The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
[tex]\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx[/tex]
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
[tex]\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7[/tex]
[tex]\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}[/tex]
