Derivatives: rules and functions

Mean value theorem

The mean value theorem says that somewhere inside [a, b][a,\ b], at some x=cx=c, the tangent line has the same slope as the average rate of change over the interval.

f(b)−f(a)b−a=f′(c)(a<c<b)\begin{gathered} \frac{f(b)-f(a)}{b-a} \\[4pt] =f'(c)\quad(a<c<b) \end{gathered}

If f(x)f(x) is continuous on the closed interval [a, b][a,\ b] and differentiable on the open interval (a, b)(a,\ b), there is a cc such that

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

The left side is the slope of the line through the two endpoints (the average rate of change); the right side is the slope of the tangent at x=cx=c. For f(x)=x2f(x)=x^2 on [1, 3][1,\ 3], the average rate of change is 9−13−1=4\frac{9-1}{3-1}=4, and f′(c)=2c=4f'(c)=2c=4 gives c=2c=2.

Examples

  1. 01What is the average rate of change on this interval?f(x)=x2,[1, 3]\displaystyle f(x)=x^{2}, \quad [1,\ 3]AnswerHide
    Answer4\displaystyle 4

    Steps

    1. Divide the rise between the ends by the width of the interval
    2. 9−13−1\displaystyle \frac{9-1}{3-1}=4\displaystyle = 4
  2. 02What goes in □?f′(c)=f(b)−f(a)□\displaystyle f'(c) = \frac{f(b)-f(a)}{\square}AnswerHide
    Answerb−a\displaystyle b-a

    Steps

    1. The average rate is the rise divided by the width
    2. f(b)−f(a)b−a\displaystyle \frac{f(b)-f(a)}{b-a}
    3. Using a−ba-b makes the width negative and flips the sign
  3. 03What is this cc?f(x)=x2,f′(c)=8\displaystyle f(x)=x^{2}, \quad f'(c)=8AnswerHide
    Answer4\displaystyle 4

    Steps

    1. Differentiating f(x)=x2f(x)=x^2 gives 2x2x; set that equal at x=cx=c
    2. 2c=8\displaystyle 2c = 8⇒c=4\displaystyle \Rightarrow c = 4
  4. 04By the mean value theorem, such a cc exists. Which equation does cc satisfy?f(b)−f(a)b−a=5,a<c<b\displaystyle \begin{gathered} \displaystyle \frac{f(b)-f(a)}{b-a} = 5, \\[6pt] \displaystyle a<c<b \end{gathered}AnswerHide
    Answerf′(c)=5\displaystyle f'(c)=5
  5. 05What is the average rate of change on this interval?f(x)=x2,[2, 5]\displaystyle f(x)=x^{2}, \quad [2,\ 5]AnswerHide
    Answer7\displaystyle 7

    Steps

    1. Divide the rise between the ends by the width of the interval
    2. 25−45−2\displaystyle \frac{25-4}{5-2}=7\displaystyle = 7
  6. 06What is the average rate of change on this interval?f(x)=x2,[3, 6]\displaystyle f(x)=x^{2}, \quad [3,\ 6]AnswerHide
    Answer9\displaystyle 9

    Steps

    1. Divide the rise between the ends by the width of the interval
    2. 36−96−3\displaystyle \frac{36-9}{6-3}=9\displaystyle = 9

FAQ

Q1Where is cc when f(x)=x2f(x)=x^2?AnswerHide
A
The average rate of change is b2−a2b−a=a+b\frac{b^2-a^2}{b-a}=a+b and f′(c)=2cf'(c)=2c, so c=a+b2c=\frac{a+b}{2}, the midpoint of the interval. This holds for every quadratic function.
Q2When can the theorem not be used?AnswerHide
A
When the function is not continuous, or not differentiable, somewhere in the interval. tan⁡x\tan x is undefined at x=π2x=\frac{\pi}{2}, so the theorem does not apply on [0, π][0,\ \pi].

Published 2026-10-14 · Updated 2026-10-14