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Thread: Derivates Question

  1. #1
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    Derivates Question

    #1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?


    #2. The following values are given:
    f(2.5) = 31.25
    f(2.8) = 39.20
    f(3.0) = 45
    f(3.1) = 48.05
    For the function f which is differentiable. Both f and f' are strictly increasing on the interval 0 < x < 5. Which of the following could be the value of f'(3)?
    (A) 20
    (B) 27.5
    (C) 29
    (D) 30
    (E) 30.5
    - originally, I tried to approximate f' by using the secant line and taking f(3.1) - f(2.8)/3.1 - 2.8 but ended up getting 29.5 as an answer..
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  2. #2
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    Quote Originally Posted by xxlvh View Post
    #1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?
    g(f(3))=g(8)

    but

    g(f(3))=3

    so:

    g(8)=3.

    From the chain rule we have:

    g'(f(x))f'(x)=1

    So:

    g'(f(3))f'(3)=1

    CB
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  3. #3
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    Quote Originally Posted by xxlvh View Post
    #1. The functions f and g are differentiable. For all x, f(g(x)) = x and g(f(x)) = x. If f(3) = 8 and f'(3) = 9, what are the values of g(8) and g'(8)?


    #2. The following values are given:
    f(2.5) = 31.25
    f(2.8) = 39.20
    f(3.0) = 45
    f(3.1) = 48.05
    For the function f which is differentiable. Both f and f' are strictly increasing on the interval 0 < x < 5. Which of the following could be the value of f'(3)?
    (A) 20
    (B) 27.5
    (C) 29
    (D) 30
    (E) 30.5
    - originally, I tried to approximate f' by using the secant line and taking f(3.1) - f(2.8)/3.1 - 2.8 but ended up getting 29.5 as an answer..
    As $\displaystyle f$ and $\displaystyle f'$ are strictly increasing we have for all $\displaystyle h,g>0$:

    $\displaystyle \frac{f(x)-f(x-h)}{h}>f'(x)>\frac{f(x+g)-f(x)}{g}$

    Now use $\displaystyle h=0.2\ x=3\ g=0.1$

    CB
    Last edited by CaptainBlack; Dec 20th 2008 at 10:11 AM.
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    I'm still a little bit lost...
    Well, when I solved for g'(f(3))f'(3)=1 I had g'(8) equalling 1/9. (Hopefully I did that correctly) I'm just not sure where the "1" came from for that equation.
    And for question 2..where did the values of h, x, and g come from?
    Last edited by xxlvh; Dec 20th 2008 at 09:12 AM.
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    Quote Originally Posted by xxlvh View Post
    I'm still a little bit lost...
    Well, when I solved for g'(f(3))f'(3)=1 I had g'(8) equalling 1/9. (Hopefully I did that correctly) I'm just not sure where the "1" came from for that equation.
    we have:

    $\displaystyle g(f(x))=x$

    diffenetiate both sides to get:

    $\displaystyle g'(f(x))f'(x)=\frac{d}{dx}x=1$

    CB
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  6. #6
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    Quote Originally Posted by xxlvh View Post
    And for question 2..where did the values of h, x, and g come from?
    The inequality is true for arbitary positive $\displaystyle h$ and $\displaystyle g$, and arbitary $\displaystyle x$.

    Now we want to know about the derivative at $\displaystyle x=3$, which determines the value of $\displaystyle x$ of interest. In the tabulated data the previous $\displaystyle x$ is $\displaystyle 2.8$ or $\displaystyle 0.2$ less than $\displaystyle 3$, and the next point is for $\displaystyle 3.1$, or $\displaystyle 0.1$ greater than $\displaystyle 3$.

    So we have:

    $\displaystyle
    \frac{f(3)-f(2.8)}{0.2}>f'(3)>\frac{f(3.1)-f(3)}{0.1}
    $

    CB
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