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Math Help - Problem with proving that function is not monotonic

  1. #1
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    Problem with proving that function is not monotonic

    Hello,

    Given two function f:\mathbb{R}\to\mathbb{R} which is monotonic increasing and g:\mathbb{R}\to\mathbb{R} which is monotonic decreasing I need to prove that h(x)=f(x)g(x) is not monotonic.
    I started my proof with assuming that h(x) is monotonic increasing so for every x_1<x_2 I know that h(x_1)<h(x_2) which means that f(x_1)g(x_1)<f(x_2)g(x_2) but I'm stuck here and don't how to continue. I know that f(x_1)<f(x_2) and g(x_1)>g(x_2) but I don't know how to connect between them.

    Thank you for your help.
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  2. #2
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    Re: Problem with proving that function is not monotonic

    Kaze1

    you have to reconsider your atempt to prove that f(x)g(x) is not monotonic.....

    I will let you find yourself what I mean

    Example 1 f(x)=x^5....and g(x) =1/x then f(x)g(x)=x^4 which is of course monotonic( increasing/decreasing)

    Example 2 f(x)=x and g(x)=1/x^2 then f(x)g(x)=1/x which is again monotonic ( decreasing)

    example 3 f(x)=x and g(x) =1/x then f(x)g(x)=1 constant function....therefore not monotonic...

    therefore the product of two functions are either monotonic or not.....either increasing or decreasing.....
    consequently your primay concept is wrong.

    meanwhile you cannot use the definition of the monotone increasing or decreasing to prove that a function f(x) is increasing or decreasing because you have to prove this for a large number of points (values) ...
    Consult the differential calculus to find a theorem that is associated with the monotone increasing functions and simply says that :if the derivative of a function is positive then the function is a monotone increasing function and if the derivative is negative then the function decreases..... a very well known theorem that is widely used to determine the nature of the stationary points of a function.

    MINOAS
    Last edited by MINOANMAN; March 9th 2013 at 08:24 AM.
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  3. #3
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    Re: Problem with proving that function is not monotonic

    Minoas: I'm very confused by your response. In example 1, you state that x^4 is monotonic increasing which is not true on (-\infty, 0). In example 2 you state that g(x)=1/x^2, but this function is not monotonic decreasing on (-\infty, 0). To the second part of your response: is it possible that f(x) and g(x) are not differentiable?

    Marian
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    Re: Problem with proving that function is not monotonic

    Quote Originally Posted by MINOANMAN View Post
    Example 1 f(x)=x^5....and g(x) =1/x then f(x)g(x)=x^4 which is of course monotonic( increasing) MINOAS
    @MINOAS, I really do not see the point of that example.
    First, g(x)=\frac{1}{x} is not a function \mathbb{R}\to\mathbb{R} because it is not defined at x=0.

    Moreover, \frac{1}{x} is certainly not monotonic (neither increasing nor decreasing).

    And x^4 is certainly not monotonic increasing, as you state above.


    Quote Originally Posted by MINOANMAN View Post
    Consult the differential calculus to find a theorem that is associated with the monotone increasing functions and simply says that :if the derivative of a function is positive then the function is a monotone increasing function and if the derivative is negative then the function decreases
    There is absolutely nothing in the given to suggest that f~\&~g have derivatives.
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  5. #5
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    Re: Problem with proving that function is not monotonic

    I haven't considered functions defined strictly R-R.... or not differentiable ..
    what I wanted to express is that most of the problems regarding monotonic functions are solved using derivatives.
    Last edited by MINOANMAN; March 9th 2013 at 08:42 AM.
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  6. #6
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    Re: Problem with proving that function is not monotonic

    I somehow solved this problem. The answer is that this fact is not true. for example: f(x) = e^{x} is strictly monotonic increasing and g(x) = -e^{x} is strictly monotonic decreasing then h(x) = f(x)g(x) = -e^{2x} which is strictly monotonic decreasing too, so that fact is not true.

    Thank for your help.
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