Answer
(a) We can see a sketch of one possible graph for $f$ below.
(b) There is one solution for the equation $f(x) = 0$ since the graph crosses the x-axis at exactly one point.
(c) It is not possible that $f'(2) = \frac{1}{3}$ since the graph is concave down and $f'(3) = \frac{1}{2}$
Work Step by Step
(a) $f(3) = 2$
$f'(3) = \frac{1}{2}$
The slope of the graph at $x = 3$ is $\frac{1}{2}$
$f'(x) \gt 0$ for all $x$
The graph is increasing on the interval $(-\infty, \infty)$
$f''(x) \lt 0$ for all $x$
The graph is concave down on the interval $(-\infty, \infty)$
We can see a sketch of one possible graph for $f$ below.
(b) There is one solution for the equation $f(x) = 0$ since the graph crosses the x-axis at exactly one point.
(c) It is not possible that $f'(2) = \frac{1}{3}$
The graph is concave down for all values of $x$.
Therefore $f'(2) \gt f'(3) = \frac{1}{2}$
