Question

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

Answer 1

Here
$f\left(x\right)=\frac{1}{8}\log x-bx+x^2$ is defined and continuous for all $x >
0$
Then $f^{\prime }\left(x\right)=\frac{1}{8x}-b+2x$
or $f^{\prime }\left(x\right)=\frac{16x^2-8bx+1}{8x}$
For extrema, let $f^{\prime }\left(x\right)=0$
$\Rightarrow 16x^2-8bx+1=0$
so, $x=\frac{8b\pm \sqrt{64(b^2-1})}{2\times 16}$ or
$x=\frac{b\pm \sqrt{b^2-1}}{4}$
Obviously the roots are real if $b^2-1\ge 0$
$\Rightarrow b>1$
Sign scheme of $f^{\prime }\left(x\right)$ as shown in Fig.
$f^{\prime }\left(x\right)$ changes sign from $+ve$ to $-ve$ at
$x=\frac{b-\sqrt{b^2-1}}{4}$
$\therefore f(x{)}_{max}$ at
$x=\frac{b-\sqrt{b^2-1}}{4}$
and $f^{\prime }\left(x\right)$ changes sign from $-ve$ to $+ve$ at
$x=\frac{b+\sqrt{b^2-1}}{4}$
$\therefore f(x{)}_{min}$ at
$x=\frac{b+\sqrt{b^2-1}}{4}$
also if $b=1$
$f^{\prime }\left(x\right)=\frac{16x^2-8x+1}{x}=\frac{(4x-1{)}^2}{x}$ no
changes in sign.
$\therefore$ neither maximum or minimum if $b=1$
Thus $f\left(x\right)$
$\{\begin{array}{c}f(x{)}_{max},whenx=\frac{b-\sqrt{b^2-1}}{4}andb>1\\ f(x{)}_{min},whenx=\frac{b+\sqrt{b^2-1}}{4}andb>1\\ f\left(x\right)neithermaximumnorminimum,whenb=1\end{array}$