# Question #59139

Sep 20, 2017

${f}^{- 1} \left(x\right) = \pm \sqrt{x + 3} + 2$

#### Explanation:

A function $f \left(x\right)$ has an inverse if it is "one-to-one" , i.e. for every value of $x$ there is a unique value of $f \left(x\right)$.

Quadratic functions are not one-to-one, so they do not have inverses.

But if we restrict the domain, we can define an inverse function for it.

In this case, there is no specific domain provided.

We can still algebraically come up with a general "inverse function" for this quadratic.

We have: $f \left(x\right) = {x}^{2} - 4 x + 7$.

First, we need to write $y$ in place of $f \left(x\right)$:

$R i g h t a r r o w y = {x}^{2} - 4 x + 7$

Then, let's interchange the variables:

$R i g h t a r r o w x = {y}^{2} - 4 y + 7$

To find the "inverse", we need to solve for $y$.

So, let's complete the square:

$R i g h t a r r o w x = {y}^{2} - 4 y + 4 + 7 - 4$

$R i g h t a r r o w x = \left({y}^{2} - 4 y + 4\right) + 3$

$R i g h t a r r o w x = \left({y}^{2} - 2 y - 2 y + 4\right) + 3$

$R i g h t a r r o w x = y \left(y - 2\right) - 2 \left(y - 2\right) + 3$

$R i g h t a r r o w x = \left(y - 2\right) \left(y - 2\right) + 3$

$R i g h t a r r o w x = {\left(y - 2\right)}^{2} + 3$

$R i g h t a r r o w x + 3 = {\left(y - 2\right)}^{2}$

$R i g h t a r r o w \pm \sqrt{x + 3} = y - 2$

$R i g h t a r r o w \pm \sqrt{x + 3} + 2 = y$

Now, let's replace $y$ with ${f}^{- 1} \left(x\right)$:

$R i g h t a r r o w \pm \sqrt{x + 3} + 2 = {f}^{- 1} \left(x\right)$

$\therefore {f}^{- 1} \left(x\right) = \pm \sqrt{x + 3} + 2$

This can be the inverse of our quadratic $f \left(x\right)$, if we restrict its domain to, for example, $x \ge 0$.