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Algebra Help Inverse One-to-one

posted by on .

I need to know if I have done this correctly.

Problem
If f is one-to-one, find an equation for its inverse.
f(x)=x^3-5
y=x^3-5
x=y^3-5
5+x=y^3-5+5
5+x=y^3
My answer:
y^3=5+x
3ãy=3ã5+x
y=3ã5+x
f^-1(x)3ã5+x

  • Algebra Help Inverse One-to-one - ,

    y = (5+x)^(1/3)
    is the easy way to write that.
    Now lets say x = 3
    then
    y = 8^1/3 = 2
    but what if y = (-1 + i sqrt 3)
    square that
    1 -2 i sqrt 3 - 3 = (-2 - 2i sqrt 3)
    multiply that by (-1+ i sqrt 3)
    2 -2 i sqrt 3 + 2 i sqrt 3 + 6
    = 8 !!!!!
    I also claim that the complex conjugate
    y = (-1 -i sqrt 3)
    cubed is also 8
    therefore I claim that y = 2 or y = -1+i sqrt 3 or y = -1 -i sqrt 3
    are all values of y for x = 3
    So if your algebra subject is about complex as well as real numbers, there are three values of y for any value of x in the inverse problem.

  • Algebra Help Inverse One-to-one - ,

    Sorry, my answer is:

    y^3=5+x
    3ãy=3ã5+x
    y=3ã5+x
    f^-1(x)=3ã5+x

  • Algebra Help Inverse One-to-one - ,

    but whatever your symbol is for the cube root, you have to take the cube root of the whole right side, need parentheses
    (5+x)

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