How do I find the inverse of f(x) when the equation is: f(x)=(x+3)^2, x

Question

How do I find the inverse of f(x) when the equation is: f(x)=(x+3)^2, x<-3

anonymous · Answer

The answer is supposed to be f-1(x) = -√x - 3

anonymous · Answer

You just replace the viable x with y, and then rearrange the equation.

So you have:$$y=(x+3)^{2}$$

Replace x with y:

$$x=(y+3)^{2}$$

rearrange the equation.

$$\pm \sqrt{x}=y+3$$$$y=\pm \sqrt{x}-3$$

Check to see if the equation is an inverse by substitution. If the solution equals x then it is an inverse.

Substituting$$y=-\sqrt{x}-3$$ into$$f(x)$$ So $$f(-\sqrt{x}-3)=(-\sqrt{x}-3+3)^{2}$$
y=x so $$y=-\sqrt{x}-3$$ is an inverse.

Substituting $$y=\sqrt{x}-3$$into $$f(\sqrt{x}-3)=(\sqrt{x}-3+3)^{2}$$

y=x So

$$f(\sqrt{x}-3) $$ is also an inverse

anonymous · Answer

Interestingly, when you take both inverses, you get another parabola that fails the verticle line test.

So it looks like you take one or the other inverses, but not both.