Answer
No maximum.
Local minimum: $f(x)=0$ and $x=1$ or $x=-1$
Work Step by Step
Given $y=({(x^2-1})^{(\frac{1}{2})}$
The critical point is where the first derivative is zero:
so $\frac{dy}{dx}=\frac{d{({x^2}-1)^{(\frac{1}{2})})}}{dx}$
${\frac{dy}{dx}}=\frac{1}{2({({x^2}-1)}^{(\frac{1}{2})}}{({{2x}})}$
so the critical point is: ${\frac{dy}{dx}}=0$
$f'(x)=\frac{1}{2({({x^2}-1)}^{(\frac{1}{2})}}{({{2x}})}=0$
$x=0$ and ${({x^2}-1)}^{(\frac{1}{2})}=0$
$x=0,x=1,x=-1$, but note that zero is not in the domain of the function
the critical value of y can be obtained by putting in the value of x:
$f(-1)=f(1)=0$
Thus we have:
No maximum.
Local minimum: $f(x)=0$ and $x=1$ or $x=-1$
