"Consider the function f : R → R defined as $f(x) =\frac{x}{1+ x^2}$. Range of the function f is "
Answer & explanation
Correct answer: option 3
$f(x)=\frac{x}{1+x^2}$
considering denominator ≠ 0 for any x
so $f'(x)=\frac{(1+x^2)-2x^2}{(1+x^2)^2}=0⇒x^2=1⇒x=±1$
$=\frac{1+x^2}{(1+x^2)^2}=0$
$±1$ → point of centreing
$f(-1) = -\frac{1}{2}$ minimum value
$f(1) = \frac{1}{2}$ maximum value
Range = $[-\frac{1}{2},\frac{1}{2}]$