Question

Find g^-1(x) A B C D

Answer 1

Answer:

$a. \ g^{-1}=-1 \pm\sqrt{\frac{3}{x}+1}$

Step-by-step explanation:

A function has an inverse function if and only if passes the Horizontal Line Test for Inverse Functions. This test tells us that a function $f$ has an inverse function if and only if there is no any horizontal line that intersects the graph of $f$ at more than one point. So the function is called one-to-one. The graph of $g$ is shown below. As you can see, this function does not pass the Horizontal Line Test, therefore the inverse is not a function. However, let's find $g^-{1}(x)$:

$g(x)=\frac{3}{x^2+2x} \\ \\ Substitute \ g(x) \ by \ y \\ \\y=\frac{3}{x^2+2x} \\ \\ Interchange \ x \ and \ y: \\ \\ x=\frac{3}{y^2+2y} \\ \\ Solve \ for \ y: \\ \\y^2+2y=\frac{3}{x} \\ \\ Completing \ square \\ \\y^2+2y\mathbf{+ 1}\mathbf{-1}=\frac{3}{x} \\ \\(y+1)^2=\frac{3}{x}+1 \\ \\y+1=\pm\sqrt{\frac{3}{x}+1} \\ \\ y=-1 \pm\sqrt{\frac{3}{x}+1}$

Finally, substitute $y \ by \ g^{-1}$:

$boxed{g^{-1}=-1 \pm\sqrt{\frac{3}{x}+1}}$