Finding critical numbers

Differentiate: $f'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1)$. This is a polynomial, so it is defined everywhere, meaning the only critical numbers come from setting $f'(x) = 0$.

Solving $15x^2(x^2-1) = 0$ gives $x = 0, 1, -1$. All three values lie within $(-2, 2)$, so these are the critical numbers of $f$ on the given interval.

Test signs of $f'(x) = 15x^2(x^2-1)$ on each subinterval: for $x < -1$, $f'(x) > 0$; for $-1 < x < 0$, $f'(x) < 0$; for $0 < x < 1$, $f'(x) < 0$; for $x > 1$, $f'(x) > 0$.

At $x = -1$, $f'$ changes from positive to negative, so $f(-1) = 2$ is a relative maximum. At $x = 1$, $f'$ changes from negative to positive, so $f(1) = -2$ is a relative minimum. At $x = 0$, $f'$ is negative on both sides (no sign change), so $x = 0$ is neither a maximum nor a minimum.

Evaluate $f$ at all critical numbers and endpoints: $f(-2) = 3(-32) - 5(-8) = -96 + 40 = -56$, $f(-1) = 2$, $f(0) = 0$, $f(1) = -2$, and $f(2) = 3(32) - 5(8) = 96 - 40 = 56$.

Comparing all five values $-56, 2, 0, -2, 56$, the largest is $56$ at $x = 2$ and the smallest is $-56$ at $x = -2$. So the absolute extrema occur at the endpoints, not at the relative extrema from part (b); the relative max/min at $x=\pm1$ are only local, not global.