Let $f(x),g(x)$ be positive strictly increasing function.
When can I assure that $h(x)=\dfrac{f(x)}{g(x)}$ is increasing or decreasing?
I'm trying to prove that a function is increasing, but is too complicated so i'm looking for some criteria that would help.
I tried the derivative, but that does not work because the function I'm dealing with are too complicated to derive, so I'm asking for the case when the functions are nondifferentiable, thanks.
Best Answer
to find how is the behavior of h(x) we have to calculate it's derivative so:
$$h' = \frac{f'g-g'f}{g^2}$$
in order that this fraction be positive(so h be increasing) we must have:
$$f'g-g'f\ge0 \;\;\;\;\; \; \Rightarrow \frac{f'}{f} \ge \frac{g'}{g}$$ now integrating yields: $$ln(f)\ge ln(g) + constant \Rightarrow f \ge constant \times g $$
so you must identify that f is always greater than g and since g and f are both positive this test can be done easily. on the other hand if $$g \ge constant \times f $$ then h is decreasing.
this just popped into my head now but there must be more sophisticated ways.
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After discussion with S.harp he proved to me the above statement is wrong and the counter example is the one in the comments below. so I edit this answer this way: that constant is an arbitrary positive number. so for above solution to be true there must be an interval of the form $(\beta,+ \infty) \; $ in which the inequality $$f \ge constant \times g $$ or the other one holds for any positive arbitrary constant . this happens when the growth rate of f is greater than g other whise the claim could be wrong. for example: $f= 100+x \; and \; g = x$ although $f>g$ but $h = f/g $ is decreasing. so if growth rate of f is greater than g then the claim is right.
I apologize for wrong response. also thanks to S.harp for revealing that first claim was wrong.