Question

I need help finding the inverse of this problem

Answer 1

$y=\sqrt{x-1}+6$

(Notice that we always have $y\ge6[tex].)Swap [tex]x$ and $y$, then solve for $y$:

$x=\sqrt{y-1}+6$

$x-6=\sqrt{y-1}$

$(x-6)^2=(\sqrt{y-1})^2$

$(x-6)^2=y-1$

$y=(x-6)^2+1$ (this is the inverse)

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This inverse is valid only for $x\ge6$. Why?

Suppose we take $x=0$. Then

$y=(0-6)^2+1=37$

This would suggest that in the original equation, we should get $x=37$ when $y=0$. But when we check this, we end up with

$0=\sqrt{37-1}+6=\sqrt{36}+6=6+6=12$

which is clearly not true.