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

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

In Words

There is at least one point where the instantaneous rate of change (derivative) equals the average rate of change (slope of secant line).

Visual Understanding

The Secant Line

The secant line connects $(a, f(a))$ to $(b, f(b))$.

Its slope is: $\frac{f(b) - f(a)}{b - a}$

The Tangent Line

The MVT says there's a point $c$ where the tangent line is parallel to the secant line.

Geometric interpretation: Somewhere between $a$ and $b$, the curve has a tangent line with the same slope as the overall average slope.

Rolle's Theorem (Special Case)

Rolle's Theorem is a special case of MVT when $f(a) = f(b)$.

Statement

If $f$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $f(a) = f(b)$, then there exists $c$ in $(a, b)$ such that:

$f'(c) = 0$

In Words

If a function starts and ends at the same height, somewhere in between it has a horizontal tangent!

Example: If you hike up and back down to your starting elevation, at some point you reached the top (where slope = 0).

How to Apply the MVT

Step-by-Step Process

Step 1: Verify the hypotheses

• Is $f$ continuous on $[a, b]$?
• Is $f$ differentiable on $(a, b)$?

Step 2: Calculate the average rate of change $m = \frac{f(b) - f(a)}{b - a}$

Step 3: Find $f'(x)$

Step 4: Solve $f'(c) = m$ for $c$

Step 5: Verify that $c$ is in $(a, b)$

Example 1: Finding the Point

Find all values of $c$ that satisfy the MVT for $f(x) = x^2$ on $[1, 3]$.

Step 1: Check hypotheses

$f(x) = x^2$ is a polynomial:

• Continuous everywhere ✓
• Differentiable everywhere ✓

MVT applies!

Step 2: Average rate of change

$\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4$

Step 3: Find derivative

$f'(x) = 2x$

Step 4: Solve $f'(c) = 4$

$2c = 4$ $c = 2$

Step 5: Verify

Is $c = 2$ in $(1, 3)$? Yes! ✓

Answer: $c = 2$ satisfies the MVT.

At $x = 2$, the instantaneous rate equals the average rate!

Example 2: MVT Doesn't Apply

Why doesn't MVT apply to $f(x) = |x|$ on $[-1, 1]$?

$f$ is continuous on $[-1, 1]$ ✓

But $f$ is NOT differentiable at $x = 0$ (corner/cusp) ✗

Since $0$ is in $(-1, 1)$, the MVT does not apply.

Why the Hypotheses Matter

Continuity Required

If $f$ has a jump discontinuity, there may be no point where the tangent slope equals the average slope.

Differentiability Required

If $f$ has a corner (like $|x|$ at $x=0$), the derivative doesn't exist there, so we can't apply MVT.

On $(a, b)$, not $[a, b]$

The function only needs to be differentiable on the open interval. It's okay if $f'(a)$ or $f'(b)$ don't exist!

Applications of MVT

Application 1: Proving Functions are Equal

If $f'(x) = g'(x)$ for all $x$ in an interval, then: $f(x) = g(x) + C$

for some constant $C$.

Why: If $h(x) = f(x) - g(x)$, then $h'(x) = 0$ everywhere. By MVT, $h$ must be constant!

Application 2: Estimating Values

If we know $f'(x)$ is bounded, we can estimate how much $f$ can change.

Example: If $|f'(x)| \leq 2$ for all $x$ in $[0, 5]$, then: $|f(5) - f(0)| \leq 2 \cdot 5 = 10$

Application 3: Speed Limits

If a car travels 120 miles in 2 hours (average 60 mph), by MVT the car must have been going exactly 60 mph at some instant!

If the speed limit is 55 mph, the driver was definitely speeding at some point. 🚗

Increasing and Decreasing Functions

Important Corollary

Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$.

Increasing Function Test:

• If $f'(x) > 0$ for all $x$ in $(a, b)$, then $f$ is increasing on $[a, b]$

Decreasing Function Test:

• If $f'(x) < 0$ for all $x$ in $(a, b)$, then $f$ is decreasing on $[a, b]$

Constant Function Test:

• If $f'(x) = 0$ for all $x$ in $(a, b)$, then $f$ is constant on $[a, b]$

Proof idea: Uses MVT repeatedly!

⚠️ Common Mistakes

Mistake 1: Wrong Interval

The value $c$ must be in the open interval $(a, b)$, not including endpoints!

Mistake 2: Not Checking Hypotheses

Always verify continuity and differentiability before applying MVT!

Mistake 3: Expecting Unique $c$

MVT says "at least one" $c$ exists. There might be multiple values!

Mistake 4: Confusing with Extreme Value Theorem

• EVT: Guarantees max and min exist
• MVT: Guarantees a point where instantaneous rate = average rate

MVT vs. Other Theorems

Extreme Value Theorem (EVT)

EVT: Continuous function on $[a, b]$ has absolute max and min

MVT: Connects average and instantaneous rates

Intermediate Value Theorem (IVT)

IVT: Continuous function takes all values between $f(a)$ and $f(b)$

MVT: About derivatives, not function values

Extended Mean Value Theorem (Cauchy)

There's a more general version for two functions:

If $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists $c$ in $(a, b)$ such that:

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

This is used in proving L'Hôpital's Rule!

Proof Sketch of MVT

Idea: Create a function that measures the vertical distance from the curve to the secant line, then apply Rolle's Theorem.

Define: $h(x) = f(x) - \left[f(a) + \frac{f(b)-f(a)}{b-a}(x-a)\right]$

Then $h(a) = h(b) = 0$, so by Rolle's Theorem, there exists $c$ where $h'(c) = 0$.

Working out $h'(c) = 0$ gives $f'(c) = \frac{f(b)-f(a)}{b-a}$ ✓

📝 Key Takeaways

1. MVT connects average and instantaneous rates - somewhere the tangent is parallel to the secant

2. Requires continuity on $[a, b]$ and differentiability on $(a, b)$

3. Guarantees existence of at least one $c$, but doesn't tell you how many

4. Applications: Proving functions equal, bounding changes, theoretical foundation

5. Geometric meaning: Tangent line parallel to secant line

6. Rolle's Theorem is the special case when $f(a) = f(b)$

Practice Tips

1. Always check hypotheses before applying MVT
2. Calculate average rate first: $\frac{f(b)-f(a)}{b-a}$
3. Set $f'(c)$ equal to the average rate and solve
4. Verify $c$ is in the open interval $(a, b)$
5. Understand geometrically - it's about parallel slopes!

$f$ is NOT continuous at $x = 0$ ✗

Since $0 \in [-1, 1]$, the function is not continuous on the entire interval.

Conclusion

The Mean Value Theorem does not apply because $f$ is not continuous on $[-1, 1]$.

Additional observation:

Even if we tried to apply it:

$f(-1) = -1$ and $f(1) = 1$

Average rate: $\frac{1 - (-1)}{1 - (-1)} = \frac{2}{2} = 1$

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

Setting $-\frac{1}{c^2} = 1$ gives $c^2 = -1$, which has no real solution!

This makes sense because the function is discontinuous - there's no point where the tangent slope equals the average slope.

Answer: MVT does not apply because $f$ is not continuous on $[-1, 1]$ (discontinuity at $x = 0$).