= \frac{6 - 0}{3}$

$= 2$

The derivative of $f (x)$ is

$f^{'} (x) = 3 x^{2} - 6 x + 2$

Set the derivative equal to the average rate of change:

$3 c^{2} - 6 c + 2 = 2$

$3 c^{2} - 6 c = 0$

$3 c (c - 2) = 0$

$c = 0, 2$

Because $c$ must be in the open interval $(0, 3)$ , we eliminate $c = 0$ . So $c = 2$ is the only value that satisfies the MVT.

c) $f (x) = 1 - x^{2}, [- 1, 0]$

(spoiler)

$f (x)$ is the top half of the unit circle. It is continuous on the interval $[- 1, 0]$ even though at the endpoint $x = - 1$ , it doesn’t have a two-sided limit.

This is still fine, because continuity on a closed interval $[a, b]$ only requires:

continuity from the right at the left endpoint: $x \to a^{+} lim f (x) = f (a)$
continuity from the left at the right endpoint: $x \to b^{-} lim f (x) = f (b)$

The average rate of change over $[- 1, 0]$ is

$\frac{f ( 0 ) - f ( - 1 )}{0 - ( - 1 )}$

$= \frac{1 - 0}{1}$

$= 1$

The derivative of $f (x)$ is

$f^{'} (x) = - \frac{x}{1 - x ^{2}}$

Set the derivative equal to the average rate of change:

$- \frac{c}{1 - c ^{2}} = 1$

$- c = 1 - c^{2}$

$c^{2} = 1 - c^{2}$

$2 c^{2} = 1$

$c = \pm \frac{2}{2}$

Because we squared both sides, we may have introduced an extraneous solution. Checking both values in the original equation shows that $c = - \frac{2}{2}$ is the only solution that satisfies the MVT.

A car drives 100 miles between 9 AM and 11 AM on a long stretch of road with a posted speed limit of 35 mph. Has the driver broken the speed limit laws at any point?

Solution

(spoiler)

The average speed is distance divided by time:

$\frac{100 miles}{2 hours} = 50 mph$

Rolle’s theorem

Rolle’s theorem:

If a function $f (x)$ satisfies these 3 conditions:

Continuity: $f (x)$ is continuous on the closed interval $[a, b]$
Differentiability: $f (x)$ is differentiable on the open interval $(a, b)$
$f (a) = f (b)$

Then there exists at least one point $c$ in $(a, b)$ such that

$f^{'} (c) = 0$

Examples

Determine if Rolle’s theorem can be applied to $f (x)$ on the closed interval given and if so, find the values of $c$ that satisfy the theorem.

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$
b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$
c) $f (x) = ∣ x ∣, [- 2, 2]$

Solutions

a) $f (x) = x^{2} - 5 x + 1, [0, 5]$

$f (x)$ is a polynomial, so it’s continuous and differentiable everywhere.

Also,

$f (0) = 1 and f (5) = 25 - 25 + 1 = 1,$

so $f (0) = f (5)$ . Rolle’s theorem applies.

The derivative is

$f^{'} (x) = 2 x - 5.$

Set $f^{'} (c) = 0$ :

$2 c - 5 = 0$

$c = \frac{5}{2}$

So $c = \frac{5}{2}$ . This matches the graph: there is a horizontal tangent at $x = \frac{5}{2}$ .

b) $f (x) = \frac{x ^{2} + 9}{x - 3}, [- 1, 6]$

(spoiler)

c) $f (x) = ∣ x ∣, [- 2, 2]$

(spoiler)

Extreme value theorem

Extreme value theorem

If $f (x)$ is continuous on a closed interval $[a, b]$ , then it must attain both an absolute maximum and an absolute minimum value on that interval.

These extrema occur at either critical points (where $f^{'} (x) = 0$ or is undefined) or at the endpoints ( $a$ and $b$ ) of the interval.

To use the extreme value theorem to find the absolute maximum and minimum on an interval:

Verify continuity on the interval. If there are any breaks, jumps, or asymptotes, the EVT is not applicable.
Find critical points: Solve $f^{'} (x) = 0$ or determine where it is undefined within $[a, b]$ .
Evaluate function values: Compute $f (x)$ at the critical points and at the endpoints - $f (a)$ and $f (b)$ .
Compare function values: the largest function value is the absolute maximum and the smallest is the absolute minimum on that interval.

Examples

Find the absolute extrema of $f (x) = x^{2} - 4 x + 3$ on $[0, 5]$ .

Solution

(spoiler)

1. Verify continuity

$f (x)$ is a polynomial, so it’s continuous.

2. Find critical points

$f^{'} (x) = 2 x - 4$

Solve $f^{'} (x) = 0$ :

$2 x - 4 = 0$

$x = 2$

3. Evaluate function values

At the critical point $x = 2$ ,

$f (2) = 2^{2} - 4 (2) + 3 = - 1$

At the endpoints,

$f (0) = 0^{2} - 4 (0) + 3 = 3$

$f (5) = 5^{2} - 4 (5) + 3 = 8$

4. Compare function values

The absolute maximum is $8$ , which occurs at $x = 5$ .

The absolute minimum is $- 1$ , which occurs at $x = 2$ .

Find the absolute extrema of

$f (x) = \frac{x}{x + 1}$

on $[0, 3]$ .

Solution

(spoiler)

1. Verify continuity

This rational function has a vertical asymptote at $x = - 1$ , but $- 1$ is not in $[0, 3]$ . So $f$ is continuous on the given interval.

2. Find critical points

$f^{'} (x) = \frac{( x + 1 ) ( 1 ) - ( x ) ( 1 )}{( x + 1 ) ^{2}}$

$= \frac{1}{( x + 1 ) ^{2}}$

Since $f^{'} (x) > 0$ for all $x$ in $[0, 3]$ , there are no critical points in the interval.

3. Evaluate function values

With no critical points, only evaluate the endpoints:

$f (0) = \frac{0}{0 + 1}$

$= 0$

$f (3) = \frac{3}{3 + 1}$

$= \frac{3}{4}$

4. Compare function values.

The absolute minimum is $0$ at $x = 0$ .

The absolute maximum is $\frac{3}{4}$ at $x = 3$ .