5.1 - Using the Mean Value Theorem

Formal Definition

If: -

f

is continuous on the closed interval

[a, b]

f

is differentiable on the open interval

(a, b)

Then:

There exists at least one point

c

(a, b)

such that:

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

In words: the instantaneous rate of change (the derivative) at some point

c

equals the average rate of change over the interval.

Visual Understanding

In this diagram: - The secant line from A to B has a slope equal to the average rate of change - At points C₁ and C₂, the tangent lines are parallel to the secant line - That means

f'(c_1) = f'(c_2) = \frac{f(b) - f(a)}{b - a}

Important note: the Mean Value Theorem guarantees at least one such point exists. There might be more than one, as shown here, or exactly one.

Example 1: Finding the MVT Point

Given:

f(x) = -x^2 + 3x + 10

on the interval

[2, 6]

. Find all values of

c

that satisfy the Mean Value Theorem.

Step 1: Calculate the average rate of change

\frac{f(6) - f(2)}{6 - 2}

Find

f(6)

f(6) = -(6)^2 + 3(6) + 10 = -36 + 18 + 10 = -8

Find

f(2)

f(2) = -(2)^2 + 3(2) + 10 = -4 + 6 + 10 = 12

Average rate of change:

\frac{-8 - 12}{6 - 2} = \frac{-20}{4} = -5

Step 2: Find where the instantaneous rate equals the average rate

f'(x) = -2x + 3

Set it equal to the average rate:

f'(c) = -5

-2c + 3 = -5

-2c = -8

\boxed{c = 4}

Conclusion: at

x = 4

, which is inside the interval

(2, 6)

, the instantaneous rate of change equals the average rate of change of

-5

When the Conditions Fail

The Mean Value Theorem only promises a point

c

when both conditions hold. If the graph has a corner, a cusp, or a break inside the interval, there might be no tangent line parallel to the secant line at all.

Example: Does the Mean Value Theorem apply to

f(x) = x^{2/3}

[-1, 8]

Try it anyway. The average rate of change is:

\frac{f(8) - f(-1)}{8 - (-1)} = \frac{4 - 1}{9} = \frac{1}{3}

The derivative is

f'(x) = \frac{2}{3}x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}

. Set it equal to the average rate:

\frac{2}{3\sqrt[3]{c}} = \frac{1}{3} \quad \Rightarrow \quad \sqrt[3]{c} = 2 \quad \Rightarrow \quad c = 8

But

c = 8

is an endpoint, not a point inside

(-1, 8)

. On

(-1, 0)

the derivative is negative, so it can’t equal

\frac{1}{3}

there either. No value of

c

works.

Why it fails:

f

is continuous on

[-1, 8]

, but

f'(0)

doesn’t exist because the graph has a cusp at the origin. So

f

isn’t differentiable on

(-1, 8)

, and the theorem doesn’t apply.

Justification to write: “The Mean Value Theorem does not apply because

f

is not differentiable at

x = 0

, which is in the interval

(-1, 8)

.” When a condition fails, name the condition and the point where it fails.

A failed condition means the theorem makes no promise. It doesn’t mean a point

c

can never exist. Some functions that break a condition still happen to have one.

Mean Value Theorem vs. Intermediate Value Theorem

These two get confused a lot. Here’s how to tell them apart.

Mean Value Theorem (MVT)

About: rates of change, meaning slopes and derivatives

Statement: if you draw a secant line between two points on a continuous, differentiable function, there’s at least one point where the tangent line is parallel to that secant line.

Requires: - Continuity on

[a, b]

- Differentiability on

(a, b)

Guarantees: a point

c

where

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

Intermediate Value Theorem (IVT)

About: function values, meaning

y

-values

Statement: if a function is continuous on an interval, it takes on every value between

f(a)

and

f(b)

Requires: - Continuity on

[a, b]

Guarantees: for any value

k

between

f(a)

and

f(b)

, there’s a point

c

where

f(c) = k

Quick Comparison

Feature	MVT	IVT
Focus	Slopes and rates	Function values
Requires	Differentiable	Continuous
Compares	Derivative to average rate	Function value to target
Finds	Where slope equals average	Where function equals target

Example 2: Hot Air Balloon (IVT and MVT)

A hot air balloon’s height in feet is given by

h(t)

, where

t

is time in minutes. The function

h(t)

is twice differentiable.

$t$ (min)	0	5	10	15	20	25	30
$h(t)$ (ft)	0	40	55	70	65	75	80

Part A: Is there a time between 5 and 15 minutes when the balloon is at 50 feet?

This one needs the Intermediate Value Theorem, since it’s asking about a specific height value.

Check conditions:

h(t)

is continuous on

[5, 15]

, since twice differentiable implies continuous.

Identify endpoints:

h(5) = 40

and

h(15) = 70

Apply IVT: since

h

is continuous on

[5, 15]

and

40 < 50 < 70

, the Intermediate Value Theorem guarantees there’s a value

c

(5, 15)

such that

h(c) = 50

Justification to write: “According to the Intermediate Value Theorem, since

h

is continuous on

[5, 15]

and

50

is between

h(5) = 40

and

h(15) = 70

, there must exist a time

c

in the interval

(5, 15)

where

h(c) = 50

feet.”

Part B: Is there a time between 20 and 30 minutes when the velocity is 1.5 ft/min?

This one needs the Mean Value Theorem, since it’s asking about a rate of change.

Check conditions:

h(t)

is differentiable on

[20, 30]

, since it’s twice differentiable.

Calculate the average velocity:

\frac{h(30) - h(20)}{30 - 20} = \frac{80 - 65}{10} = \frac{15}{10} = 1.5 \text{ ft/min}

Apply MVT: since

h

is differentiable on

[20, 30]

and the average rate of change is 1.5 ft/min, the Mean Value Theorem guarantees there’s a value

c

(20, 30)

such that

h'(c) = 1.5

ft/min.

Justification to write: “According to the Mean Value Theorem, since

h(t)

is differentiable on

[20, 30]

and the average velocity is

\frac{h(30) - h(20)}{30 - 20} = 1.5

ft/min, there must exist a time

c

in the interval

(20, 30)

where the instantaneous velocity

h'(c) = 1.5

ft/min.”

Example 3: Application with Temperature

The temperature

T

in degrees Fahrenheit of a room over a 6-hour period is modeled by a differentiable function. At

t = 1

hour,

T = 65°F

, and at

t = 5

hours,

T = 73°F

Question: must there be a time in the interval

(1, 5)

when the temperature is increasing at exactly 2°F per hour?

Calculate the average rate of temperature change:

\frac{T(5) - T(1)}{5 - 1} = \frac{73 - 65}{4} = \frac{8}{4} = 2 \text{ °F/hour}

Apply MVT: since

T

is differentiable, by the Mean Value Theorem there has to be a time

c

(1, 5)

where:

T'(c) = 2 \text{ °F/hour}

Answer: yes. The Mean Value Theorem guarantees at least one time when the instantaneous rate of temperature increase is exactly 2°F per hour.

Writing Proper Justifications

When you use MVT or IVT on exams, follow this structure.

For IVT

State the condition: “ $f$ is continuous on $[a, b]$ ”

Show the endpoint values: “ $f(a) = \_\_\_$ and $f(b) = \_\_\_$ ”

Apply the theorem: “By the Intermediate Value Theorem, there exists $c$ in $(a, b)$ such that $f(c) = k$ ”

For MVT

State the condition: “ $f$ is differentiable on $[a, b]$ ”

Calculate the average rate: “ $\frac{f(b) - f(a)}{b - a} = \_\_\_$ ”

Apply the theorem: “By the Mean Value Theorem, there exists $c$ in $(a, b)$ such that $f'(c) = \_\_\_$ ”

Common Mistakes

❌ Confusing MVT and IVT. MVT is about rates and slopes, IVT is about function values

❌ Forgetting to state the conditions. You have to explicitly say “continuous” for IVT or “differentiable” for MVT

❌ Using closed vs. open intervals incorrectly. Continuity is needed on $[a, b]$ (closed), differentiability on $(a, b)$ (open)

❌ Not showing the calculation. You have to show the average rate calculation for MVT, or the endpoint values for IVT

❌ Incomplete justification. You have to reference the theorem by name and show all the required conditions

When to Use Which Theorem

Use IVT when: - You’re asked if a function reaches a specific value - The question uses phrases like “equals,” “reaches,” “is at height,” or “passes through” - Only continuity is mentioned

Use MVT when: - You’re asked about rates of change, velocity, speed, or slopes - The question uses phrases like “rate of change equals,” “velocity is,” or “increasing at” - Differentiability is mentioned

Key Takeaways

Mean Value Theorem: the instantaneous rate equals the average rate at some point

MVT formula: $f'(c) = \frac{f(b) - f(a)}{b - a}$

Conditions for MVT: continuous on $[a, b]$ , differentiable on $(a, b)$

Geometric meaning: a tangent line parallel to the secant line exists

IVT vs. MVT: IVT is about values, MVT is about rates

At least one point. MVT guarantees at least one $c$ exists, and there could be more

Proper justification required. State the conditions, show your work, reference the theorem

Real-world applications: speed cameras, toll booths, any average vs. instantaneous rate scenario

Twice differentiable implies both differentiable AND continuous

Complete justifications. AP exams require stating conditions and showing calculations

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