Find the Critical Points: $f(x) =(x^2-1)^3$ This question probably has more to do with my Algebra skills than Calculus. Nonetheless, can someone explain why the factored "term" is not set to zero (0) [second picture]. Thanks in advance. 



$f(x)=(x^2-1)^3$ is a sixth-degree polynomial with $x=\pm 1$ being triple zeroes.
Moreover, $f(x)$ is an even function, hence $f'(x)$ (that is a fifth-degree polynomial) has double zeroes in $x=\pm 1$ and a simple zero in $x=0$. The critical points of $f(x)$ may occur only at the zeroes of the derivative or at the endpoints of the given interval $I=[-1,5]$, hence to compute $\max_{x\in I}f(x)$ and $\min_{x\in I}f(x)$ it is enough to compute $f(-1),f(0),f(1),f(5)$.
The same approach works for the second exercise, too, by considering $F(t)^2$ in place of $F(t)$, getting a third-degree polynomial.
