I'm stuck on this question help!

Mathematics

1 answer:

Vadim26 [7]3 years ago

6 0

1500 is your answer hope it helps

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

Katyanochek1 [597]

<h2>Answer:</h2>

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

<h2>Step-by-step explanation:</h2>

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 $g^-{1}(x)$ :

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