Answer:
4th choice [tex]\bold{f^{-1}(x) = \sqrt{x} + 4}[/tex]
Step-by-step explanation:
Definition of the inverse of a function
A function g is the inverse of a function f if whenever y=f(x) then x=g(y). In other words, applying f and then g is the same thing as doing nothing. We can write this in terms of the composition of f and g as g(f(x))=x. The domain of f becomes the range of g and the range of f becomes the domain of g
To solve for the inverse of the function [tex]f(x) =\left(x-4\right)^2[/tex]
Let [tex]y=\left(x-4\right)^2[/tex]
[tex]\mathrm{Replace}\:x\:\mathrm{with}\:y[/tex] [tex]\text{ and replace }\:y\:\mathrm{with}\:x[/tex]
[tex]x=\left(y-4\right)^2[/tex]
Switch sides
[tex]\left(y-4\right)^2=x[/tex]
Take square roots on both sides
[tex]y-4=\pm\sqrt{x}[/tex]
Add 4 on both sides to solve for y
[tex]y = \pm\sqrt{x} + 4[/tex]
We have two solutions
[tex]y=\sqrt{x}+4,\:y=-\sqrt{x}+4[/tex]
To determine which one of these to be chosen not that in the given choices we can eliminate the first two since x cannot be negative
The third choice can also be eliminated since
[tex]-\sqrt{x} + 4[/tex] is a decreasing function for [tex]x \ge 0[/tex]
So the last answer choice is correct and the inverse of[tex]f(x) = (x-4)^2[/tex]
is given by [tex]f^{-1}(x) = \sqrt{x} + 4[/tex]
Answer:4th choice [tex]\bold{f^{-1}(x) = \sqrt{x} + 4}[/tex]
Note
Domain of (x-4)² is [4, ∞) since x ≥ 4 and (x-4)² cannot be negative
Range of (x-4)² is [0, ∞)
Domain of [tex]\sqrt{x}\:+\:4[/tex] is [0, ∞)
Range of [tex]\sqrt{x}\:+\:4[/tex] is [4, ∞)
so indeed the domain of (x-4)² has become the range of [tex]\sqrt{x}\:+\:4[/tex] and the range of (x-4)² has become the domain of [tex]\sqrt{x}\:+\:4[/tex]
