Question

Determine the points of maxima and minima of the function $f\left(x\right)=\frac{1}{8}\ln x-bx+x^2,x>0$ where $b\ge 0$ is a constant.

• A. $Min.atx=\frac{1}{4}(b+\sqrt{b^2-1}),max.atx=\frac{1}{4}(b-\sqrt{b^2-1})$
• B. $Min.atx=\frac{1}{4}(b-\sqrt{b^2-1}),max.atx=\frac{1}{4}(b+\sqrt{b^2-1})$
• C. $Min.atx=\frac{1}{4}(b+\sqrt{b^2+1}),max.atx=\frac{1}{4}(b-\sqrt{b^2-1})$
• D. $Min.atx=\frac{1}{4}(b+\sqrt{b^2-1}),max.atx=\frac{1}{4}(b-\sqrt{b^2+1})$

Answer 1

Answer: (A)

$Min.atx=\frac{1}{4}(b+\sqrt{b^2-1}),max.atx=\frac{1}{4}(b-\sqrt{b^2-1})$

To determine the critical points,
$f^{\prime }\left(x\right)=0$
Or
$\frac{1}{8x}-b+2x=0$
Or
$16x^2-8bx+1=0$
Or
$x=\frac{8b\pm \sqrt{64b^2-64}}{32}$
$=\frac{b\pm \sqrt{b^2-1}}{4}$
Hence the crirical points are $x_1=\frac{b+\sqrt{b^2-1}}{4}$ and $x_2=\frac{b-\sqrt{b^2-1}}{4}$.
Now $f^{\prime \prime }\left(x\right)=1-\frac{1}{8x^2}$
Hence $f^{\prime \prime }\left(x_2\right)<0$ and $f^{\prime \prime }\left(x_1\right)>0$
Thus $f\left(x\right)$ has a maxima at $x_2$ and minima at $x_1$.