University Calculus: Early Transcendentals (3rd Edition)

No maximum. Local minimum: $f(x)=0$ and $x=1$ or $x=-1$
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 $x=-1$