# Monotonicity

• Sep 19th 2009, 12:11 PM
Kasper
Monotonicity
Hey guys, I've been working on the topic of monotonicity, and I've got a bit of a grasp on using the definitions of strictly/monotonically increasing/decreasing to show which category a function falls into, but this one is slipping me, because it's not so much a composite function as a modification of an existing function...

Quote:

Given that $\displaystyle g(x)$ is negative and monotonically decreasing, what can be said about the monotonicity of: $\displaystyle h(x)=\frac{1}{g(x)}$?
I want to say that the monotonicity is reverse and that $\displaystyle h(x)$ is monotonically increasing, because with reciprocal functions, where $\displaystyle g(x)$ was small, $\displaystyle h(x)$ will be large, and where $\displaystyle g(x)$ was large, $\displaystyle h(x)$ will be small.

If that's even right, I'm not sure how to write the proof when I only have the definition in terms of 1 function, $\displaystyle g(x)$.

Any ideas?
• Sep 19th 2009, 12:54 PM
Krizalid
$\displaystyle h(x)=\frac{1}{g(x)}\implies h'(x)=-\frac{g'(x)}{g^{2}(x)}.$

As $\displaystyle g(x)$ being decreasing, it verifies that $\displaystyle g'(x)<0$ and $\displaystyle g^2(x)>0$ thus $\displaystyle h(x)$ is increasing.
• Sep 20th 2009, 07:25 AM
Kasper
Quote:

Originally Posted by Krizalid
$\displaystyle h(x)=\frac{1}{g(x)}\implies h'(x)=-\frac{g'(x)}{g^{2}(x)}.$

As $\displaystyle g(x)$ being decreasing, it verifies that $\displaystyle g'(x)<0$ and $\displaystyle g^2(x)>0$ thus $\displaystyle h(x)$ is increasing.

Cool, thanks a lot!
• Sep 20th 2009, 07:45 AM
bkarpuz
Krizalid already given the proof, but I have to mention that the $\displaystyle g$ function need not to be differentiable.
Therefore, we can also use the following.
For any $\displaystyle x,y\in\mathbb{R}$ with $\displaystyle y\geq x$, we have $\displaystyle 0>g(x)\geq g(y)$ (i.e. $\displaystyle g(x)-g(y)\geq0$), which implies
$\displaystyle h(y)-h(x)=\frac{1}{g(y)}-\frac{1}{g(x)}=\frac{g(x)-g(y)}{g(x)g(y)}\geq0$ (i.e. $\displaystyle h(y)\geq h(x)$).
• Sep 20th 2009, 10:41 AM
Kasper
Quote:

Originally Posted by bkarpuz
Krizalid already given the proof, but I have to mention that the $\displaystyle g$ function need not to be differentiable.
Therefore, we can also use the following.
For any $\displaystyle x,y\in\mathbb{R}$ with $\displaystyle y\geq x$, we have $\displaystyle 0>g(x)\geq g(y)$ (i.e. $\displaystyle g(x)-g(y)\geq0$), which implies
$\displaystyle h(y)-h(x)=\frac{1}{g(y)}-\frac{1}{g(x)}=\frac{g(x)-g(y)}{g(x)g(y)}\geq0$ (i.e. $\displaystyle h(y)\geq h(x)$).

Awesome, I think this is what my professor prefers at this stage in the course; though to be honest, the differentiation proof makes more sense to me. I need to become more acquainted with using the definitions in proofs, rather than my usual worded rationales.

Thanks again guys. :)