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The Mean Value
Theorem

Discover how the instantaneous rate of change must equal the average rate of change at least once on any smooth curve.

AP Calculus AB

What Is the Average Rate of Change?

Given a function \( f \) on the interval \( [a, b] \), the average rate of change is the slope of the secant line connecting the endpoints:

This tells you how fast \( f \) changes on average across the entire interval: it's the "big picture" slope.

What Is the Instantaneous Rate?

The instantaneous rate of change at a point \( x = c \) is the derivative \( f'(c) \): the slope of the tangent line at that exact point.

Secant Line
Connects two points on the curve. Slope = average rate of change over \( [a, b] \).
Tangent Line
Touches the curve at one point. Slope = instantaneous rate of change \( f'(c) \).

Why Does the MVT Matter?

The Mean Value Theorem is one of the most powerful results in calculus. It connects the local behavior (derivative) of a function to its global behavior (net change).

Proofs
It's used to prove that a function with \( f'(x) = 0 \) everywhere is constant.
Applications
Speed traps: if average speed exceeds the limit, you exceeded it at some instant.
Estimation
It guarantees bounds on function values when you only know the derivative's range.

The Formal Statement

In words: somewhere between \( a \) and \( b \), the tangent line is parallel to the secant line connecting \( (a, f(a)) \) and \( (b, f(b)) \).

Visualizing the MVT

A continuous route through hills: elevation, grade, and intermediate height

A walking path crosses a ridge and hollow beside a staff-bearing observer.Continuity on the closed interval and differentiability in its interior guarantee a tangent parallel to the endpoint secant.

Mean Value Theorem: match a slope

Continuity on the closed interval and differentiability in its interior guarantee a tangent parallel to the endpoint secant.Continuity on the closed interval and differentiability in its interior guarantee a tangent parallel to the endpoint secant.

The Two Hypotheses

Both conditions are essential. If either fails, the MVT conclusion may not hold.

Continuous on \( [a, b] \)
The function has no jumps, holes, or vertical asymptotes on the closed interval. Endpoints are included.
Differentiable on \( (a, b) \)
The function has no sharp corners or vertical tangents on the open interval. Endpoints are excluded.

MVT vs Rolle's Theorem

Mean Value Theorem Rolle's Theorem
Hypotheses Continuous on \( [a,b] \), differentiable on \( (a,b) \) Same + \( f(a) = f(b) \)
Conclusion \( f'(c) = \frac{f(b)-f(a)}{b-a} \) \( f'(c) = 0 \)
Secant slope Any value Exactly \( 0 \)
Relationship General case Special case of MVT

How to Apply the MVT

1
Check continuity on \( [a, b] \). Is \( f \) defined and continuous at every point, including endpoints?
2
Check differentiability on \( (a, b) \). Does \( f'(x) \) exist at every interior point?
3
Compute the average rate: \( \displaystyle\frac{f(b) - f(a)}{b - a} \).
4
Set \( f'(c) \) equal to that rate and solve for \( c \) in \( (a, b) \).
5
Verify that each solution \( c \) lies in the open interval \( (a, b) \).

Example 1: \( f(x) = x^2 \)

Find the value of \( c \) guaranteed by the MVT for \( f(x) = x^2 \) on \( [1, 3] \).

1
Hypotheses: \( f(x) = x^2 \) is a polynomial: continuous and differentiable everywhere. ✓
2
Average rate: \( \displaystyle\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \) 4
3
Set \( f'(c) = 4 \): Since \( f'(x) = 2x \), we get \( 2c = 4 \), so \( c = 2 \).
4
Verify: \( c = 2 \in (1, 3) \). ✓

Example 1: Visual

Example 2: \( f(x) = x^3 - 3x \)

Find all values of \( c \) guaranteed by the MVT for \( f(x) = x^3 - 3x \) on \( [-1, 2] \).

1
Hypotheses: Polynomial: continuous and differentiable everywhere. ✓
2
Compute endpoints: \( f(-1) = -1 + 3 = 2 \), \( f(2) = 8 - 6 = 2 \).
3
Average rate: \( \displaystyle\frac{2 - 2}{2 - (-1)} = \frac{0}{3} = \) 0
4
Solve \( f'(c) = 0 \): \( 3c^2 - 3 = 0 \implies c^2 = 1 \implies c = \pm 1 \). Only \( c = 1 \) lies in \( (-1, 2) \).

Example 3: \( f(x) = \sqrt{x} \)

Find \( c \) for \( f(x) = \sqrt{x} \) on \( [0, 4] \).

1
Continuous on \( [0,4] \)? Yes. Differentiable on \( (0,4) \)? Yes (\( f'(x) = \frac{1}{2\sqrt{x}} \) exists for \( x > 0 \)). ✓
2
Average rate: \( \displaystyle\frac{\sqrt{4} - \sqrt{0}}{4 - 0} = \frac{2}{4} = \) \( \frac{1}{2} \)
3
Solve: \( \frac{1}{2\sqrt{c}} = \frac{1}{2} \implies \sqrt{c} = 1 \implies \) \( c = 1 \)
4
Verify: \( c = 1 \in (0, 4) \). ✓

Common Mistakes

✕ Forgetting Hypotheses
Applying the MVT without verifying that \( f \) is continuous on \( [a,b] \) and differentiable on \( (a,b) \).
✕ Wrong Interval for \( c \)
The MVT guarantees \( c \) in the open interval \( (a,b) \). Answers at the endpoints are invalid.
✕ Confusing MVT with IVT
The IVT is about function values between \( f(a) \) and \( f(b) \). The MVT is about derivative values.

Summary Table

MVT Rolle's IVT
About Derivatives Derivatives Function values
Requires Cont. + Diff. Cont. + Diff. + \( f(a)=f(b) \) Continuity only
Guarantees \( f'(c) = \frac{f(b)-f(a)}{b-a} \) \( f'(c) = 0 \) \( f(c) = k \)
Key idea Tangent ∥ secant Horizontal tangent Hits every value

Question 1 of 5

The MVT guarantees that for \( f(x) = x^2 + 1 \) on \( [0, 2] \), there exists \( c \in (0,2) \) such that \( f'(c) \) equals:

Question 2 of 5

Which condition is not required for the Mean Value Theorem to apply?

Question 3 of 5

For \( f(x) = x^3 \) on \( [0, 3] \), what is the value of \( c \) guaranteed by the MVT?

Question 4 of 5

The MVT cannot be applied to which function on \( [-1, 1] \)?

Question 5 of 5

A car travels 150 miles in 2 hours. By the MVT, which statement is guaranteed?