Mean Value Theorem

Type f and the interval. The dashed chord joins the two ends; drag c along the curve, or with the slider, until the tangent at c runs parallel to it. The lower panel draws f′ against the chord's slope m, so every c is a point where the two meet.

Mean value theorem	$f'(c)=\frac{f(b)-f(a)}{b-a},\qquad a<c<b$
Rolle's theorem	$f(a)=f(b)\ \Rightarrow\ f'(c)=0,\qquad a<c<b$

Worked example — Mean Value Theorem

f at the left end	1
f at the right end	9
Slope of the chord, m	4
Continuous on [a, b], differentiable on (a, b)	Both hold
Your c	1.6
Slope of the tangent at your c	3.2

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Mean Value Theorem

The mean value theorem: if f is continuous on [a, b] and differentiable on (a, b), then at some c strictly between a and b the tangent is parallel to the chord, f′(c) = (f(b) − f(a))/(b − a). Rolle's theorem is the case f(a) = f(b), where f′(c) = 0.

Where the formula comes from

Check the hypotheses. f must have no break anywhere on [a, b], ends included, and a derivative at every point strictly inside. Polynomials, eˣ, sin and cos pass on any interval; look for division by zero, a root of a negative, or a corner such as |x|.
$\begin{aligned}&\lim_{x\to x_0}f(x)=f(x_0)\quad(a\le x_0\le b)\\&f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\quad(a<x<b)\end{aligned}$
Find the chord's slope. m is the rise from f(a) to f(b) over the run from a to b, the dashed line on the picture.
$m=\frac{f(b)-f(a)}{b-a}$
Why a c exists. The gap g between the curve and the chord is 0 at both ends. A continuous function on a closed interval has a largest and a smallest value; if g is not 0 everywhere, one of them is inside, where the curve is farthest from the chord, and there g′ = 0. That is f′(c) = m. On the picture it is the point where the tangent runs parallel.
$\begin{aligned}&g(x)=f(x)-f(a)-m\,(x-a)\\&g(a)=g(b)=0\\&\Rightarrow\ g'(c)=f'(c)-m=0\end{aligned}$
Solve f′(c) = m and keep only the roots strictly between a and b. An answer outside the interval, or at an end, is not a c.
$f'(c)=m,\qquad a<c<b$

Problems solved in full

Find c, and reject the root outside 2 steps
Show that f(x) = x³ satisfies the hypotheses of the mean value theorem on [0, 3], and find every c it promises.

Open this problem in the tool
1. f is a polynomial, so it is continuous and differentiable everywhere. The chord's slope:
  
  $m=\frac{f(3)-f(0)}{3-0}=\frac{27-0}{3}=9$
2. Solve f′(c) = m:
  
  $3c^{2}=9\ \Rightarrow\ c=\pm\sqrt{3}$
Answer

c = √3 ≈ 1.7321. The root −√3 also solves 3c² = 9 but lies outside (0, 3).

$c=\sqrt{3}\approx1{.}7321$
Rolle's theorem: equal ends 2 steps
Verify Rolle's theorem for f(x) = sin x + cos x on [0, π/2].

Open this problem in the tool
Show solution
1. sin and cos are continuous and differentiable everywhere. The ends agree:
  
  $\begin{aligned}&f(0)=0+1=1\\&f\!\left(\tfrac{\pi}{2}\right)=1+0=1\end{aligned}$
2. So the chord is flat and some c has f′(c) = 0:
  
  $\begin{aligned}&\cos c-\sin c=0\\&\Rightarrow\ \tan c=1\ \Rightarrow\ c=\frac{\pi}{4}\end{aligned}$
Answer

c = π/4 ≈ 0.7854, the top of the arch.

$c=\frac{\pi}{4}\approx0{.}7854$
Two values of c 2 steps
Find every c promised by the mean value theorem for f(x) = x³ − 3x on [−2, 2].

Open this problem in the tool
Show solution
1. A polynomial, so both hypotheses hold. The chord's slope:
  
  $m=\frac{f(2)-f(-2)}{2-(-2)}=\frac{2-(-2)}{4}=1$
2. Solve f′(c) = m:
  
  $3c^{2}-3=1\ \Rightarrow\ c=\pm\frac{2}{\sqrt{3}}$
Answer

c = −2/√3 ≈ −1.1547 and c = 2/√3 ≈ 1.1547. Both lie in (−2, 2), so the theorem's "at least one" is two here.

$c=\pm\frac{2}{\sqrt{3}}\approx\pm1{.}1547$
An exponential: c by a logarithm 2 steps
Find the value of c given by the mean value theorem for f(x) = eˣ on [0, 1].

Open this problem in the tool
Show solution
1. eˣ is continuous and differentiable everywhere. The chord's slope:
  
  $m=\frac{e^{1}-e^{0}}{1-0}=e-1$
2. Solve f′(c) = eᶜ = m by taking logarithms:
  
  $e^{c}=e-1\ \Rightarrow\ c=\ln(e-1)$
Answer

c = ln(e − 1) ≈ 0.5413, which lies in (0, 1).

$c=\ln(e-1)\approx0{.}5413$
A corner inside: no c 2 steps
Show that there is no c in (−1, 2) with f′(c) = (f(2) − f(−1))/3 for f(x) = |x|. Why does this not contradict the mean value theorem?

Open this problem in the tool
Show solution
1. The chord's slope:
  
  $m=\frac{|2|-|-1|}{2-(-1)}=\frac{1}{3}$
2. Away from 0 the slope of |x| is −1 or 1, never 1/3:
  
  $\begin{aligned}&f'(x)=-1\quad(x<0)\\&f'(x)=1\quad(x>0)\end{aligned}$
Answer

No c exists. |x| has no derivative at 0, which lies inside (−1, 2), so the hypothesis of differentiability fails and the theorem promises nothing.

$f'(c)\neq\tfrac{1}{3}\ \ \forall c\in(-1,2)$
A break inside: Rolle's conclusion fails 2 steps
f(x) = 1/x² has f(−1) = f(1). Is there a c in (−1, 1) with f′(c) = 0? Explain.

Open this problem in the tool
Show solution
1. The ends agree, so the chord is flat:
  
  $f(-1)=\frac{1}{(-1)^{2}}=1=f(1),\qquad m=0$
2. But f′ is never 0:
  
  $f'(x)=-\frac{2}{x^{3}}\neq0\ \ (x\neq0)$
Answer

No. f is not defined at 0, so it is not continuous on [−1, 1], Rolle's theorem does not apply, and its conclusion fails.

$f'(c)\neq0\ \ \forall c\in(-1,1)$
No derivative at an end: the theorem still applies 2 steps
Does the mean value theorem apply to f(x) = √x on [0, 4]? If so, find c.

Open this problem in the tool
Show solution
1. √x is continuous on [0, 4] and differentiable on (0, 4); f′(0) does not exist, but an end is not required to have one. The chord's slope:
  
  $m=\frac{\sqrt{4}-\sqrt{0}}{4-0}=\frac{1}{2}$
2. Solve f′(c) = m:
  
  $\frac{1}{2\sqrt{c}}=\frac{1}{2}\ \Rightarrow\ \sqrt{c}=1\ \Rightarrow\ c=1$
Answer

Yes; c = 1.

$c=1$
Average speed between two cameras 2 steps
A car passes one camera and, 6 minutes later, a second camera 12 km further along the road. The limit is 100 km/h. Show that at some instant between them the car was doing exactly 120 km/h.
Show solution
1. Let s(t) be the distance travelled in km after t hours. The car moves without jumps and has a speed at every instant, so s is continuous and differentiable. The average speed over the 0.1 h is the chord's slope:
  
  $\frac{s(0{.}1)-s(0)}{0{.}1-0}=\frac{12}{0{.}1}=120\ \mathrm{km/h}$
2. By the mean value theorem some instant c between the cameras has s′(c) equal to it:
  
  $s'(c)=120\ \mathrm{km/h},\qquad 0<c<0{.}1$
Answer

At some instant the speed was exactly 120 km/h, so the car broke the limit even if no single camera caught it.

$s'(c)=120\ \mathrm{km/h}>100\ \mathrm{km/h}$
A bound from a bound on f′ 2 steps
f is differentiable, f(1) = 10 and f′(x) ≤ 3 for every x. How large can f(5) be?
Show solution
1. On [1, 5] the mean value theorem gives some c with:
  
  $\frac{f(5)-f(1)}{5-1}=f'(c),\qquad 1<c<5$
2. Since f′(c) ≤ 3:
  
  $f(5)=f(1)+4f'(c)\le10+4\cdot3=22$
Answer

f(5) ≤ 22, and f(x) = 3x + 7 reaches it.

$f(5)\le22$

Supplementary problems, answers only

Find c for f(x) = x² − 4x + 1 on [0, 5].

Show answer

c = 2.5

$c=\frac{5}{2}$
Verify Rolle's theorem for f(x) = x³ − 4x on [0, 2] and find c.

Show answer

c = 2/√3 ≈ 1.1547

$c=\frac{2}{\sqrt{3}}\approx1{.}1547$
Find c for f(x) = ln x on [1, e].

Show answer

c = e − 1 ≈ 1.7183

$c=e-1\approx1{.}7183$
Find c for f(x) = 1/x on [1, 3].

Show answer

c = √3 ≈ 1.7321

$c=\sqrt{3}\approx1{.}7321$
Does the mean value theorem apply to f(x) = |x − 2| on [0, 3]? Is there a c?

Show answer

No: a corner at x = 2; the chord's slope is −1/3 and f′ is only ±1, so there is no c

$f'(x)=\pm1\neq-\frac{1}{3}$
f(0) = 5 and f′(x) ≥ −2 for every x. What is the least possible value of f(6)?

Show answer

−7

$f(6)\ge-7$

The ends may misbehave; the inside may not

Press √x on [0, 4]. At 0 the curve leaves the axis vertically and f′(0) does not exist, yet c = 1 is there, a tangent of slope 1/2 parallel to the chord. Now press |x| on [−1, 2]. The chord's slope is 1/3, every tangent has slope −1 or 1, and no c exists. The difference is where the bad point sits. The theorem only ever promises a c strictly between a and b, so it needs a derivative only there; an end may have none. What an end may not do is break away from the curve, because then the chord joins the wrong point. Continuity is asked of the closed interval, differentiability of the open one, and that difference is the whole reason the statement is written with two kinds of bracket.

The theorem behind every + C

Few exam questions ask for c. The theorem earns its place by what it proves. If f′(x) = 0 at every point of an interval, every chord there has the slope of some tangent, which is 0, so f takes the same value at any two points: f is constant. Two antiderivatives of the same function have a difference whose derivative is 0, so on an interval they differ by a constant. That is the + C, and it is a theorem, not a convention. On an interval is part of it: ln|x| + 1 for x < 0 together with ln|x| − 3 for x > 0 is also an antiderivative of 1/x, because no interval crosses 0. The same argument with f′ > 0 says every chord rises, so f is increasing, which is the test the next page is built on.

For a parabola, c is always the midpoint

Press x² on [1, 3]: c = 2, halfway. That is no accident of the numbers. For f(x) = px² + qx + r the chord's slope works out to p(a + b) + q, which is exactly f′ at (a + b)/2, on every interval. Read as motion it says that under constant acceleration the average velocity over a stretch of time equals the velocity at its midpoint. William Heytesbury stated that rule in Oxford in the 1330s, and Galileo built on it three centuries later for falling bodies. A cubic breaks it at once: press x³ on [0, 3] and c = √3 ≈ 1.7321, not 1.5.

Learning path

What f′ and f″ say about a graph

References (2)

Rolle's theorem (Theorem 4.4) and the mean value theorem (Theorem 4.5) with their hypotheses, continuity on [a, b] and differentiability on (a, b); and the corollaries insight block 2 is about: a zero derivative on an interval means a constant (4.6), equal derivatives differ by a constant (4.7), and f′ > 0 means increasing (4.8): G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 4.4, “The Mean Value Theorem”.
Insight block 3: William Heytesbury formulated the middle degree theorem (the mean speed theorem) for uniformly accelerated motion in his Regulae solvendi sophismata; his extant writings date from the 1330s, and the theorem was later developed by Galileo: M. Hanke and E. Jung, “William Heytesbury”, The Stanford Encyclopedia of Philosophy, first published 19 January 2018, revised 19 June 2026.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.

Mean Value Theorem

Where the formula comes from

Supplementary problems, answers only

The ends may misbehave; the inside may not 🖖

The theorem behind every + C 🖖

For a parabola, c is always the midpoint 🖖

The ends may misbehave; the inside may not

The theorem behind every + C

For a parabola, c is always the midpoint