Question

The following function has a local minima at which value of $x$?
$f\left(x\right)=x\sqrt{5-x^2}$

• A. $-\frac{\sqrt{5}}{2}$
• B. $\sqrt{5}$
• C. $\sqrt{\frac{5}{2}}$
• D. $-\sqrt{\frac{5}{2}}$

Answer 1

The correct option is D $-\sqrt{\frac{5}{2}}$

$f\left(x\right)=x\sqrt{5-x^2}$
$f^{\prime }\left(x\right)=\frac{x(-2x)}{2\sqrt{5-x^2}}+\sqrt{5-x^2}$
$=\frac{-x^2+5-x^2}{\sqrt{5-x^2}}$
$f\left(x\right)=\frac{5-2x^2}{\sqrt{5-x^2}}$
For critical points $f^{\prime }\left(x\right)=0$
$\Rightarrow 5-2x^2=0$
$x=\pm \sqrt{\frac{5}{2}}$
$f^{\prime \prime }\left(x\right)=\frac{x(2x^2-15)}{(5-x^2{)}^{3/2}}$
$f^{\prime \prime }\left(\frac{\sqrt{5}}{2}\right)=\frac{\sqrt{\frac{5}{2}}(-10)}{{(5-\frac{5}{2})}^{3/2}}$
and $f^{\prime \prime }(-\sqrt{\frac{5}{2}})>0$
$\Rightarrow$ At $x=\sqrt{\frac{5}{2}}f\left(x\right)$ has local maxima.
At $x=-\sqrt{\frac{5}{2}}f\left(x\right)$ has local minima.