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Mean Value Theorem

$$f \text{ continuous on } [a,b],\; f \text{ differentiable on } (a,b)$$
$$\exists c\in(a,b) \text{ such that } f'(c) = \frac{f(b)-f(a)}{b-a}$$

Rolle's Theorem: if f(a) = f(b), then there exists a number c with $f'(c)=0$.

Applying the MVT — Quadratic (CE Board)

Verify the Mean Value Theorem for $f(x) = x^2 - 3x + 2$ on $[1,\,3]$ and find all values of $c$.

$f$ is a polynomial — continuous on $[1,3]$ and differentiable on $(1,3)$ ✓

$f(1) = 1 - 3 + 2 = 0,\quad f(3) = 9 - 9 + 2 = 2$
$\dfrac{f(3)-f(1)}{3-1} = \dfrac{2-0}{2} = 1$

Find $c$ such that $f'(c) = 1$:

$f'(x) = 2x - 3 = 1 \implies x = 2$

$c = 2 \in (1,3)$ ✓

Rolle's Theorem — Cubic (CE Board)

Verify Rolle's Theorem for $f(x) = x^3 - 3x$ on $[-\sqrt{3},\,\sqrt{3}]$ and find all values of $c$.

$f(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0,\quad f(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0$

Since $f(-\sqrt{3}) = f(\sqrt{3})$, Rolle's Theorem applies.

$f'(x) = 3x^2 - 3 = 0 \implies x = \pm 1$

Both $c = -1$ and $c = 1$ lie in $(-\sqrt{3},\,\sqrt{3})$ ✓

MVT for Square Root Function — CE Board

Find the value of $c$ guaranteed by the Mean Value Theorem for $f(x) = \sqrt{x}$ on $[1,\,4]$.

$\dfrac{f(4) - f(1)}{4 - 1} = \dfrac{2 - 1}{3} = \dfrac{1}{3}$

Find $c$ such that $f'(c) = \dfrac{1}{3}$:

$f'(x) = \dfrac{1}{2\sqrt{x}} = \dfrac{1}{3} \implies 2\sqrt{c} = 3 \implies \sqrt{c} = \dfrac{3}{2}$
$\boxed{c = \dfrac{9}{4} = 2.25 \in (1,\,4)\ ✓}$

MVT for Sine Function — CE Board

Find the value of $c$ in the Mean Value Theorem for $f(x) = \sin x$ on $[0,\,\pi]$.

$\dfrac{f(\pi) - f(0)}{\pi - 0} = \dfrac{0 - 0}{\pi} = 0$

Find $c$ such that $f'(c) = 0$:

$f'(x) = \cos x = 0 \implies x = \dfrac{\pi}{2}$

$c = \dfrac{\pi}{2} \approx 1.57 \in (0,\pi)$ ✓

MVT — Physical Interpretation (CE Board)

A car travels 150 km in exactly 1.5 hours. Using the Mean Value Theorem, show that the car's instantaneous speed must have equaled exactly 100 km/h at some point during the trip.

Let $s(t)$ be the car's position (in km) at time $t$ (in hours), with $s(0) = 0$ and $s(1.5) = 150$.

$s(t)$ is continuous on $[0, 1.5]$ and differentiable on $(0, 1.5)$ (assuming smooth motion). By the MVT, there exists $c \in (0, 1.5)$ such that:

$s'(c) = \dfrac{s(1.5) - s(0)}{1.5 - 0} = \dfrac{150}{1.5} = \boxed{100\ \text{km/h}}$

Since $s'(c)$ is the instantaneous speed, the car was traveling exactly 100 km/h at time $c$.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t827

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

A box is to be constructed from a piece of zinc 80 cm square by cutting equal squares from each corner and turning up the zinc to form the sides. What is the volume of the largest box that can be so constructed in liters?

  1. 42.15
  2. 37.93
  3. 35.24
  4. 28.74
Cutting squares of side $x$ gives $V = x(80 - 2x)^2$.
$V' = (80 - 2x)(80 - 6x) = 0 \Rightarrow x = \frac{40}{3} = 13.33$ cm
$V = 13.33(80 - 26.67)^2 = 37{,}926$ cm3
$\boxed{37.93 \text{ liters}}$

Question Bank: t866

MSTE - Differential Calculus / Differential Calculus / Gemini mapped Chapter 4 to 6

The length of arc of the function $f(x) = x^{2/3}$ from $x = 0$ to $x = 9$ is:

  1. 12.47
  2. 8.54
  3. 10.13
  4. 15.36
Arc length $L = \int_0^9 \sqrt{1 + (f')^2}\,dx$ with $f' = \frac{2}{3}x^{-1/3}$.
$L = \int_0^9 \frac{\sqrt{9x^{2/3} + 4}}{3x^{1/3}}\,dx$
Substitute $u = 9x^{2/3} + 4$ ($du = 6x^{-1/3}dx$):
$L = \frac{1}{18}\int_4^{42.94}\sqrt{u}\,du = \frac{1}{27}\left[42.94^{3/2} - 4^{3/2}\right]$
$\boxed{L = 10.13}$

Question Bank: w191

MSTE - Differential Calculus / Differential Calculus / MSTE May 2022

Find the value of $x_0$ prescribed by the extended law of the mean, if $f(x)=3x+2$ and $g(x)=x^2+1$ on $[1,4]$.

  1. 1.5
  2. 2.5
  3. 2
  4. 3
The extended mean-value theorem gives $$\frac{f'(x_0)}{g'(x_0)}=\frac{f(4)-f(1)}{g(4)-g(1)}.$$ Here $f'(x_0)=3$, $g'(x_0)=2x_0$, $f(1)=5$, $f(4)=14$, $g(1)=2$, and $g(4)=17$. Hence $$\frac{3}{2x_0}=\frac{14-5}{17-2}=\frac35,$$ so $$x_0=\frac52=2.5.$$

Question Bank: w309

MSTE - Differential Calculus / Differential Calculus / MSTE April 2025

THE LAW OF THE MEAN. If $f(x)$ is continuous on the interval $a \le x \le b$ and if $f'(x)$ exists everywhere on the interval except possibly at the endpoints, then there is at least one value of x, say $x = x_o$, between a and b such that $\dfrac{f(b) - f(a)}{b - a} = f'(x_o)$. Find the value of $x_o$ prescribed by the law of the mean when $f(x) = 3x^{2} + 4x - 3$ and $a = 1$, $b = 3$.

  1. 1
  2. 3
  3. 2
  4. 4
$f(x) = 3x^{2} + 4x - 3$
$f'(x) = 6x + 4$

Endpoint values:
$f(1) = 3(1)^{2} + 4(1) - 3 = 4$
$f(3) = 3(3)^{2} + 4(3) - 3 = 36$

Apply the law of the mean:
$\dfrac{f(3) - f(1)}{3 - 1} = f'(x_o)$
$\dfrac{36 - 4}{3 - 1} = 6x + 4$
$16 = 6x + 4$
$\boxed{x = x_o = 2}$