Let f(x) and g(x) be real functions defined for x>0 such that f is negative and increasing, whiles g is positive and decreasing. Can we conclude that the product f(x)g(x) is decreasing? or What condition(s) will make this product decreasing? Your comments will be appreciated.
Thank you.
If your f and g value variations (increasing and decreasing respectively) are variations in real value (not absolute value) like I think, then the product f(x)g(x) is increasing because it is becoming less negative.
Another way of putting it is that the variation (derivative) of f(x)g(x) is given by:
[f(x)g(x)]'= f '(x)g(x) + f(x)g'(x) = ag(x)+f(x)b > 0
a and b are the derivatives of f and g. I do not know if they exist for every x, but considering f and g variations, we take a as positive and b as negative. Since f is negative and g is positive then [f(x)g(x)]' is always positive and f(x)g(x) is increasing, not decreasing.
Kind regards.
Fernando.
Dear Prof. Fernando Lobo Pimentel,
Thank you for your comment. I am most grateful.
Following up on Prof Fernando's comment: If you want the product of f and g to be decreasing then some possibilities are f and its derivative are both positive while g and its derivative are both negative OR f and its derivative are both negative while g and its derivative are both positive etc.
Dear Prof. Francis Tabi Oduro,
It is true that the product f(x)g(x) is increasing. Thank you for the helpful comments. I am grateful.
Dear Prof. Elman Shahverdiev,
I agree that the product f(x)g(x) is increasing. Thank you for the helpful comment. I am grateful.
Dear Prof. Elman Shahverdiev,
I agree that the product f(x)g(x) is increasing. Thank you for the helpful comment. I am grateful.
Since the minus sign of the increasing function will make the positive decreasing function increasing, the product in this case will be always increasing.
The product fg is always increasing, you don't have to make any assumptions on the regularity of f and g, in particular you don't have to assume that f and g are differentiable.
Suppose x > y, then by assumption, f(x) > f(y) and f(x)0, g(y) > 0. Hence
f(x)g(x) - f(y)g(y) = f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)
= f(x)(g(x) - g(y)) + g(y)(f(x) - f(y)) > 0
because each of the two terms is positive.
Dear Dr. Rogier Brussee,
Thank you for your helpful comments. I appreciate your contribution.
The product will be definitely negative now, fg will be increasing and decreasing as per the orders of the function of f(x) and g(x).
For example, if O(f(x))>O(g(x)), then the product fg will be decreasing and accordingly we may derive for increasing.
f(x)=1/x^2, g(x)=x. f(x)g(x)=1/x is the decreasing function in ]0,+infinity[ but
f(x)=1/x, g(x)=x^2, f(x)g(x)=x is yhe increasing function. in the cases, f is the decreasing function in ]0,+infinity[ and g is the increasing function in ]0,+infinity[.
Generaly, we don't havefg the increasing or decreasing function when f is decreasing function and g is increasing function.
Dear Salifou Mboutngam,
Note that f(x) is negative and increasing whiles g(x) is positive and decreasing. Following the previous comments, it can be established that the product f(x)g(x) is increasing.
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