Question

Range of the function $f\left(x\right)=\frac{x^2}{x^2+1}$ is.

• A. $(-1,0)$
• B. $(-1,1)$
• C. $[0,1)$
• D. $(1,1)$

Answer 1

Answer: (C)

$[0,1)$

Clearly $f\left(x\right)$ is defined for all real $x$

The function will always return positive values very clearly.

Also $f^{\prime }\left(x\right)=\frac{(x^2+1)2x-x^2\left(2x\right)}{(x^2+1{)}^2}=\frac{2x}{(x^2+1{)}^2}$

When x is negative the function is monotonically decreasing while it changes monotonicity at $x=0$ and increases thereon.

To check the range we need to check for minima at $x=0$ and as x tends to largely negative or positive values

$f\left(0\right)=0$

${lim}_{x\to \infty }f\left(x\right)=\frac{1}{1+\frac{1}{x^2}}=1$

${lim}_{x\to -\infty }f\left(x\right)=\frac{1}{1+\frac{1}{x^2}}=1$

Hence range is $[0,1)$ . 1 is not included in the interval as it is the limiting value of the function at very large values of x. It is never attained.