5.1 Important theorems

Achievable AP Calculus AB

5. Analytical uses

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Important theorems

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What you’ll learn

Three major existence theorems in calculus:

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

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

As shown in the figure below, function $f$ satisfies the Mean value theorem at two values, $c_{1}$ and $c_{2}$ .

At these points, the slopes of the tangent lines, $f^{'} (c_{1})$ and $f^{'} (c_{2})$ , are equal to the slope of the secant line passing through the endpoints of the interval $x = a$ and $x = b$ .

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

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

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

Solutions

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

(spoiler)

$f (x)$ is a polynomial, which is continuous and differentiable for all $x$ , so 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:

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

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

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

(spoiler)

$f (x)$ is a polynomial, so 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:

$3 c^{2} - 6 c + 2 3 c^{2} - 6 c 3 c (c - 2) c = 2 = 0 = 0 = 0, 2$

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

(spoiler)

$f (x)$ is the upper half of a circle of radius $1$ .

1. Continuity

Although the semicircle is not defined past the left endpoint for $x < - 1$ , 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: $x \to b^{-} lim f (x) = f (b)$

So the 1st condition of the MVT is satisfied.

2. Differentiability

Even though $f^{'} (- 1)$ does not exist, the MVT only requires differentiability on the open interval $(- 1, 0)$ . Since $f (x)$ is differentiable on the entire interval $(- 1, 0)$ , the 2nd condition is also satisfied.

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:

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

$c = - \frac{2}{2}$ is the only solution that satisfies the MVT because it lies within the open interval $(- 1, 0)$ .

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 (a derivative of $0$ ).

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

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

c) $f (x) = ∣ x ∣$ on $[- 2, 2]$

Solutions

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

(spoiler)

$f (x)$ is a polynomial so the first two conditions of Rolle’s theorem are met.

For the 3rd condition, check:

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

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

Next, differentiate $f (x)$ :

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

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

$2 c - 5 c = 0 = \frac{5}{2}$

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

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

(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 in the interval, but the theorem does not guarantee one.

c) $f (x) = ∣ x ∣$ on $[- 2, 2]$

(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 on this graph 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.

AP tip:

If a problem contains the key phrase “absolute extrema” (minima or maxima), the Extreme value theorem is likely to be relevant.

Examples

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

(spoiler)

1. Verify continuity

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

2. Find critical points

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

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

$2 x - 4 x = 0 = 2$

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, evaluate only 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

Three major existence theorems in calculus:

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

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

As shown in the figure below, function $f$ satisfies the Mean value theorem at two values, $c_{1}$ and $c_{2}$ .

At these points, the slopes of the tangent lines, $f^{'} (c_{1})$ and $f^{'} (c_{2})$ , are equal to the slope of the secant line passing through the endpoints of the interval $x = a$ and $x = b$ .

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

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

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

Solutions

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

(spoiler)

$f (x)$ is a polynomial, which is continuous and differentiable for all $x$ , so 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:

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

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

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

(spoiler)

$f (x)$ is a polynomial, so 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:

$3 c^{2} - 6 c + 2 3 c^{2} - 6 c 3 c (c - 2) c = 2 = 0 = 0 = 0, 2$

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

(spoiler)

$f (x)$ is the upper half of a circle of radius $1$ .

1. Continuity

Although the semicircle is not defined past the left endpoint for $x < - 1$ , 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: $x \to b^{-} lim f (x) = f (b)$

So the 1st condition of the MVT is satisfied.

2. Differentiability

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:

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

$c = - \frac{2}{2}$ is the only solution that satisfies the MVT because it lies within the open interval $(- 1, 0)$ .

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

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

c) $f (x) = ∣ x ∣$ on $[- 2, 2]$

Solutions

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

(spoiler)

$f (x)$ is a polynomial so the first two conditions of Rolle’s theorem are met.

For the 3rd condition, check:

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

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

Next, differentiate $f (x)$ :

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

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

$2 c - 5 c = 0 = \frac{5}{2}$

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

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

(spoiler)

c) $f (x) = ∣ x ∣$ on $[- 2, 2]$

(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.

AP tip:

If a problem contains the key phrase “absolute extrema” (minima or maxima), the Extreme value theorem is likely to be relevant.

Examples

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

(spoiler)

1. Verify continuity

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

2. Find critical points

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

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

$2 x - 4 x = 0 = 2$

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, evaluate only 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 now in "early access" - get 50% off for a limited time.