Answer
(a). $f'(x) = 3 cos 3x$, which is zero when $3x =
. . . − π/2, π/2, 3π/2, . . .$, so when $x = . . . −
π/6, π/6, π/2, . . .$. The only such values on the
given interval are $x = −π/6$ and $x = π/6$.
(b). We have $f(−π/4) = -\frac{1}{\sqrt{2}}≈ −7.07$, $f(−π/6) =
−1$, $f(π/6) = 1$, and $f(π/3) = 0$, so the absolute
maximum of $f$ is $1$ and the absolute minimum is
$−1$.
Work Step by Step
See the explanation from the graph.
