anonymous
  • anonymous
How do I find the inverse of f(x) when the equation is: f(x)=(x+3)^2, x<-3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The answer is supposed to be f-1(x) = -√x - 3
anonymous
  • anonymous
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
  • anonymous
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.

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