Find a valueof"a" such that the function

f(x) = x^2e^ax has a critical point at x = -1.

I have no clue how to do this??

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- Oct 27th 2008, 04:52 PMab32solving critical point problems??
**Find a value****of**"**a" such that the function**

**f(x) = x^2e^ax has a critical point at x = -1.**

**I have no clue how to do this??**

- Oct 27th 2008, 05:10 PMKrizalid
Critical points are found by makin' $\displaystyle f'(x)=0,$ do it.

- Oct 27th 2008, 05:15 PMab32
i dont know how f! when theres a e????

- Oct 28th 2008, 12:02 PMab32
can any one else tell me how to slove this? Please

- Oct 28th 2008, 01:37 PMCakecake
First you find the derivative of it, by using the product rule. If you don't know how to do that then you should start paying attention in class.

$\displaystyle x^2e^{ax}$

$\displaystyle 2xe^{ax} + ae^{ax}x^2$ I'm almost positive this is right, the derivative of the exponent of e may be wrong. So if it is, can someone kindly point that out.

You can carry on from here. Clean it up a bit to make life easier, then set to equal zero and substitute x then solve for a. - Oct 28th 2008, 02:10 PM11rdc11
- Oct 28th 2008, 02:46 PMab32
thanks for your help i understand how to get the dervi , now how do i use that -1 critical point?

- Oct 28th 2008, 03:27 PM11rdc11
You know you have a critical point when the derivative equals 0. The problem told you the critical point was x = -1 so plug all this into your equation

$\displaystyle 0= 2(-1)e^{a(-1)} + (-1)^2ae^{a(-1)}$

Now just solve for a