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Math Help - Difficult BC Composition Problem

  1. #1
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    Difficult BC Composition Problem

    Let f and g be differentiable functions with the following properties:

    I. f(x) < 0 for all x
    II. g(5)=2

    If h(x) = f(x)/g(x) and h'(x)= f'(x)/g(x), then g(x)

    I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

    Thanks!
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  2. #2
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    Quote Originally Posted by r2d2 View Post
    Let f and g be differentiable functions with the following properties:

    I. f(x) < 0 for all x
    II. g(5)=2

    If h(x) = f(x)/g(x) and h'(x)= f'(x)/g(x), then g(x)

    I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

    Thanks!
    h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}=\frac{f'(x)}{g(x)}

    And we can use property 2 to say

    h'(5)=\frac{2f'(x)-f(x)g'(x)}{4}=\frac{f'(x)}{2}

    So, 4f'(x)-2f(x)g'(x)=4f'(x)

    So, 2f(x)g'(x)=0

    But f(x)<0 so f(x)\not =0 and therefore
    g'(x)=0

    So g(x)=2
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  3. #3
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    Quote Originally Posted by r2d2 View Post
    Let f and g be differentiable functions with the following properties:

    I. f(x) < 0 for all x
    II. g(5)=2

    If h(x) = f(x)/g(x) and h'(x)= f'(x)/g(x), then g(x)

    I started by taking the derivative of h(x) and set it equal to h'(x) but I cant seem to get g(x) by itself. Been working on it for hours, and can't seem to get it right. Anyone know how to solve this tricky problem?

    Thanks!
    I don't really understand what you want to do... do you want to find g(x) ?

    If so:

    h'(x) = \frac{f'(x)}{g(x)} = \frac{f'(x)g(x) -f(x)g'(x)}{(g(x))^2} \Rightarrow \frac{f'(x)}{g(x)} - \frac{f'(x)g(x)}{(g(x))^2} + \frac{f(x)g'(x)}{(g(x))^2}=0

    \Rightarrow \frac{f'(x)}{g(x)} - \frac{f'(x)}{g(x)} + \frac{f(x)g'(x)}{(g(x))^2} = 0 \Rightarrow f(x)g'(x) = 0

    But we know that f(x) < 0 \Rightarrow f(x) \neq 0 and therefore g'(x) = 0 \Rightarrow g(x) = C where C is a constant.

    We know that g(5) = 2 \Rightarrow g(x) = 2
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  4. #4
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    Thank you for borth of your responses. I think where I got confused was whether or not g(x) had to be a function or a number, so at first I disregarded the second information.

    Looking at it again, this problem seems so simple if I think of the right processes.

    Thanks again!

    R2
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