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Math Help - Strange problem on inverse function =S

  1. #1
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    Strange problem on inverse function =S

    hi

    am solving past exams to be prepared for my exam next week
    i stopped on this problem :

    let f(x)=x^4 + x^3 + 1 , where x in [0,2] let g(x) = f^-1 (x)

    if h(x) = f(2 g(x)) , find h`(3)??

    its easy problem, i know all what i know is how to find h`(x) ??
    then it will be easy

    help plz !!
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  2. #2
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    Quote Originally Posted by TWiX View Post
    hi

    am solving past exams to be prepared for my exam next week
    i stopped on this problem :

    let f(x)=x^4 + x^3 + 1 , where x in [0,2] let g(x) = f^-1 (x)

    if h(x) = f(2 g(x)) , find h`(3)??

    its easy problem, i know all what i know is how to find h`(x) ??
    then it will be easy

    help plz !!
    I presume that h`(x) is the derivative. (I would have used h'(x).)

    Use the chain rule:
    h'(x)= 2g'(x) f '(2g(x)) and g'(x)= 1/f '(x)

    It is easy to calculate that f'(x)= 4x^3+ 3x^2

    The only "hard" part, perhaps, is finding g(3). To do that you have to solve f(x)= x^4+ ^3+ 1= 3 so x^3+ x^2= 2. Fortunately that has an easy soution in [0, 1]!
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    it seems we will have to use the composite rule to differentiate it !?!
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  4. #4
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    Quote Originally Posted by TWiX View Post
    it seems we will have to use the composite rule to differentiate it !?!
    Yes, of course! It is a composite function.
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    What is the formula of the composite rule ??
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  6. #6
    MHF Contributor Amer's Avatar
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    h(x) = f(2 g(x))

    h'(x) =\left( \frac{d}{dx}(2g(x))\right)f'(2g(x))

    h'(x) = 2g'(x)f'(2g(x))
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  7. #7
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    Thank you ;]]
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  8. #8
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    What is the domain of the composites Function ??

    suppose the doman of f = A
    domain of g = B

    what is the doman of f( g(x) ) , g( f(x) ) ?!
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  9. #9
    MHF Contributor Amer's Avatar
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    let ...f:A\rightarrow B...G:C\rightarrow D

    so

    let..h=f(g)...h:A\rightarrow D

    so

    f(g(x)) domain f but range of g
    g(f(x)) domain g but range of f

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  10. #10
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    if g(x) = f^{-1}(x) , then ...

     <br />
f[g(x)] = g[f(x)] = x<br />

     <br />
f'[g(x)] \cdot g'(x) = g'[f(x)] \cdot f'(x) = 1<br />

    so ...

     <br />
f'[g(x)] = \frac{1}{g'(x)}<br />

    and

     <br />
g'[f(x)] = \frac{1}{f'(x)}<br />


    f(x) = x^4 + x^3 + 1 and f'(x) = 4x^3 + 3x^2

    note that f(1) = 3 , so g(3) = 1

     <br />
g'(3) = g'[f(1)] = \frac{1}{f'(1)} = \frac{1}{7}<br />


     <br />
h'(x) = f'[2g(x)] \cdot 2g'(x)<br />

     <br />
h'(3) = f'[2g(3)] \cdot 2g'(3) = f'[2(1)] \cdot 2\left(\frac{1}{7}\right) = \frac{88}{7}<br />
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