SOLUTION: Given the function f(x)=3x^3-1, find f^(-1) f(x)=3x^3-1 find f^-1(x) Given the function f(x)=2x^2+1 and g(x)=x-4, find (fog)(x). Restated that is f(x)=2x^2+1 and g(x)=x-4 find (

Algebra ->  Algebra  -> Functions -> SOLUTION: Given the function f(x)=3x^3-1, find f^(-1) f(x)=3x^3-1 find f^-1(x) Given the function f(x)=2x^2+1 and g(x)=x-4, find (fog)(x). Restated that is f(x)=2x^2+1 and g(x)=x-4 find (      Log On

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Question 2095: Given the function f(x)=3x^3-1, find f^(-1) f(x)=3x^3-1 find f^-1(x)
Given the function f(x)=2x^2+1 and g(x)=x-4, find (fog)(x). Restated that is f(x)=2x^2+1 and g(x)=x-4 find (fog) (x)
Solve the following x: log base 2(3x-5)=3log2 (3x-5)=3
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
Not too sure of your notation here, sorry...
inverse of a function is purely, re-write it as x=, that is all...
so, y=3x%5E3-1 is x%5E3+=+%28%28y%2B1%29%2F%283%29%29
so x = cube root of %28%28y%2B1%29%2F%283%29%29
so f%5E%28-1%29%28x%29 = cube root of %28%28x%2B1%29%2F%283%29%29

2. f(x) = 2x%5E2%2B1 and g(x)=x-4
Now, i am assuming that your fog(x) is just the same as fg(x)???. If so, fg(x) is f(x-4)...now just put this "x" value in the f(x) equation....
ie f(x-4) = 2%28x-4%29%5E2%2B1
If you want to expand then do so...f(x-4) = 2(x^2-8x+16)+1 so this becomes
f(x-4) = 2x^2-16x+33.

jon