Question

# The range of the function $$f(x)=\dfrac{x^{2}}{1+x^{2}}$$ is

Solution

## $$\begin{array}{l} Here \\ { x^{ 2 } }\ge 0 \\ { x^{ 2 } }+1>0 \end{array}$$Since, the denominator is always greater than the numerator for all values other than 0 and at $$\begin{array}{l} 0,f\left( x \right) =0 \\ \therefore Range=\left[ { 0,1 } \right) \end{array}$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More