# Thread: Problem with proving that function is not monotonic

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

Hello,

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

2. ## 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

3. ## Re: Problem with proving that function is not monotonic

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

Marian

4. ## Re: Problem with proving that function is not monotonic

Originally Posted by MINOANMAN
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, $\displaystyle g(x)=\frac{1}{x}$ is not a function $\displaystyle \mathbb{R}\to\mathbb{R}$ because it is not defined at $\displaystyle x=0$.

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

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

Originally Posted by MINOANMAN
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 $\displaystyle f~\&~g$ have derivatives.

5. ## 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.

6. ## 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: $\displaystyle f(x) = e^{x}$ is strictly monotonic increasing and $\displaystyle g(x) = -e^{x}$ is strictly monotonic decreasing then $\displaystyle h(x) = f(x)g(x) = -e^{2x}$ which is strictly monotonic decreasing too, so that fact is not true.