Local/global maximum

Question

Suppose I have the following function $f(x)=x^2(1-x^2)$. Now this function has maxima at $\pm\frac1{\sqrt2}$ and they are equal. Is this maximum absolute or relative? It is absolute right?

Answer 1

Let $f(x)=x^2(1-x^2)$.

Note that $f$ is an even function, so has $y$-axis symmetry. Thus, it suffices to analyze the behavior of $f$ on $[0,\infty)$.

Computing $f'(x)$, we get $f'(x)=2x(1-2x^2)$.

• On the interval $(0,\frac{1}{\sqrt{2}})$, we have $f'(x) > 0$, so $f$ is strictly increasing.
• On the interval $(\frac{1}{\sqrt{2}},\infty)$, we have $f'(x) < 0$, so $f$ is strictly decreasing.

It follows that on the interval $[0,\infty)$, $f$ achieves a global maximum at $x={\large{\frac{1}{\sqrt{2}}}}$.

By symmetry, $f$ also achieves a global maximum at $x=-\large{\frac{1}{\sqrt{2}}}$.

Answer 2

Alternately, write $f(x)=\frac14-(\frac12-x^2)^2$. As squares are non-negative, it is easier to see why the maximum is when $x^2=\frac12$ and why it is global.