The function [tex]f(x)=(x-2)^3+8[/tex] is one-to-one and the inverse of [tex]f(x)[/tex] is [tex]f^{-1} (x)=(x-8)^\frac{1}{3}+2[/tex].
A function is one-to-one if for [tex]f(a)=f(b)[/tex], we show that [tex]a=b[/tex].
[tex]f(a)=(a-2)^3+8[/tex]
[tex]f(b)=(b-2)^3+8[/tex]
Take [tex]f(a)=f(b)[/tex]
[tex](a-2)^3+8=(b-2)^3+8[/tex]
[tex](a-2)^3=(b-2)^3[/tex]
Let [tex]A=a-2[/tex] and [tex]B=b-2[/tex]
If [tex]A^3=B^3[/tex], then [tex]A=B[/tex].
So, [tex]a-2=b-2[/tex]
[tex]a=b[/tex].
So, the function is one-to-one.
For the inverse of [tex]f(x)[/tex], replace [tex]x[/tex] with [tex]y[/tex] and solve for [tex]y[/tex].
[tex]x=(y-2)^3+8[/tex]
[tex](y-2)^3=x-8[/tex]
[tex]y-2=(x-8)^\frac{1}{3}[/tex]
[tex]y=(x-8)^\frac{1}{3}+2[/tex]
So, the inverse of [tex]f(x)[/tex] is [tex]y=(x-8)^\frac{1}{3}+2[/tex].
Learn more about and inverse of a function here:
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