Question

[thegersters]

I have this problem

If f(x) = (2x + 2)1/3, then f -1(x) =

I think the answer is (x3 + 2)/2

AM I WRONG?

 

[soroban]

Hello, thegersters!

Your notation is puzzling . . . but I'll take a guess.

If \(\displaystyle f(x) = (2x\ +\ 2)^{\frac{1}{3}}\), then \(\displaystyle f^{-1}(x)\ =\)

I think the answer is: \(\displaystyle \L \frac{x^2\ +\ 2}{2}\)

AM I WRONG? . . . . close, but wrong

We have: . \(\displaystyle \L(2x\ +\ 2)^{\frac{1}{3}} \;= \;y\)

Switch variables: . \(\displaystyle \L(2y\ +\ 2)^{\frac{1}{3}} \;= \;x\)

Cube both sides: . \(\displaystyle \L2y\ +\ 2 \;= \;x^3\)

Subtract 2: . \(\displaystyle \L2y \;= \;x^3\ -\ 2\)

Divide by 2: . \(\displaystyle \L y \;= \;\frac{x^3\ -\ 2}{2}\)


Therefore: . \(\displaystyle \L f^{-1}(x) \;= \;\frac{x^3\ -\ 2}{2}\)