How do you find the inverse of $f (x) = {(x + 2)}^{2}$ $f(x) = (x + 2)^2$ and is it a function?

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How do you find the inverse of $f (x) = {(x + 2)}^{2}$ $f(x) = (x + 2)^2$ and is it a function?

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Answer 1 · 2016-04-01T08:57:34

$x = - 2 \pm \sqrt{y}$ $x = -2+-sqrty$

Explanation:

The inverse $f^{- 1}$ $f^-1$ of f is defined by $x = f^{- 1} f (x)$ $x = f^-1f(x)$ .
$f^{- 1}$ $f^-1$ might be single-valued or multi-valued.

Here, if y = f(x) = ${(x + 2)}^{2}$ $(x+2)^2$
$y \geq 0 .$ $y>=0.$ ,
$\sqrt{y} = \pm (x + 2)$ $sqrty = +-(x+2)$
x + 2 = $\pm \sqrt{y}$ $+-sqrty$
$x = - 2 \pm \sqrt{y}$ $x =-2+-sqrty$ .

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How do you find the inverse of $f (x) = {(x + 2)}^{2}$ $f(x) = (x + 2)^2$ and is it a function?

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How do you find the inverse of f(x) = (x + 2)^2f(x)=(x+2)2f(x) = (x + 2)^2 and is it a function?

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How do you find the inverse of $f (x) = {(x + 2)}^{2}$ $f(x) = (x + 2)^2$ and is it a function?