If [tex]h^{-1}(x)[/tex] is the inverse of [tex]h(x)[/tex], then
[tex]h\left(h^{-1}(x)\right) = x[/tex]
Given that [tex]h(x)=-(x-2)^2+7[/tex], we have
[tex]h\left(h^{-1}(x)\right) = -\left(h^{-1}(x)-2\right)^2 + 7 = x[/tex]
Solve for the inverse :
[tex]-\left(h^{-1}(x)-2\right)^2 + 7 = x \\\\ -\left(h^{-1}(x)-2\right)^2 = x-7 \\\\ \left(h^{-1}(x)-2\right)^2 = 7-x \\\\ \sqrt{\left(h^{-1}(x)-2\right)^2} = \sqrt{7-x} \\\\ \left|h^{-1}(x)-2\right| = \sqrt{7-x}[/tex]
At this point, we have two possible solutions:
• If [tex]h^{-1}(x)\ge2[/tex], then [tex]\left|h^{-1}(x)-2\right| = h^{-1}(x)-2[/tex]. Continuing with the equation, we have
[tex]h^{-1}(x) - 2 = \sqrt{7-x} \\\\ h^{-1}(x) = 2 + \sqrt{7-x}[/tex]
• Otherwise, if [tex]h^{-1}(x)<2[/tex], then [tex]\left|h^{-1}(x)-2\right|=-\left(h^{-1}(x)\right)=2-h^{-1}(x)[/tex], in which case
[tex]2-h^{-1}(x) = \sqrt{7-x} \\\\ h^{-1}(x) = 2 - \sqrt{7-x}[/tex]
Both solutions are simultaneously incompatible. For instance, when x = 0 we have [tex]h^{-1}(0)=2+\sqrt7[/tex] using the first solution, and [tex]h^{-1}(0)=2-\sqrt7[/tex] using the second one. Choosing one over the other depends on how you restrict the domain.
(A) If we want to stick with the first solution, we would required that x ≥ 2. In other words, we would have [tex]h(x)[/tex] defined only when x ≥ 2; then [tex]h\left(h^{-1}(x)\right)[/tex] is defined only when [tex]h^{-1}(x)\ge2[/tex], so that ...
(B) ... the inverse is
[tex]\boxed{h^{-1}(x)=2+\sqrt{7-x}}[/tex]
(C) The domain of the inverse is the same as the range of the original function, and vice versa.
• Domain: we have for all x that
[tex]-(x-2)^2 \le 0[/tex]
so
[tex]-(x-2)^2 + 7 \le 7[/tex]
which means the range of [tex]h(x)[/tex], and hence the domain of [tex]y=h^{-1}(x)[/tex], is
[tex]\left\{y \mid y \le 7\}[/tex]
• Range: we get this immediately from the domain restriction we chose earlier,
[tex]\left\{x \mid x \ge 2\}[/tex]
See the attached plot - [tex]h(x)[/tex] is shown as a dashed orange curve; [tex]h(x)[/tex] with the restricted domain is shown in blue; the inverse [tex]h^{-1}(x)[/tex] corresponding to the restricted [tex]h(x)[/tex] is shown in green; the other inverse is shown with a dashed red curve. You can see a geometrical properties of inverses: if you mirror the blue curve along the dotted line (y = x), you would get the green curve.
