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=-\frac{x^2}2+C`

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

`7=-\frac{0^2}2+C\implies C=7`

so that on [-3, 0] we have

`f(x)=-\frac{x^2}2+7`

and

`f(-3)=\frac52=2.5`