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Thread: Monotonicity

  1. #1
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    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...

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

    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.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle h(x)=\frac{1}{g(x)}\implies h'(x)=-\frac{g'(x)}{g^{2}(x)}.$

    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!
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  4. #4
    Senior Member bkarpuz's Avatar
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    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. $\displaystyle h(y)\geq h(x)$).
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  5. #5
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    Quote Originally Posted by bkarpuz View Post
    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.
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