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Math Help - Showing that a particular function is quasiconcave

  1. #1
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    Showing that a particular function is quasiconcave

    Hi

    I'm not sure if this is the right subforum to be posting this question in as it might seem too simple.

    Show that f(x) = x^2 for x>0 is a quasi-concave function.

    First of all, I understand that e^x which is a convex function, is quasi-concave since any monotone function is quasi-concave if the domain is a convex subset of \mathbb{R}.

    My question however, is how do I show that the function f(x) = x^2 is quasi-concave graphically? Is there some way in which a line can be drawn to show that all the points within line are a convex set and therefore the function is quasi-concave?

    Thanks for the help
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Hweengee;612209My question however, is how do I show that the function [tex
    f(x) = x^2[/tex] is quasi-concave graphically? Is there some way in which a line can be drawn to show that all the points within line are a convex set and therefore the function is quasi-concave?

    If I\subset \mathbb{R} is an interval, then f:I\subset \mathbb{R}\rightarrow{\mathbb{R}} is quasi-concave iff

    f[\alpha x+(1-\alpha y) ]\geq \min \left\{{f(x),f(y)}\right\},\quad \forall x\;\forall y\in I,\; \forall \alpha \in [0,1]

    So, in the particular case

    0,+\infty) \rightarrow{\mathbb{R}},\quad f(x)=x^2" alt="f0,+\infty) \rightarrow{\mathbb{R}},\quad f(x)=x^2" />

    analyze graphically that for all [x,y]\subset (0,+\infty) the image of any z\in [x,y] is greater or equal than f(x) .


    Fernando Revilla
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