Question

The range of the function y = $\frac{x}{1+x^2}$ is

Answer 1

To find :

Range of function $y=\frac{x}{1+x^2}$

Range is given by 'R'

Say f(x) = y

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

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

$y+y{x}^2-x=0$

$y{x}^2-x+y=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$[\because$ quadratic formula

and the $so{l}^n$ of x are real when

$b^2-4ac⩾0$

so like wire

${}_{(}-1)-4(y\left)\right(y)⩾0$

we solve the inequality for the values set of value of y.

$1-4y^2⩾0$

$1-4y^2⩾0$

$1^2-(2y{)}^2⩾0$

$(1-2y)(1+2y)⩾0$

$(2y+1)(2y-1)⩽0$ [Taking negative common]

$(2y+1)⩾0$ & $(2y-1)⩽0$

$y⩾-1/2y⩽1/2$

$-1/2⩽y⩽1/2$

so the range of y is $[\frac{1}{2},\frac{-1}{2}].$