How do you find the inverse of $g (x) = {(x - 5)}^{2}$ $g(x) = (x -5)^2$ ?

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Answer 1 · 2018-05-30T22:06:33

$g^{- 1} (x) = 5 + \sqrt{x}$ $g^-1(x) =5+sqrt(x )$

or

$g^{- 1} (x) = 5 - \sqrt{x}$ $g^-1(x) =5-sqrt(x )$

$g (x) = {(x - 5)}^{2}$ $g(x) = (x -5)^2$

$y = {(x - 5)}^{2}$ $y = (x -5)^2$

Switch the $x$ $x$ and $y$ $y$ :

$x = {(y - 5)}^{2}$ $x = (y -5)^2$

solve for $x$ $x$ :

$\pm \sqrt{x} = \sqrt{{(y - 5)}^{2}}$ $+-sqrt(x )= sqrt((y -5)^2)$

$\pm \sqrt{x} = y - 5$ $+-sqrt(x )= y -5$

$y = 5 \pm \sqrt{x}$ $y=5+-sqrt(x )$

Now you have the problem that this inverse is not a function unless you restrict its range.

$g^{- 1} (x) = 5 + \sqrt{x}$ $g^-1(x) =5+sqrt(x )$

or

$g^{- 1} (x) = 5 - \sqrt{x}$ $g^-1(x) =5-sqrt(x )$

How do you find the inverse of g(x) = (x -5)^2g(x)=(x−5)2g(x) = (x -5)^2?