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Function inverse problem

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    I have been given:
    h(x) is the inverse for f(x) = x^3+x
    They want to know h'(2).

    3. The attempt at a solution

    I know that since h(x) and f(x) are inverses:
    f(h(x)) = x

    differentiating with respect to x gives
    f'(h(x))h'(x) = 1

    So h'(x) = 1/f'(h(x))

    Therefore h'(2) = 1/f'(h(2))

    to find h(2)
    x^3+x = 2

    since f(h(x)) = x
    x^3+x-2 = 0

    so h(2) = 1

    f(x) = x^3+x
    f'(x) = 3x^2+1

    h'(2) = 1/f'(h(2))
    h'(2) = 1/3(1)^2+1
    h'(2) = 1/4

    My question is: How do you find the inverse of y = x^3+x ?
    You interchange x and y then solve for y: y^3+y=x But I don't know how to solve that for y. Any help would be appreciated.
     
  2. jcsd
  3. Nov 29, 2009 #2

    jgens

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    Gold Member

    We're not supposed to give complete solutions, so here's a hint:
    [tex]y = w - \frac{1}{3w}[/tex]
     
  4. Nov 29, 2009 #3
    I am sorry but the above doesn't help me solve: y^3+y=x for y. If you will see above, I have already got the solution to the problem using some algebraic gymnastics, I am just curious as to how to find the inverse for y=x^3+x.
     
  5. Nov 29, 2009 #4

    jgens

    User Avatar
    Gold Member

    I know! You want to solve the cubic equation [itex]y^3 + y - x = 0[/itex] and the substitution [itex]y = w + \frac{1}{3w}[/itex] allows you to solve that cubic equation quite nicely.
     
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