f(x)=4-2(x-6)^1/3 evaluate when f(x)=5
so I did this:
5=4-2(x-6)^1/3
1=-2(x-6)^1/3
(-1/2)=(x-6)^1/3
How do I get rid of the exponent 1/3? I am stumped.
cube both sides
(an exponent of 1/3 means cube root, so what is the inverse of taking the cube root ?)
(-1/2)^3=[(x-6)^1/3]^3
-1/8 = x-6
x = 6-1/8
x = 47/8