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Math Help - Finding the inverse of a function

  1. #1
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    Finding the inverse of a function

    Find the inverse of the function:

    f(x) = (3/4) x^5 + 5

    read "three-fourths x to the fifth plus 5"


    Graphing my progress on my calculator as I go, the last point where I get an inverse is at

    5^[sqrt] ((4/3)x - (20/3))
    ^(root with an index of 5)

    If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by

    5^[sqrt](3^4)
    ^again, root with an index of 5

    to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?

    Help is much appreciated.
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  2. #2
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    Quote Originally Posted by happydino1 View Post
    Find the inverse of the function:

    f(x) = (3/4) x^5 + 5

    read "three-fourths x to the fifth plus 5"


    Graphing my progress on my calculator as I go, the last point where I get an inverse is at

    5^[sqrt] ((4/3)x - (20/3))
    ^(root with an index of 5)

    If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by

    5^[sqrt](3^4)
    ^again, root with an index of 5

    to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?

    Help is much appreciated.
    Analytically to find the inverse of a function y = f(x) we switch the roles of x and y to get x = f(y), then solve for y: y = g(x). g(x) is our inverse function.

    So
    y = f(x) = \frac{3}{4}x^5 + 5

    Goes to
    x = \frac{3}{4}y^5 + 5

    x - 5 = \frac{3}{4}y^5

    y^5 = \frac{4}{3}(x - 5)

    y = \sqrt[5]{\frac{4}{3}(x - 5)}

    -Dan
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