# Thread: Question on Inferior Good Question, requires verifying?

1. ## Question on Inferior Good Question, requires verifying?

The function used is:
g(x)=[e^(Uo/A)][e^(-x^2)/(2A)]

where x is a quantity of a good, Uo is a constant that represents utility, and A is a positive constant.

We have to prove that the graph of g(x) is concave down for x < √(A) and concave up for x > √(A).

How do we verify this?

2. Originally Posted by k101
The function used is:
g(x)=[e^(Uo/A)][e^(-x^2)/(2A)]

where x is a quantity of a good, Uo is a constant that represents utility, and A is a positive constant.

We have to prove that the graph of g(x) is concave down for x < √(A) and concave up for x > √(A).

How do we verify this?

Second derivative test.

3. Yes, I am familiar with that. But I am confused on how to find the derivative considering that both Uo and A are constants, and isn't the derivative of a constant 0?

4. I have to admit that I wondered what an "inferior good question" would be. If it's a "good question", how can it be "inferior"?

You need to differentiate with respect to x, not $U_0$ or A! What you have is essentially
$Ce^{-x^2}$ where $C= \frac{e^{U_0/A}}{2A}$, a constant.

By the chain rule, the first derivative is $-2xCe^{-x^2}$ and the second derivative is $-2Ce^{-x^2}+ 4x^2Ce^{-x^2}$.

5. What is C?

6. Originally Posted by HallsofIvy
I have to admit that I wondered what an "inferior good question" would be. If it's a "good question", how can it be "inferior"?

You need to differentiate with respect to x, not $U_0$ or A! What you have is essentially
$Ce^{-x^2}$ where $C= \frac{e^{U_0/A}}{2A}$, a constant.

By the chain rule, the first derivative is $-2xCe^{-x^2}$ and the second derivative is $-2Ce^{-x^2}+ 4x^2e^{-x^2}$.
Inferior goods has to do with economics. Inferior goods demand decreases when consumers income increases.

For instance, if you make more money, you wouldn't be buying generic peanut butter. You would by Peter Pan or Jiffy

7. Originally Posted by k101
What is C?
A constant because you have e^{a constant}= C

9. Originally Posted by k101
$C\in\mathbb{R}$

You say A and Uo is a constant. For simplicity, let's assume they are both one.

$\displaystyle \frac{e^{1/1}}{2}=\frac{e}{2}$

e is nothing more than just a number so it is a constant.

Any constant multiplied by a constant or raised to a constant is always a constant.

Instead of saying $\displaystyle\frac{e^{Uo/A}}{2A}$, we can call it C or K or M or N or anything you want to.

Also, Halls told you what C is already.

Originally Posted by HallsofIvy
$C= \frac{e^{U_0/A}}{2A}$, a constant.

10. Originally Posted by k101
What is C?
I told you that in my post!

$C= \frac{e^{U_0/A}}{2A}$

I just replace that expression from your post with the single letter "C" to emphasize that A and $U_0$ as simply fixed numbers (constants) and so is any combination of them.