Answer:
Option D is correct.
Range of the function h(x)
[tex](-\infty, 1) \cup [0, \infty)[/tex]
Step-by-step explanation:
The range of the function is the set of all dependent variables for which function is defined
Given the function:
[tex]h(x) = \frac{x^2}{1-x^2}[/tex]
Let y = h(x)
then;
[tex]y= \frac{x^2}{1-x^2}[/tex]
By cross multiply we have;
[tex]y(1-x^2) = x^2[/tex]
Using distributive property: [tex]a\cdot (b+c) = a\cdot b + a\cdot c[/tex]
[tex]y-yx^2 = x^2[/tex]
Add both sides [tex]yx^2[/tex] we get;
[tex]y= x^2+yx^2[/tex]
or
[tex]y= x^2(1+y)[/tex]
Divide both sides (1+y) we get;
[tex]\frac{y}{y+1} = x^2[/tex]
or
[tex]x = \sqrt{\frac{y}{1+y}}[/tex]
or
[tex]h^{-1}(y)= \sqrt{\frac{y}{1+y}}[/tex]
interchange x and y value we have;
[tex]h^{-1}(x)= \sqrt{\frac{x}{1+x}}[/tex]
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
[tex](-\infty, 1) \cup [0, \infty)[/tex]
