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Thread: f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

  1. #1
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    f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

    This one also comes from Purcell and Varberg, fifth edition, chapter 4, section 2 problem 41b.

    Given f'(x) > 0 and g'(x) > 0, what simple additional conditions would guarantee that f(x)g(x) increases for all x?

    The answer key has f(x) > 0 and g(x) > 0, but I have a counter example: f(x) = x and g(x) = x^2 + 1. f(x) < 0 for x < 0 but the derivative of f(x)g(x) > 0, and therefore f(x)g(x) increases for all x.

    I got only the following condition from the product rule: f(x)/f'(x) > -g(x)/g'(x).
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    Re: f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

    Quote Originally Posted by Zexuo View Post
    This one also comes from Purcell and Varberg, fifth edition, chapter 4, section 2 problem 41b.

    Given f'(x) > 0 and g'(x) > 0, what simple additional conditions would guarantee that f(x)g(x) increases for all x?

    The answer key has f(x) > 0 and g(x) > 0, but I have a counter example: f(x) = x and g(x) = x^2 + 1. f(x) < 0 for x < 0
    but the derivative of f(x)g(x) > 0, and therefore f(x)g(x) increases for all x.

    I got only the following condition from the product rule: f(x)/f'(x) > -g(x)/g'(x).
    Your alleged counterexample does not meet the given conditions.

    While the derivative for your f(x), which is 1, is greater than zero, the derivative for your g(x), which is 2x, is not always positive.
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    Re: f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

    Quote Originally Posted by Zexuo View Post
    This one also comes from Purcell and Varberg, fifth edition, chapter 4, section 2 problem 41b.
    Given f'(x) > 0 and g'(x) > 0, what simple additional conditions would guarantee that f(x)g(x) increases for all x?
    The answer key has f(x) > 0 and g(x) > 0.
    If we know that $(\forall x)[f>0,~g>0,~f'>0,~\&~g'>0]$
    then $(\forall x)[D_x(f(x)\cdot g(x))=\{f'(x)\cdot g(x)+f(x)\cdot g'(x)\}>0]$

    "DONE!" said the queen with the crown. "How can that be?" asked Alice.
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    Re: f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

    Quote Originally Posted by Zexuo View Post
    This one also comes from Purcell and Varberg, fifth edition, chapter 4, section 2 problem 41b.

    Given f'(x) > 0 and g'(x) > 0, what simple additional conditions would guarantee that f(x)g(x) increases for all x?

    The answer key has f(x) > 0 and g(x) > 0, but I have a counter example: f(x) = x and g(x) = x^2 + 1. f(x) < 0 for x < 0 but the derivative of f(x)g(x) > 0, and therefore f(x)g(x) increases for all x.

    I got only the following condition from the product rule: f(x)/f'(x) > -g(x)/g'(x).
    In addition to what greg1313 said, you seem to be thinking that if you found some setup where $f(x)g(x)$ increases for all $x$, but not both $f(x)>0$ and $g(x)>0$ that you would have a counterexample. But the problem does not say that those conditions are necessary. Just that they are sufficient. So you wouldn't have had a counterexample anyway.
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    Re: f'(x) > 0, g'(x) > 0; additional condition for f(x)g(x) increasing everywhere?

    Thanks for the catch, greg1313. In looking for a counterexample I lost sight of the given conditions.

    Walagaster, I indeed sought necessary conditions instead of sufficient ones.
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