## Calculus (3rd Edition)

The critical points are $x=0, \sqrt {\frac{3}{5}}$ and $-\sqrt {\frac{3}{5}}$. The second derivative test fails at $x=0$ $f(\sqrt {\frac{3}{5}})$ is a local minimum. $f(-\sqrt {\frac{3}{5}})$ is a local maximum.
$f'(x)=5x^{4}-3x^{2}$ is defined everywhere. The critical points of $f(x)$ are the solutions of $5x^{4}-3x^{2}=0$ $5x^{4}-3x^{2}=x^{2}(5x^{2}-3)=0$ $\implies x^{2}=0$ or $5x^{2}-3=0$ $\implies x=0$ or $x=±\sqrt {\frac{3}{5}}$ $f''(x)=20x^{3}-6x$ $f''(0)= 0$ (the second derivative test failed at the critical point 0) $f''(\sqrt \frac{3}{5})=20(\sqrt \frac{3}{5})^{3}-6(\sqrt \frac{3}{5})$$=12\sqrt \frac{3}{5}-6\sqrt \frac{3}{5}=positive \implies f(\sqrt \frac{3}{5}) is a local minimum. f''(-\sqrt \frac{3}{5})=20(-\sqrt \frac{3}{5})^{3}-6(-\sqrt \frac{3}{5})$$=-12\sqrt \frac{3}{5}+6\sqrt \frac{3}{5}=negative$ $\implies f(-\sqrt \frac{3}{5})$ is a local maximum.