Question

The graph of f ′ (x), the derivative of f of x, is continuous for all x and consists of five line segments as shown below. Given f (0) = 7, find the absolute minimum value of f (x) over the interval [–3, 0].
0
2.5
4.5
11.5

Answer 1

$f'(x)\ge0$ for all $x$ in [-3, 0], so $f(x)$ is non-decreasing over this interval, and in particular we know right away that its minimum value must occur at $x=-3$.

From the plot, it's clear that on [-3, 0] we have $f'(x)=-x$. So

$f(x)=\displaystyle\int(-x)\,\mathrm dx=-\dfrac{x^2}2+C$

for some constant $C$. Given that $f(0)=7$, we find that

$7=-\dfrac{0^2}2+C\implies C=7$

so that on [-3, 0] we have

$f(x)=-\dfrac{x^2}2+7$

and

$f(-3)=\dfrac52=2.5$