Answer
(a). $f(x) = \frac{3}{8}x^2 − \frac{1}{2}$ , which is zero when $3x^2 − 4 = 0$,
which occurs for $x = ±\frac{2}{\sqrt{3}}$.
(b). There is a local maximum at $x = -\frac{2}{\sqrt{3}}$ and a local minimum at $x = = +\frac{2}{\sqrt{3}}$, but only the local minimum occurs on the given interval.
Work Step by Step
See the graph of the function for explanation.
