1. ## Problem

I can't solve this problem:

We have A(x)= (x-1) (2-x) (6+2x)
a. For which prices of x, A(x)=0
b. Find the prices of x, to give A(x)= is not 0
c. Find the prices of x, to has A(x) a inverse/reverse.

I hope you understand this problem, and solve it.
I am not good in English.

2. Originally Posted by gdespina
I can't solve this problem:

We have A(x)= (x-1) (2-x) (6+2x)
a. For which prices of x, A(x)=0
b. Find the prices of x, to give A(x)= is not 0
c. Find the prices of x, to has A(x) a inverse/reverse.

I hope you understand this problem, and solve it.
I am not good in English.
I think "price" = value

a) A(x) = 0 if any of the 3 factors is zero.
So,
x-1 = 0 <----x = 1

2-x = 0 <----x = 2

6 +2x = 0
2x = 0 -6
x = -6/2 = -3

Therefore, A(x) = 0 if x=1, or x=2, or x = -3. -------------answer.

------------------------------------------------
b)
To give A(x) not equal to zero is to not let x equal to 1 or 2 or -3.
Or x is not equal to 1, not equal to 2, or not equal to -3. And x can be any number except those 3 numbers. ------------answer.

------------------
c)
So that A(x) has an inverse or reverse?

Sorry, I don't understand that.

3. For Part C I'm pretty sure you are asking for where the domain must be restricted such that the function has an inverse... In this case its pretty easy to tell if you look at the graph first...

For a function to have an inverse it must be one-to-one... For the purposes of testing this on a graph it simply means that no vertical or horizontal line crosses the graph of the function in more than one place.

In this case you have a cubic function and so it is one-to-one because the preceding is true.

Therefore, there is no need to restrict the domain based on the function itself (it will always have an inverse). Since we are talking about the prices of items, it doesn't make sense for the price to fall below zero so in this context the function has an inverse when x >= 0.

Hope that helps