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 [tex]g^{-1}(x)[/tex] be the inverse of g(x). Then by definition of inverse function,
[tex]g\left(g^{-1}(x)\right) = 4 \left(g^{-1}(x) + 3\right)^2 - 8 = x[/tex]
Solve for the inverse:
[tex]4 \left(g^{-1}(x) + 3\right)^2 - 8 = x[/tex]
[tex]4 \left(g^{-1}(x) + 3\right)^2 = x + 8[/tex]
[tex]\left(g^{-1}(x) + 3\right)^2 = \dfrac{x + 8}4[/tex]
[tex]\sqrt{\left(g^{-1}(x) + 3\right)^2} = \sqrt{\dfrac{x + 8}4}[/tex]
[tex]\left| g^{-1}(x) + 3 \right| = \dfrac{\sqrt{x+8}}2[/tex]
Recall the definition of absolute value:
[tex]|x| = \begin{cases}x & \text{if }x\ge0\\-x&\text{if }x<0\end{cases}[/tex]
This means there are two possible solutions for the inverse of g(x) :
• if [tex]g^{-1}(x) + 3 \ge 0[/tex], then
[tex]g^{-1}(x) + 3 = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2[/tex]
• otherwise, if [tex]g^{-1}(x)+3<0[/tex], then
[tex]-\left(g^{-1}(x) + 3\right) = \dfrac{\sqrt{x+8}}2 \implies g^{-1}(x) = -3-\dfrac{\sqrt{x+8}}2[/tex]
Which we choose as the inverse depends on how we restrict the domain of g(x). For example:
Remember that the inverse must satisfy
[tex]g\left(g^{-1}(x)\right) = x[/tex]
In the first case above, [tex]g^{-1}(x) + 3 \ge 0[/tex], or [tex]g^{-1}(x) \ge -3[/tex]. This suggests that we could restrict the domain of g(x) to be [tex]x \ge -3[/tex].
Then as long as [tex]x \ge -3[/tex], the inverse is
[tex]g^{-1}(x) = -3+\dfrac{\sqrt{x+8}}2[/tex]
