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Find the inverse of y=x^3 + x +4

  1. May 7, 2008 #1
    1. The problem statement, all variables and given/known data


    I'm helping my friend review for his exam in precalc and I can't seem to get this inverse.

    Find the inverse of y=x^3 + x +4
    2. Relevant equations



    3. The attempt at a solution
    I switched the variables and then tried to isolate the y but I failed.
    I ended with x-4=y^3+y I don't know how to isolate the y from here how would you go about it?
     
  2. jcsd
  3. May 7, 2008 #2

    symbolipoint

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    you would obtain x=y^3 +y +4, but unless you can find a way to produce an equivalent cubed binomial for the right side, you are stuck. If there are more than one y value for any single x value, then that is not an inverse function.
     
  4. May 7, 2008 #3
    Given f (x) = x3 + x + 4 , find f -1 (1) accurate to 2 decimal places. Is the exact question I thought I would make it easier by just asking to find the inverse on here...
     
  5. May 7, 2008 #4

    symbolipoint

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    If you are asking for how to evaluate the inverse of f(x) at x=1, then you want to start with the inverse relationship, x= y^3 + y + 4 [I assume you still meant that the original function was a cubic], and substitute 1=x; and then find the possible values for y. If you obtain two values of y, then the inverse of f(x) is not a function.

    1 = y^3 + y + 4
    y^3 + y + 3 = 0
    Complete the square or use quadratic formula solution;
    What is y?
     
  6. May 7, 2008 #5

    symbolipoint

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    What I said in here (below) is ridiculous. You can't complete the square for a cubic; although you can look for the presence of any x which gives more than one y value.

     
  7. May 8, 2008 #6

    HallsofIvy

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    That's a completely different question! In order to solve the equation y3+ y+ 4= x, for any x, you would have to use Cardano's cubic formula- and that's extremely complex!

    What you are asked to do here is to use some numerical method to solve the equation y3+ y+ 4= 1 or y3+ y+ 3= 0. A simple way to solve such an equation is "bijection". If y= -2, (-2)3+ (-2)+ 3= -8- 2+ 3= -7< 0. If y= -1, (-1)3+ (-1)+ 3= 1> 0 Since the value at x= -2 is negative and the value at x= -1 is positive, there must be a solution between -2 and -1. Try half way between: -3/2. Is (-3/2)3+ (-3/2)+ 3 positive or negative? Whichever it is, there must be a root between -3/2 and either -2 or -1 depending on which gave the opposite sign. Try half-way between just because it is convenient.
     
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