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Math Help - 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 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?
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  2. #2
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    Krizalid's Avatar
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    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.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    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!
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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 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)).
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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 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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