Question

Calculus. Please answer question attached.

Answer 1

(a) Given $f(x) = x^3$, the derivative is

$f'(x)=3x^2$

which is exists for all $x$ in the domain of $f$, so $]f$ is differentiable everywhere and satisfies the mean value theorem. There is some number $c$ in the open interval (0, 2) such that

$f'(c) = \dfrac{f(2) - f(0)}{2-0} \iff 3c^2 = \dfrac{8-0}2 = 4$

Solve for $c$ :

$3c^2 = 4 \implies c^2 = \dfrac43 \implies \boxed{c = \dfrac2{\sqrt3}}$

We omit the negative square root since it doesn't belong to (0, 2). Graphically, the MVT tells us the tangent line to the curve $f(x)=x^3$ at $x=\frac2{\sqrt3}$ is parallel to the secant line through the endpoints of the given interval.

(b) $f(x)=1+x+x^2$ has derivative

$f'(x)=1+2x$

By the MVT,

$f'(c) = \dfrac{f(2)-f(0)}{2-0} \iff 1+2c = \dfrac{7-1}2 \implies \boxed{c = 1}$

(c) $f(x) = \cos(2\pi x)$ has derivative

$f'(x) = -2\pi \sin(2\pi x)$

By the MVT,

$f'(c) = \dfrac{f(2)-f(0)}{2-0} \iff -2\pi \sin(2\pi c) = \dfrac{0-0}2 \implies \sin(2\pi c) = 0 \\\\ \implies 2\pi c = n\pi \implies c = \dfrac n2$

where $n$ is any integer. There are 3 solutions in the interval (0, 2),

$\boxed{c = \dfrac12, c = 1, c = \dfrac32}$

(Pictured is the situation with $c=\frac12$)