Show that the maximum of $f(x)= \sqrt{x^{2}+a} - |x| $ on $[-1,1]$ for some real $a>0$ is $\sqrt{a}$.

Question

Show that the maximum of $f(x)= \sqrt{x^{2}+a} - |x| $ on $[-1,1]$ for some real $a>0$ is $\sqrt{a}$.

Plotting the graph of both $\sqrt{x^{2}+a}$ and $|x|$, it is clear that the maximum difference is at $x=0$ and is equal to $\sqrt{a}$. However, as $f(x)= \sqrt{x^{2}+a} - |x| $ is not differentiable at $x=0$ I am unsure of how to prove this.

Any help would be greatly appreciated.

Answer 1

If we write $|x| = \sqrt{x^2}$, we obtain:

$$f(x) = \sqrt{x^2+a} - \sqrt{x^2} = \frac {(\sqrt{x^2+a} - \sqrt{x^2})(\sqrt{x^2+a} + \sqrt{x^2})}{\sqrt{x^2+a} + \sqrt{x^2}} = \frac {a}{\sqrt{x^2+a}+\sqrt{x^2}}$$

The denominator is obviously strictly increasing as $|x|$ increases.

Hence the maximum of $f$ occurs at $x=0$, with $f(0) = \sqrt a$.

Answer 2

The first derivative (where it is defined) is $$f'(x)=\frac x{\sqrt{x^2+a^2}}-\frac{x}{|x|}$$ Notice that there are no roots for this. But you have discontinuities. So you need to check these points separately. $f'>0$ so $f$ is increasing for $x<0$, and $f'<0$ for $x>0$. You need to calculate $$\lim_{x\to \pm 0}f(x)=\sqrt a$$ Notice that I've calculate the limits on both sides. It just happens that they are equal.