Answer
We can see a sketch of a possible graph below.
Work Step by Step
$f'(0) = f'(4) = 0$
The slope of the graph is zero at $x = 0$ and $x=4$
$f'(x) = 1$ if $x \lt -1$
The graph has a constant slope of 1 in the interval $(-\infty, -1)$
$f'(x) \gt 0$ if $0 \lt x \lt 2$
The graph is increasing on this interval.
$f'(x) \lt 0$ if $-1 \lt x \lt 0$ or $2 \lt x \lt 4$ or $x \gt 4$
The graph is decreasing on these intervals.
$\lim\limits_{x \to 2^-}f'(x) = \infty$
$\lim\limits_{x \to 2^+}f'(x) = -\infty$
The graph has a vertical asymptote at $x = 2$
$f''(x) \gt 0$ if $-1 \lt x \lt 2$ or $2 \lt x \lt 4$
The graph is concave up on these intervals.
$f''(x) \lt 0$ if $x \gt 4$
The graph is concave down on the interval $(4, \infty)$
