Question

$the \: range \: of \: f(x) = \sqrt{ \frac{x {}^{2} }{ {x}^{2} + 1 } }$

Answer 1

F(x) =(1/x²+x)
Is the answer hope it helps...
Plzzz mark me as brainliest

Answer 2

$0\le x^2<\infty\\\Rightarrow1\le1+x^2<\infty\\\Rightarrow1\ge\dfrac{1}{1+x^2}>0\\\Rightarrow-1\le-\dfrac{1}{1+x^2}<0\\\Rightarrow0\le1-\dfrac{1}{1+x^2}<1\\\Rightarrow0\le\dfrac{x^2}{1+x^2}<1\\\Rightarrow0\le\sqrt{\dfrac{x^2}{1+x^2}}<1\\\Rightarrow 0\le f(x)<1\\\Rightarrow Range=[0,1)$