# Thread: Showing that a particular function is quasiconcave

1. ## 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

2. 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

$f0,+\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