Question

How do I find the inverse?

Answer 1

g(x) as given has no inverse because there are instances of two x values giving the same value of g(x). For instance,

x = -1   ⇒   g(-1) = 4 (-1 + 3)² - 8 = 8

x = -5   ⇒   g(-5) = 4 (-5 + 3)² - 8 = 8

Only a one-to-one function can have an inverse. g(x) is not one-to-one.

However, if we restrict the domain of g(x), we can find an inverse over that domain. Let $g^{-1}(x)$ be the inverse of g(x). Then by definition of inverse function,

$g\left(g^{-1}(x)\right) = 4 \left(g^{-1}(x) + 3\right)^2 - 8 = x$

Solve for the inverse:

$4 \left(g^{-1}(x) + 3\right)^2 - 8 = x$

$4 \left(g^{-1}(x) + 3\right)^2 = x + 8$

$\left(g^{-1}(x) + 3\right)^2 = \dfrac{x + 8}4$

$\sqrt{\left(g^{-1}(x) + 3\right)^2} = \sqrt{\dfrac{x + 8}4}$

$\left| g^{-1}(x) + 3 \right| = \dfrac{\sqrt{x+8}}2$

Recall the definition of absolute value:

$|x| = \begin{cases}x & \text{if }x\ge0\\-x&\text{if }x$

This means there are two possible solutions for the inverse of g(x) :

• if $g^{-1}(x) + 3 \ge 0$, then

$g^{-1}(x) + 3 = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2$

• otherwise, if $g^{-1}(x)+3$, then

$-\left(g^{-1}(x) + 3\right) = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3-\dfrac{\sqrt{x+8}}2$

Which we choose as the inverse depends on how we restrict the domain of g(x). For example:

Remember that the inverse must satisfy

$g\left(g^{-1}(x)\right) = x$

In the first case above, $g^{-1}(x) + 3 \ge 0$, or $g^{-1}(x) \ge -3$. This suggests that we could restrict the domain of g(x) to be $x \ge -3$.

Then as long as $x \ge -3$, the inverse is

$g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2$