Finding critical points

Differentiate: $f'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1)$. This factors further as $f'(x) = 15x^2(x-1)(x+1)$.

Setting $f'(x) = 0$ gives $x = 0, 1, -1$. Since $f'(x)$ is a polynomial, it is defined for all $x$, so these are the only critical numbers.

Since $f'(x) = 15x^4 - 15x^2$, differentiate again to get $f''(x) = 60x^3 - 30x = 30x(2x^2 - 1)$.

Evaluate at each critical number: $f''(-1) = 30(-1)(2-1) = -30 < 0$, so $x=-1$ is a local maximum. $f''(1) = 30(1)(2-1) = 30 > 0$, so $x=1$ is a local minimum. $f''(0) = 0$, so the test is inconclusive at $x=0$; checking the sign of $f'(x) = 15x^2(x-1)(x+1)$ near $0$ shows it is negative on both sides (e.g., at $x=-0.5$ and $x=0.5$), so $x=0$ is neither a local max nor a local min.