# Critical Points of xe^x

1. Feb 11, 2008

### Mitchtwitchita

can anybody tell me what the critical point of xe^x is? When I try putting it into my calculator, it just shows a line staring at zero with an asymptote at x=1.

2. Feb 11, 2008

### HallsofIvy

Staff Emeritus
What is the definition of "critical point"?

And I can't help but wonder what you put into your calculator! y= xex does not have any asymptotes. What is the value of y= xex when x= 1?

3. Feb 11, 2008

### mrandersdk

$$\frac{\partial}{\partial x} xe^x = x*e^x + e^x$$

so you need to find

$$0 = x*e^x + e^x$$

so the critical point is

$$x = -1$$

it is the place where the function is not locally a diffeomorphism, that is where the inverse function theorem don't apply, so for higher dimensions you need to calculate the jacobian.

4. Feb 11, 2008

### Mitchtwitchita

Thanks for your help guys. I think I got it now. When x=1, the answer is -1. This is the critical point for the function. I don't what the deal was with my calculator. It could have been the scale. Anyhoo, thanks again.

5. Feb 12, 2008

### mrandersdk

not sure what you meen by:

When x=1, the answer is -1

it is not 'when x=1 or equal to anything then...'

you simply need to find where the derivative is equal to 0 or undefined, and for your function that is in x=-1.

6. Feb 12, 2008

### mathwonk

another argument against calculators. a trivial problem made hard by using and believing a calculator.

7. Feb 12, 2008

### HallsofIvy

Staff Emeritus
Then you don't got it. If "when x= 1, the answer is -1" is in response to my question "What is the value of y= xex when x= 1?" (my point being that if it has a value, x= 1 cannot be an asyptote), then when x= 1, y= xex= 1(e1)= 1. I can't imagine how you would get a negative number for that. And you have already been told that the critical point is NOT at x= 1.

8. Feb 12, 2008

### Mitchtwitchita

I apologize, I see that my answer has caused some clamor. I meant that when I differentiated and solved for x, the answer is -1...which was my initial query. Thanks again for the help guys!