Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Solving functions algebraically (cube roots)

  1. Oct 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Show f and g are inverse functions or state that they are not.
    f(x)= cube root of -8x-6 g(x)= -(x^3+6)/(8)

    2. Relevant equations
    You find inverses by plugging the equations into each other, if they are inverses then once you simplify the composed equation, it will equal x.

    3. The attempt at a solution

    cube root of {-8[-(x^3+6)/(8)]-6} I plugged g(x) into f(x) now I need to solve it, this is what I did but its not right because the answer should come out to be just x.

    I multiplied everything by 8 to get it out of the denominator and simplified the numerator which left me with

    cube root of {64x^3-432}

    then I thought maybe you could pull out the 64 and x^3 from the cube root to get

    4x times the cube root of {-432}

    which is definitely not going to = x not matter what I do to it.

    I feel a bit stupid for not being able to solve this because the solving is not supposed to be the challenge, the concept of what makes the functions of inverses is. I think I've just forgotten how to solve harder equations.
  2. jcsd
  3. Oct 31, 2007 #2


    User Avatar
    Gold Member

    I think its just a case of a mistake in your algebra. Can you show your full working perhaps?
  4. Oct 31, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    Say f(x) = y = cube root of -8x-6.

    Solve for [EDIT: x].

    Hint: y^3 = -8x-6.

    Replace x with y, and y with x. Check to see y = g(x).
    Last edited: Nov 1, 2007
  5. Nov 1, 2007 #4


    User Avatar
    Science Advisor

    You multiplied everything by 8? Wouldn't it be easier just to cancel the "8" in the numerator with the "8" in the denominator?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook