Question

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

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

Answer 1

Answer: (C)

$\frac{-1}{2}$

$f\left(x\right)=\frac{x}{1+x^2}$
$f^{\prime }\left(x\right)=\frac{1-x^2}{(1+x^2{)}^2}$
For maxima or minima ,
$f^{\prime }\left(x\right)=0$
$\Rightarrow x=-1,1$
$f^{\prime \prime }\left(x\right)=-\frac{4x^3}{(1+x^2{)}^4}$
$f^{\prime \prime }\left(1\right)<0$ and $f^{\prime \prime }(-1)>0$
So, f(x) attains minimum value at $x=-1$
$f(-1)=-\frac{1}{2}$