Answer:
[tex]f^{-1}(x) = (x + 8)^2[/tex]
[tex]x \ge -8[/tex]
Step-by-step explanation:
Given
[tex]f(x) = \sqrt x - 8[/tex]
Solving (a): [tex]f^{-1}(x)[/tex]
We have:
[tex]f(x) = \sqrt x - 8[/tex]
Express f(x) as y
[tex]y = \sqrt x - 8[/tex]
Swap x and y
[tex]x = \sqrt y - 8[/tex]
Add 8 to [tex]both\ sides[/tex]
[tex]x + 8 = \sqrt y - 8 + 8[/tex]
[tex]x + 8 = \sqrt y[/tex]
Square both sides
[tex](x + 8)^2 = y[/tex]
Rewrite as:
[tex]y = (x + 8)^2[/tex]
Express y as: [tex]f^{-1}(x)[/tex]
[tex]f^{-1}(x) = (x + 8)^2[/tex]
To determine the domain, we have:
The original function is [tex]f(x) = \sqrt x - 8[/tex]
The range of this is: [tex]f(x) \ge -8[/tex]
The [tex]domain[/tex] of the [tex]inverse[/tex] function is the [tex]range[/tex] of the [tex]original[/tex] function.
Hence, the domain is:
[tex]x \ge -8[/tex]
