Chapter 3 · Applications of Derivatives

3.5 The Mean Value Theorem

Somewhere between two points, the instantaneous rate of change equals the average rate of change — Rolle's theorem, the MVT and its corollaries.

The Mean Value Theorem is one of the most important theorems in calculus. We start with a special case, Rolle's theorem, then look at the theorem itself and three corollaries that underpin the rest of the course.

Rolle's Theorem

Informally: if a differentiable function has equal outputs at the ends of an interval, then somewhere in between its derivative is zero — the graph must level off at least once on its way back.

📐 Theorem 3.8 — Rolle's Theorem
Let $f$ be continuous over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$ such that $f(a)=f(b)$. Then there exists at least one $c\in(a,b)$ such that $f^\prime(c)=0$.
✏️ Example 1 — Using Rolle's theorem
For $f(x)=x^2+2x$ on $[-2,0]$, verify that the function satisfies the criteria of Rolle's theorem and find all values $c$ in the interval where $f^\prime(c)=0$.
Step 1 $f$ is a polynomial, so it is continuous and differentiable everywhere. Also $f(-2)=4-4=0=f(0)$. The criteria hold.
Step 2 $f^\prime(x)=2x+2=2(x+1)$, so $f^\prime(c)=0$ gives $c=-1\in(-2,0)$.

The Mean Value Theorem and Its Meaning

Rolle's theorem is the special case $f(a)=f(b)$. Tilt the picture and you get the Mean Value Theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then somewhere the tangent line is parallel to the secant line joining $(a,f(a))$ and $(b,f(b))$. There may be more than one such point.

📐 Theorem 3.9 — Mean Value Theorem
Let $f$ be continuous over $[a,b]$ and differentiable over $(a,b)$. Then there exists at least one point $c\in(a,b)$ such that $$f^\prime(c)=\frac{f(b)-f(a)}{b-a}.$$
Interactive 1 — Hunting for c

The red secant joins the endpoints. Slide the blue tangent along the curve until it becomes parallel — the two slope readouts then agree, and the verdict turns green. Two options show what goes wrong when a hypothesis fails.

$x$—
secant slope $(f(b) - f(a))/(b - a)$—
tangent slope $f'(x)$—
hypotheses—
✏️ Example 2 — Verifying that the Mean Value Theorem applies
For $f(x)=\sqrt x$ over $[0,9]$, show that $f$ satisfies the hypotheses of the Mean Value Theorem, and find the values $c$ it guarantees.
Step 1 $\sqrt x$ is continuous on $[0,9]$ and differentiable on $(0,9)$, so the theorem applies.
Step 2 $f^\prime(x)=\dfrac{1}{2\sqrt x}$ and the secant slope is $\dfrac{\sqrt9-\sqrt0}{9-0}=\dfrac13$.
Step 3 Solve $\dfrac{1}{2\sqrt c}=\dfrac13$: $\sqrt c=\tfrac32$, so $c=\dfrac94$.
✏️ Example 3 — Mean Value Theorem and velocity
A rock dropped from $100$ ft has position $s(t)=-16t^2+100$ until it hits the ground.
  • a How long before it hits the ground?
  • b Find the average velocity from release to impact.
  • c Find the time guaranteed by the Mean Value Theorem when the instantaneous velocity equals $v_{\text{avg}}$.
(a) $-16t^2+100=0\Rightarrow t=\pm\tfrac52$; taking $t\ge0$, it lands at $t=\tfrac52$ s.
(b) $v_{\text{avg}}=\dfrac{s(5/2)-s(0)}{5/2-0}=\dfrac{0-100}{5/2}=-40$ ft/s.
(c) $s$ is continuous on $[0,5/2]$ and differentiable on its interior, so some $c$ has $s^\prime(c)=-40$. Since $s^\prime(t)=-32t$: $-32c=-40\Rightarrow c=\tfrac54$ s.
Meaning Halfway through the fall in time — not in distance — the rock is moving at exactly its average speed. For a quadratic the MVT point is always the midpoint of the interval.

Corollaries of the Mean Value Theorem

📐 Corollary 1 — Functions with a derivative of zero
Let $f$ be differentiable over an interval $I$. If $f^\prime(x)=0$ for all $x\in I$, then $f(x)$ is constant for all $x\in I$.
📐 Corollary 2 — Constant Difference Theorem
If $f$ and $g$ are differentiable over an interval $I$ and $f^\prime(x)=g^\prime(x)$ for all $x\in I$, then $f(x)=g(x)+C$ for some constant $C$. This is why every antiderivative in Chapter 4 carries a "$+C$".
📐 Corollary 3 — Increasing and decreasing functions
Let $f$ be continuous over $[a,b]$ and differentiable over $(a,b)$. If $f^\prime(x)>0$ for all $x\in(a,b)$, $f$ is increasing over $[a,b]$. If $f^\prime(x)<0$ for all $x\in(a,b)$, $f$ is decreasing over $[a,b]$.
⚠️ Why the corollaries need the MVT
Corollary 1 sounds obvious, but the proof is the MVT: for any two points $x_1<x_2$ in $I$, there is a $c$ between them with $f(x_2)-f(x_1)=f^\prime(c)(x_2-x_1)=0$. Without the theorem, "zero slope everywhere" would not by itself tell us anything about the values.

Exercises

I
Verify that $f(x)=2x^2-8x+6$ defined over $[1,3]$ satisfies the conditions of Rolle's theorem. Find all points $c$ guaranteed by Rolle's theorem.
Step 1 $f$ is a polynomial — continuous on $[1,3]$ and differentiable on $(1,3)$. $f(1)=2-8+6=0$ and $f(3)=18-24+6=0$, so $f(1)=f(3)$.
Step 2 $f^\prime(x)=4x-8=0\Rightarrow c=2\in(1,3)$.
II
Determine whether the Mean Value Theorem applies for the functions over the given interval $[a,b]$. Justify your answer (and where it applies, find $c$).
  • 1 $f(x)=x^3$ over $[0,2]$
  • 2 $y=x^3+2x+1$ over $[0,6]$
  • 3 $f(x)=(x-1)^{10}$ over $[0,2]$
  • 4 $y=e^x$ over $[0,1]$
  • 5 $y=\ln(2x+3)$ over $\left[-\tfrac32,0\right]$
  • 6 $y=\sqrt{9-x^2}$ over $[-3,3]$
  • 7 $y=x+\dfrac1x$ over $\left[\tfrac12,4\right]$
Test each function
$x$—
secant slope $(f(b) - f(a))/(b - a)$—
tangent slope $f'(x)$—
hypotheses—
1 Applies (polynomial). Secant slope $\dfrac{8-0}{2}=4$; $3c^2=4\Rightarrow c=\dfrac{2}{\sqrt3}\approx1.155$.
2 Applies. Secant slope $\dfrac{229-1}{6}=38$; $3c^2+2=38\Rightarrow c=2\sqrt3\approx3.464$.
3 Applies. $f(0)=f(2)=1$, secant slope $0$; $10(c-1)^9=0\Rightarrow c=1$ (Rolle's case).
4 Applies. Secant slope $e-1$; $e^c=e-1\Rightarrow c=\ln(e-1)\approx0.5413$.
5 Does not apply. $\ln(2x+3)$ is undefined at $x=-\tfrac32$ (it $\to-\infty$ there), so it is not continuous on the closed interval $\left[-\tfrac32,0\right]$.
6 Applies. $\sqrt{9-x^2}$ is continuous on $[-3,3]$ and differentiable on $(-3,3)$ (the vertical tangents are only at the endpoints). Secant slope $0$; $f^\prime(c)=\dfrac{-c}{\sqrt{9-c^2}}=0\Rightarrow c=0$.
7 Applies ($x=0$ is outside the interval). Secant slope $\dfrac{4.25-2.5}{3.5}=\tfrac12$; $1-\dfrac{1}{c^2}=\dfrac12\Rightarrow c=\sqrt2\approx1.414$.
III
A ball is dropped from a height of $200$ ft. Its position is $s(t)=-16t^2+200$. Find the time $t$ when the instantaneous velocity of the ball equals its average velocity.
Step 1 Landing: $16t^2=200\Rightarrow t=\sqrt{12.5}=\dfrac{5\sqrt2}{2}\approx3.536$ s.
Step 2 $v_{\text{avg}}=\dfrac{0-200}{5\sqrt2/2}=-\dfrac{80}{\sqrt2}=-40\sqrt2\approx-56.57$ ft/s.
Step 3 $s^\prime(t)=-32t=-40\sqrt2\Rightarrow t=\dfrac{5\sqrt2}{4}\approx1.768$ s — exactly half the fall time, as expected for a quadratic.
IV
For each function, decide whether Rolle's theorem or the Mean Value Theorem applies over the given interval. Where it applies, find every value $c$ it guarantees; where it does not, name the hypothesis that fails.
  • 1 Rolle: $f(x)=x^3-4x$ over $[-2,2]$
  • 2 Rolle: $f(x)=\sin 2x$ over $[0,\pi]$
  • 3 MVT: $f(x)=\dfrac1x$ over $[1,4]$
  • 4 MVT: $f(x)=x^3-x$ over $[0,3]$
  • 5 Rolle: $f(x)=x^{2/3}$ over $[-1,1]$
  • 6 Rolle: $f(x)=\tan x$ over $[0,\pi]$
Hunt for c — or find the broken hypothesis
$x$—
secant slope $(f(b) - f(a))/(b - a)$—
tangent slope $f'(x)$—
hypotheses—
1 Applies (polynomial). $f(-2)=-8+8=0$ and $f(2)=8-8=0$. $f^\prime(x)=3x^2-4=0\Rightarrow c=\pm\dfrac{2}{\sqrt3}\approx\pm1.155$ — both lie in $(-2,2)$, so there are two values.
2 Applies. $\sin 2x$ is continuous and differentiable everywhere, and $f(0)=\sin0=0=\sin2\pi=f(\pi)$. $f^\prime(x)=2\cos2x=0\Rightarrow 2c=\tfrac\pi2$ or $\tfrac{3\pi}{2}$, so $c=\dfrac\pi4$ or $c=\dfrac{3\pi}4$.
3 Applies ($x=0$ is outside $[1,4]$). Secant slope $\dfrac{\tfrac14-1}{4-1}=-\dfrac14$. $f^\prime(c)=-\dfrac1{c^2}=-\dfrac14\Rightarrow c^2=4\Rightarrow c=2$ (reject $-2\notin(1,4)$).
4 Applies. $f(0)=0$, $f(3)=24$, secant slope $\dfrac{24}{3}=8$. $3c^2-1=8\Rightarrow c^2=3\Rightarrow c=\sqrt3\approx1.732$.
5 Does not apply. $f(-1)=f(1)=1$ and $f$ is continuous, but $f^\prime(x)=\dfrac{2}{3x^{1/3}}$ is undefined at $x=0$ (a cusp), so $f$ is not differentiable on $(-1,1)$. Indeed $f^\prime(x)\neq0$ for every $x\ne0$: no $c$ exists.
6 Does not apply. $\tan0=\tan\pi=0$, but $\tan x$ has a vertical asymptote at $x=\tfrac\pi2$, so it is not continuous on $[0,\pi]$. And $f^\prime(x)=\sec^2x\ge1$ is never $0$ — the conclusion fails too.
V
Use the corollaries of the Mean Value Theorem.
  • 1 Find the intervals where $f(x)=x^3-12x+1$ is increasing and where it is decreasing.
  • 2 Find the intervals where $g(x)=xe^{-x}$ is increasing and where it is decreasing.
  • 3 Show that $x^5+3x-7=0$ has exactly one real root.
  • 4 Show that $2x-\cos x=0$ has exactly one real root.
  • 5 Find every function $f$ with $f^\prime(x)=6x^2$ for all $x$ and $f(1)=5$. Explain why there is no other.
1 $f^\prime(x)=3x^2-12=3(x-2)(x+2)$. The sign is $+$ for $x<-2$, $-$ for $-2<x<2$, $+$ for $x>2$. By Corollary 3, $f$ is increasing on $(-\infty,-2]$ and $[2,\infty)$ and decreasing on $[-2,2]$.
2 $g^\prime(x)=e^{-x}-xe^{-x}=(1-x)e^{-x}$. Since $e^{-x}>0$, the sign is that of $1-x$: $g$ is increasing on $(-\infty,1]$ and decreasing on $[1,\infty)$.
3 Existence: $f(x)=x^5+3x-7$ is continuous with $f(1)=-3<0$ and $f(2)=31>0$, so by the Intermediate Value Theorem there is a root in $(1,2)$. Uniqueness: if there were two roots $r_1<r_2$, then $f(r_1)=f(r_2)=0$ and Rolle's theorem would give $c$ with $f^\prime(c)=0$. But $f^\prime(x)=5x^4+3\ge3>0$ for all $x$ — contradiction. Exactly one real root.
4 $f(x)=2x-\cos x$: $f(0)=-1<0$, $f\!\left(\tfrac\pi2\right)=\pi>0$, so a root exists in $\left(0,\tfrac\pi2\right)$. $f^\prime(x)=2+\sin x\ge1>0$, so $f$ is strictly increasing and can cross zero only once (a second root would force $f^\prime(c)=0$ by Rolle).
5 $F(x)=2x^3$ has $F^\prime(x)=6x^2=f^\prime(x)$, so by Corollary 2, $f(x)=2x^3+C$. Then $f(1)=2+C=5\Rightarrow C=3$: $f(x)=2x^3+3$. Any other solution would differ from this one by a constant, and the condition $f(1)=5$ forces that constant to be $0$.
VI
Applications and estimates.
  • 1 Speeding ticket. On an expressway a car passes one speed camera at 9:00 a.m. and a second camera $150$ km further on at 10:12 a.m. The speed limit is $120$ km/h. Explain why the driver can be fined for speeding, even though neither camera caught the car above the limit.
  • 2 Use the Mean Value Theorem to show that $|\sin a-\sin b|\le|a-b|$ for all real $a,b$.
  • 3 Suppose $f$ is differentiable, $f(1)=10$ and $f^\prime(x)\ge2$ for $1\le x\le4$. How small can $f(4)$ possibly be?
  • 4 Conceptual. A student claims: "$f(x)=\dfrac1x$ has $f^\prime(x)=-\dfrac1{x^2}<0$ everywhere on its domain, so $f(-1)>f(1)$." Is the conclusion true? What went wrong?
1 The trip took $1\text{ h }12\text{ min}=1.2$ h, so the average speed was $v_{\text{avg}}=\dfrac{150}{1.2}=125$ km/h. The position $s(t)$ is continuous and differentiable, so by the MVT there is an instant $c$ between the cameras with $s^\prime(c)=125$ km/h $>120$ km/h. At that instant the car was speeding.
2 If $a=b$ there is nothing to prove. Otherwise apply the MVT to $\sin x$ between $a$ and $b$: $\sin a-\sin b=\cos c\,(a-b)$ for some $c$ between them. Since $|\cos c|\le1$, $|\sin a-\sin b|=|\cos c|\,|a-b|\le|a-b|$.
3 By the MVT, $f(4)-f(1)=f^\prime(c)(4-1)$ for some $c\in(1,4)$. Since $f^\prime(c)\ge2$: $f(4)\ge10+2\cdot3=16$. The bound is attained by $f(x)=2x+8$, so the smallest possible value is $f(4)=16$.
4 False: $f(-1)=-1<1=f(1)$. Corollary 3 needs $f^\prime<0$ on an interval, and $[-1,1]$ contains $x=0$, where $f$ is undefined. $f$ is decreasing on $(-\infty,0)$ and on $(0,\infty)$ separately, but not across the gap.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. What does the Mean Value Theorem prove about your speed on the Nizwa–Muscat road?

Why it matters. Point-to-point (average speed) cameras use this idea: if your average speed was too high, the MVT guarantees you were over that speed at some moment.

What to do.

  1. Use sample data: a car passes a point near Nizwa at 8:00 and a point near Muscat, 160 km away, at 9:20.
  2. Compute the average speed.
  3. Explain, using the MVT, why the speedometer showed exactly this speed at least once.
  4. State what the MVT does not tell you: when this happened, or the maximum speed.

Tools: Calculator

Hand in: A half-page note: the average speed, the MVT argument in two or three sentences, and one thing the theorem does not guarantee.

Going further. Anywhere: record start and end times of any trip and apply the same argument.

Instructor note: The average speed is $120$ km/h; students should state the conditions (position is continuous and differentiable in time).

🧪 Self-check quiz

0/5 correct

Q1. Rolle's theorem needs $f$ continuous on $[a,b]$, differentiable on $(a,b)$, and:

Q2. For $\sqrt x$ on $[0,9]$ the MVT gives $c=$

Q3. Why does the MVT fail for $|x|$ on $[-1,2]$?

Q4. If $f^\prime(x)=g^\prime(x)$ on an interval, then:

Q5. For $s(t)=-16t^2+100$ on $\left[0,\tfrac52\right]$, the MVT time is:

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