High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 1 Textbook > Applications of Differentiation

Rolle's Theorem

The Extreme Value TheoremPrevious Section
Fermat's TheoremNext Section
Popular Courses
Find My Course
0:00 / 0:00

Rolle's Theorem

Continuity and Differentiability tell us many things about a function. In particular, Rolle's Theorem tells us about the existence of horizontal tangent lines.

Rolle's Theorem

Suppose that a function f(x)f(x) is continuous on the interval [a,b][a,b] and is differentiable on the interval (a,b)(a,b). If f(a)=f(b)f(a)=f(b) then there exists a value cc in the interval (a,b)(a,b) such that f′(c)=0f'(c)=0.


Note: In other words, since ff is continuous and differentiable, then in between every two points aa and bb with the same image, the function must have a maximum or a minimum.
0:00 / 0:00

Example: Rolle's Theorem

Use Rolle's theorem if f(x)=2x2−4x+3f(x) = 2x^2-4x+3 on the interval [0,2][0,2]to show that f(x)f(x)has a horizontal tangent line for some value x=cx=c. Then, determine the value of cc which Rolle's theorem predicts.

f(x)f(x)is continuous and differentiable on x in [0,2]x\text{ in }[0,2]. Also, f(0)=f(2)=3f(0)=f(2) = 3, so all of the hypotheses of Rolle's theorem are met with a=0,b=2a=0, b=2.
So, by Rolle's theorem, there must exist some value c  in [0,2]c \ \text{ in }[0,2]such that f′(c)=0f'(c) = 0.

Now, let's find the value of cc.

f′(x)=4x−4f'(x) = 4x-4, and f′(c)=0⇒4c−4=0⇒c=1f'(c) = 0 \Rightarrow 4c-4 = 0 \Rightarrow \boxed{c=1}
checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.
The function f(x)=1x2f(x) = \frac{1}{x^2} has no horizontal tangent lines on the interval [−1,1][-1,1] , even though f(−1)=f(1)f(-1)=f(1). Why doesn't this contradict Rolle's Theorem?