# How do you find the inverse of f(x) =(x + 2)^2 - 4?

Jan 25, 2016

No inverse function exists without domain restrictions.

#### Explanation:

Set $f \left(x\right) = y$:

$y = {\left(x + 2\right)}^{2} - 4$

Interchange $y$ and $x$ in your equation:

$x = {\left(y + 2\right)}^{2} - 4$

Now, you need to solve this equation for $y$.
First of all, add $4$ on both sides:

$x + 4 = {\left(y + 2\right)}^{2}$

The next step would be to draw the root. However, this will leave you with two solutions, since e.g. for $25 = {x}^{2}$, both $5 = x$ and $- 5 = x$ are solutions.

$\sqrt{x + 4} = \left\mid y + 2 \right\mid$

$\iff \pm \sqrt{x + 4} = y + 2$

Subtract $2$ on both sides:

$- 2 \pm \sqrt{x + 4} = y$

Beware that a function must have a unique value for $y$ for each unique value of $x$.

However, this is not the case here since for e.g. $x = 12$, you have both $y = - 2 + \sqrt{16} = 2$ and $y = - 2 - \sqrt{16} = - 6$.

This means that there no inverse function exists.

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Remark:

An inverse function would exist if you restricted the domain of the original function.

As you can easily see that the vertex of the function is at $x = - 2$, it would suffice to either restrain the domain to e.g. $x \le - 2$ or to $x \ge - 2$.

For example, if your original function was

$f \left(x\right) = {\left(x + 2\right)}^{2} - 4 \text{ where } x \ge - 2$

then you could continue with the calculation from above, abandoning the negative term:

$y = - 2 + \sqrt{x + 4}$

Replace $y$ with ${f}^{- 1} \left(x\right)$:

${f}^{- 1} \left(x\right) = - 2 + \sqrt{x + 4}$