Answer
(a) $f$ is differentiable for all values of $x$ except $x=-3$ and $x=3$
We can find a formula for $f'$:
$f'(x) = 2x~~~$ if $x \leq -3$
$f'(x) = -2x~~~$ if $-3 \lt x \lt 3$
$f'(x) = 2x~~~$ if $x \geq 3$
(b) We can see a sketch of $f$ and $f'$ below.
Work Step by Step
(a) $f(x) = \vert x^2-9\vert$
We can express $f(x)$ as follows:
$f(x) = x^2-9~~~$ if $x \leq -3$
$f(x) = 9-x^2~~~$ if $-3 \lt x \lt 3$
$f(x) = x^2-9~~~$ if $x \geq 3$
Clearly $f(x)$ is continuous for all values of $x$
As $x \to -3^-$, then $~~f'(x) = 2x~~$ and the derivative approaches the value of -6
As $x \to -3^+$, then $~~f'(x) = -2x~~$ and the derivative approaches the value of 6
Since the derivatives are not equal from the left and the right, $f$ is not differentiable at $-3$
As $x \to 3^-$, then $~~f'(x) = -2x~~$ and the derivative approaches the value of -6
As $x \to 3^+$, then $~~f'(x) = 2x~~$ and the derivative approaches the value of 6
Since the derivatives are not equal from the left and the right, $f$ is not differentiable at $3$
$f$ is differentiable for all values of $x$ except $x=-3$ and $x=3$
We can find a formula for $f'$:
$f'(x) = 2x~~~$ if $x \lt -3$
$f'(x) = -2x~~~$ if $-3 \lt x \lt 3$
$f'(x) = 2x~~~$ if $x \gt 3$
(b) We can see a sketch of $f$ and $f'$ below.
