Locating critical numbers

Differentiate to get $f'(x) = 5x^4 - 5$. Setting $f'(x) = 0$ gives $5x^4 - 5 = 0$, so $x^4 = 1$, which means $x = 1$ or $x = -1$ (the equation $x^4=1$ has no other real solutions).

Since $f'(x) = 5x^4 - 5$ is a polynomial, it is defined everywhere, so there are no critical numbers from an undefined derivative. Thus the critical numbers are $x = -1$ and $x = 1$.

Evaluate $f'(x) = 5x^4-5$ at values near $x=-1$: at $x=-1.1$, $f'(-1.1) = 5(-1.1)^4 - 5 \approx 2.31 > 0$; at $x=-0.9$, $f'(-0.9) = 5(-0.9)^4 - 5 \approx -1.72 < 0$.

We check the sign immediately to the left of $x=-1$ (e.g., at $x=-1.01$) where $f'(-1.01) > 0$, and immediately to the right (e.g., at $x=-0.99$) where $f'(-0.99) < 0$. Since $f'(x)$ changes sign from positive to negative as we pass through $x=-1$, the First Derivative Test conclusively shows that $x=-1$ is a local maximum.

We have $f'(x) = 5x^4 - 5$, so $f''(x) = 20x^3$. Evaluating at $x=-1$ gives $f''(-1) = 20(-1)^3 = -20$.

Since $f'(-1) = 0$ and $f''(-1) = -20 < 0$, the Second Derivative Test confirms that $x=-1$ is a local maximum of $f$.