Question

Find the domain and range of the function $f\left(x\right)=\frac{x^2}{1+x^2}$

• A. $R^{+}$ and $R$
• B. $R$ and $[0,1)$
• C. $[2,5]$ and $[0,1)$
• D. $(2,5)$ and $(3,5)$

Answer 1

The correct option is B $R$ and $[0,1)$

$f\left(x\right)=\frac{x^2}{1+x^2}$ domain is R

let $y=\frac{x^2}{1+x^2}\Rightarrow y+x^{2}y=x^2$

$x^2(y-1)+y=0$

$x^2=\frac{y}{1-y}\Rightarrow x=\pm \sqrt{\frac{y}{1-y}},$ $y\ne 1$ & $y\ngtr 1$

so range is $[0,1)$