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Math Help - equation for the tangent

  1. #1
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    equation for the tangent

    Hey guys, I completely forgot how to do this problem.

    Find an equation for the line tangent to the graph of f^{-1} at the point (3,1) if f(x)= x^3+2 x^2-x+1.

    I'm not sure if a formal proof is needed and if so how to do it.
    Last edited by mr fantastic; May 2nd 2010 at 12:44 PM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by alice8675309 View Post
    Hey guys, I completely forgot how to do this problem, and its from the same book I've been using that's I'm still posting it in this section.


    Find an equation for the line tangent to the graph of f^{-1} at the point (3,1) if f(x)= x^3+2 x^2-x+1.

    I'm not sure if a formal proof is needed and if so how to do it.
    The tangent at the point (3,1) of the graph y=f^{-1}(x) is t:\, y=f'^{-1}(3)\cdot (x-3)+1.
    The remaining problem is to determine f'^{-1}(3). Luckily there is a formula that allows you to determine f'^{-1}(x) from the derivatives of f(x) like this: f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}.
    So in your case this means that f'^{-1}(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\ldots
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  3. #3
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    Quote Originally Posted by Failure View Post
    The tangent at the point (3,1) of the graph y=f^{-1}(x) is t:\, y=f'^{-1}(3)\cdot (x-3)+1.
    The remaining problem is to determine f'^{-1}(3). Luckily there is a formula that allows you to determine f'^{-1}(x) from the derivatives of f(x) like this: f'^{-1}(x)=\frac{1}{f'(f^{-1}(x))}.
    So in your case this means that f'^{-1}(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\ldots
    Hey thanks so much, mental block there..and since I wasn't given any inital values I just plug in for 1 into f' and get 6 so that means my answer is 1/6?
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by alice8675309 View Post
    Hey thanks so much, mental block there..and since I wasn't given any inital values I just plug in for 1 into f' and get 6 so that means my answer is 1/6?
    Well, that's the slope of the tangent, but you need the entire equation of the tangent which is, as I wrote, t:\, y=\frac{1}{6}(x-3)+1, or, if you prefer, t:\, y=\frac{1}{6}x+\frac{1}{2}.
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  5. #5
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    Quote Originally Posted by Failure View Post
    Well, that's the slope of the tangent, but you need the entire equation of the tangent which is, as I wrote, t:\, y=\frac{1}{6}(x-3)+1, or, if you prefer, t:\, y=\frac{1}{6}x+\frac{1}{2}.
    Oh yea, I wrote it out like the second one lol I was just making sure it was right, I just forgot to finish on the actual post. But thank you so much!
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