High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize AP Calculus (BC) Textbook > Applications of Differentiation: Analytical

The Mean Value Theorem

Exam-Like Practice QuestionsPrevious Section
Intervals of Increase and DecreaseNext Section
Popular Courses
Find My Course
0:00 / 0:00

The Mean Value Theorem (MVT)

We can extend the idea of Rolle's Theorem with the Mean Value Theorem.

Mean Value 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)\left(a,\ b\right). Then there exists a value cc in the interval (a, b)\left(a,\ b\right) such that

f′(c)=f(b)−f(a)b−a\displaystyle \boxed{f'(c)=\frac{f(b)-f(a)}{b-a}}




Note: In other words, since ff is continuous and differentiable, then in between every two points aa and bb, there is a point cc for which the slope of the tangent line at cc is equal to the slope of the secant line linking (a,f(a))(a,f(a)) and (b,f(b)).(b,f(b)).


0:00 / 0:00

Example: Mean Value Theorem

Suppose ff is a differentiable function such that f′(x)≤5f'(x)\leq 5 for all x in [1,5]x\text{ in }[1,5] and f(1)=3f(1)=3. What is an upper bound on f(5)f(5)?

f(b)−f(a)b−a=f′(c)\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)

a=1,  b=5a=1 ,\ \ b=5
f(5)−f(1)5−1=f′(x)≤5  ⟺  f(5)−34≤5  ⟺  f(5)≤23\begin{aligned} &\frac{f(5)-f(1)}{5-1}=f'(x)\leq5\iff \frac{f(5)-3}{4}\leq 5 \iff f(5)\leq 23\\ \end{aligned}
So 2323 is an upper bound forf(5)f(5).
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x in [−1,3]x\text{ in }[-1,3], what value of cc does the mean value theorem guarantee exists?