Mean Value Theorem

Subtopic: Theorems

The Mean Value Theorem (MVT) guarantees that somewhere on a smooth curve, the instantaneous rate equals the average rate. Drive $100$ miles in $2$ hours? At some moment, your speedometer showed exactly $50$ mph. This theorem connects derivatives to function behavior and powers many proofs in analysis.

Introduction

If you drive from $A$ to $B$ and compute your average speed, somewhere along the journey your instantaneous speed must have matched that average. You can't go from slow to fast without passing through the middle.

The MVT formalizes this: somewhere between two points, the derivative equals the average slope.

The Theorem

If $ƒ$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c\in (a,b)$ such that: ${ƒ}^{\prime }\left(c\right)=\frac{ƒ\left(b\right)-ƒ\left(a\right)}{b-a}$

The left side is instantaneous rate at $c$; the right side is average rate over $[a,b]$.

Geometric Interpretation

Draw the secant line from $(a,ƒ\left(a\right))$ to $(b,ƒ\left(b\right))$. The MVT says there's a point c where the tangent line is parallel to this secant.

Visually: tilt the picture until the secant is horizontal. The highest and lowest points now have horizontal tangents—that's where ${ƒ}^{\prime }\left(c\right)=0=$ slope of secant.

Worked Example

Verify the Mean Value Theorem (MVT) for $ƒ\left(x\right)=x^2$ on $[1,3]$.

Average rate of change:

$\frac{ƒ\left(3\right)-ƒ\left(1\right)}{3-1}=\frac{9-1}{2}=4$

Find $c$ such that ${ƒ}^{\prime }\left(c\right)=4$.
Since ${ƒ}^{\prime }\left(x\right)=2x$, we solve $2c=4$, giving $c=2$.

Check: $c=2$ lies in $(1,3)$.

At $x=2$, the tangent line has slope $4$, matching the secant slope.

Rolle's Theorem (Special Case)

If $ƒ\left(a\right)=ƒ\left(b\right)$, the secant is horizontal, so the MVT becomes:

${ƒ}^{\prime }\left(c\right)=0$ for some $c\in (a,b)$

This is Rolle's Theorem: if f starts and ends at the same height, it has a horizontal tangent somewhere in between.

Proof Sketch

Define g(x) = f(x) - [secant line through (a,f(a)) and (b,f(b))].

Then g(a) = g(b) = 0. By Rolle's Theorem, g'(c) = 0 for some c.

But g'(x) = f'(x) - (slope of secant), so f'(c) = slope of secant.

Define $g\left(x\right)=ƒ\left(x\right)-$ [secant line through $(a,ƒ\left(a\right))$ to $(b,ƒ\left(b\right))$].

Then $g\left(a\right)=g\left(b\right)=0$. By Rolle's Theorem, there exists some $c\in (a,b)$ such that $g^{\prime }\left(c\right)=0.$

But $g^{\prime }\left(x\right)={ƒ}^{\prime }\left(x\right)-$ (slope of secant), so at that point, ${ƒ}^{\prime }\left(c\right)=$ slope of the secant line.

Key Applications

Constant Functions

If ${ƒ}^{\prime }\left(x\right)=0$ everywhere on an interval, then $ƒ$ is constant.

Proof: by MVT, $ƒ\left(b\right)-ƒ\left(a\right)={ƒ}^{\prime }\left(c\right)(b-a)=0$.

Monotonicity

If ${ƒ}^{\prime }\left(x\right)>0$ on an interval, then $ƒ$ is increasing.

Proof: $ƒ\left(b\right)-ƒ\left(a\right)={ƒ}^{\prime }\left(c\right)(b-a)>0$ when $b>a$.

Bounds on Function Change

If $|{ƒ}^{\prime }\left(x\right)|\leqq$M on $[a,b]$, then $|ƒ\left(b\right)-ƒ\left(a\right)|\leqq M|b-a|$.

The derivative bounds how fast the function can change.

Conditions Matter

The MVT requires continuity on $[a,b]$ and differentiability on $(a,b)$. Without these:

• $ƒ\left(x\right)=\left|x\right|$ on $[-1,1]$: continuous but not differentiable at $0$. Secant slope $=0$, but no point has derivative $0$.

• A step function: not continuous, MVT fails.

Summary

The Mean Value Theorem says ${ƒ}^{\prime }\left(c\right)=\frac{ƒ\left(b\right)-ƒ\left(a\right)}{b-a}$ for some $c\in (a,b)$, provided $ƒ$is continuous on $[a,b]$ and differentiable on $(a,b)$. Geometrically: some tangent is parallel to the secant. Special case: Rolle's Theorem (when $ƒ\left(a\right)=ƒ\left(b\right)$). Applications: proving functions with zero derivative are constant, establishing monotonicity, bounding function changes.