Help @ Akshay, Amit Sir

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Help @ Akshay, Amit Sir

ashutosh


Why is the answer a. and why not c ?

Chiang says that y= x^2 is both quasiconcave as well as quasi convex. So by the same logic why is e^-x not quasi convex. ?

 

Please help.
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Re: Help @ Akshay, Amit Sir

ashutosh
P.S
The last picture is from chaing Pg 392. S=> stands for Upper level set and S=< for lower level set.
sam
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Re: Help @ Akshay, Amit Sir

sam
Yes, both a and c are correct :)
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Re: Help @ Akshay, Amit Sir

kangkan
In reply to this post by ashutosh
@ashotosh...f(x)=x^2 is quasi convex becuase the -x^2 is quasi concave.



The pruple curve is for -x^2 and black one fopr -e^-x

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Re: Help @ Akshay, Amit Sir

Amit Goyal
Administrator
In reply to this post by ashutosh
Hi Ashutosh,

In this problem, the domain for these functions is R (the entire real line). Both the functions f(x) = x^2 and g(x) = e^{-x} are convex and hence quasiconvex. However, f(x) = x^2 is not quasiconcave.

But if the domain of f is R+ (non-negative real numbers) and not R then f(x) = x^2 is both quasiconcave and quasiconvex.
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Re: Help @ Akshay, Amit Sir

ashutosh
Thanks a lot amit sir :). :)
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Re: Help @ Akshay, Amit Sir

ashutosh
In reply to this post by kangkan
thanks for the graph @ kangkan. :)

But which software is this ?
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Re: Help @ Akshay, Amit Sir

kangkan
this is called graph..its a free download....by the way,what is the final answer?i cudnt get it
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Re: Help @ Akshay, Amit Sir

Akshay Jain
In reply to this post by ashutosh
there are some good android applications also for plotting 2-D and 3-D graphs....
Akshay Jain
Masters in Economics
Delhi School of Economics
2013-15
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Re: Help @ Akshay, Amit Sir

ashutosh
thanks pals!
" Graph App is working :)
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Re: Help @ Akshay, Amit Sir

ashutosh
In reply to this post by kangkan
@kangkan
final answer is thus a. x^2 which is only quasi convex. And c.) will not be the answer because it is both quasi convex and quasi concave.
You can refer Chiang, its beautifully explained there. :)
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Re: Help @ Akshay, Amit Sir

kangkan
which page number please?