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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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