Monotonicity

**Kasper** · September 19th 2009, 12:11 PM

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...

Given that $g(x)$ is negative and monotonically decreasing, what can be said about the monotonicity of: $h(x)=\frac{1}{g(x)}$ ?

I want to say that the monotonicity is reverse and that $h(x)$ is monotonically increasing, because with reciprocal functions, where $g(x)$ was small, $h(x)$ will be large, and where $g(x)$ was large, $h(x)$ 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, $g(x)$ .

Any ideas?

**Krizalid** · September 19th 2009, 12:54 PM

$h(x)=\frac{1}{g(x)}\implies h'(x)=-\frac{g'(x)}{g^{2}(x)}.$

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

**Kasper** · September 20th 2009, 07:25 AM

Originally Posted by Krizalid

$h(x)=\frac{1}{g(x)}\implies h'(x)=-\frac{g'(x)}{g^{2}(x)}.$

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

Cool, thanks a lot!

**bkarpuz** · September 20th 2009, 07:45 AM

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

**Kasper** · September 20th 2009, 10:41 AM

Originally Posted by bkarpuz

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

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.

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