Question

Find the range of the function

Answer 1

Answer:

Option D is correct.

Range of the function h(x)

$(-\infty, 1) \cup [0, \infty)$

Step-by-step explanation:

The range of the function is the set of all dependent variables for which function is defined

Given the function:

$h(x) = \frac{x^2}{1-x^2}$

Let y = h(x)

then;

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

By cross multiply we have;

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

Using distributive property: $a\cdot (b+c) = a\cdot b + a\cdot c$

$y-yx^2 = x^2$

Add both sides $yx^2$ we get;

$y= x^2+yx^2$

or

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

Divide both sides (1+y) we get;

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

or

$x = \sqrt{\frac{y}{1+y}}$

or

$h^{-1}(y)= \sqrt{\frac{y}{1+y}}$

interchange x and y value we have;

$h^{-1}(x)= \sqrt{\frac{x}{1+x}}$

To find the excluded value of the inverse function of h(x)

Equate denominator of the inverse function to 0.

x +1 = 0

x = -1

so, the domain of the inverse function is the set of all real values except -1

therefore, the function h(x) is the set of all real values except -1 i.e

$(-\infty, 1) \cup [0, \infty)$