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Find the inverse of the polynomial.

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Find [itex](f^{-1})'(a)[/itex] of: [itex]f(x)=\sqrt{x^{3}+x^{2}+x+22}[/itex] ; a=5.

    2. Relevant equations


    3. The attempt at a solution

    Well, I know to find an inverse: I need to set the equation equal to y, solve for x, then swap x and y. But I don't know how to solve this for x.
    I have been searching the internet for a lesson on this type of problem but I am not finding anything.
    I can find the derivative, but I don't think that helps me here.
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2


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    Don't try to find a general formula for the inverse. It's way too hard. You only need the value of the inverse at a=5. What's f^(-1)(5)? It's perfectly ok to guess.
  4. Jan 17, 2012 #3
    Well, [tex]f(5)=\sqrt{177}\approx13.304[/tex]
    So, is the inverse 1/13.304, but then what is the derivative of the inverse?

    Or the inverse of the derivative??? at a=5[tex]f'(x)=\frac{1}{2}({x^{3}+x^{2}+x+22})^{-\frac{1}{2}}(3x^{2}+2x+1)[/tex]
    I'm not really sure of the correct guessing process.
  5. Jan 17, 2012 #4
    My choices were 2/3, 5/3, 4/3, or 7/3. I guessed 2/3, but was wrong and the correct answer was 5/3. My guessing needs work.
  6. Jan 17, 2012 #5


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    To find f^(-1)(5) you want to find a value of x such that f(x)=5. As you said, f(5)=sqrt(177). So that's not it. f(0)=sqrt(22). That's not 5 either. Put some more values into f(x) and see if you can get it to come out 5.
  7. Jan 17, 2012 #6
  8. Jan 17, 2012 #7


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    Ok, so f^(-1)(5)=1. That's the f^(-1)(a) in your formula, right?
  9. Jan 17, 2012 #8
    In this formula?[tex](f^{-1})(a)=\frac{1}{f'((f^{-1})(a))}
    equals 3/5, and the inverse is 5/3.

    So, is this the way I should go about these kind of problems?
  10. Jan 17, 2012 #9


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    Yes. That's the way to go. If a function is hard or impossible to invert, they'll probably make it easy to guess a specific value of the inverse, like in this problem.

    BTW your formula should read (you're missing a prime)

  11. Jan 17, 2012 #10
    I am still a little confused, but thank you very much for your help.

    I need to find and practice another one of these.
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