Question

The local maximum value of $f\left(x\right)=\frac{x}{1+4x+x^2}$ is

• A. $\frac{1}{6}$
• B. $-\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{1}{6}$

$f\left(x\right)=\frac{x}{1+4x+x^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x^2+4x+1)\cdot 1-x(2x+4)}{(1+4x+x^2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=-\frac{(x^2-1)}{(1+4x+x^2{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=-\frac{(x+1)(x-1)}{(1+4x+x^2{)}^2}$
From $f^{\prime }\left(x\right)=0\Rightarrow x=-1,+1$

Since, sign of $f^{\prime }\left(x\right)$ changes from positive to negative as $x$ crosses $1$ from left to right, therefore $x=1$ is a point of local maxima.

therefore $f\left(x\right)$ has local maximum value at $x=1,$
$\Rightarrow f\left(1\right)=\frac{1}{1+4+1}=\frac{1}{6}$