# solving critical point problems??

• Oct 27th 2008, 04:52 PM
ab32
solving 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 PM
Krizalid
Critical points are found by makin' $f'(x)=0,$ do it.
• Oct 27th 2008, 05:15 PM
ab32
i dont know how f! when theres a e????
• Oct 28th 2008, 12:02 PM
ab32
can any one else tell me how to slove this? Please
• Oct 28th 2008, 01:37 PM
Cakecake
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.

$x^2e^{ax}$

$2xe^{ax} + ae^{ax}x^2$ 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 PM
11rdc11
Quote:

Originally Posted by ab32
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??

$f(x) = x^2e^{ax}$

Use product rule

$f'(x)= 2xe^{ax} + x^2ae^{ax}$

The derivative of e is

$f(x)=e^x$

$f'(x)=e^xdx$
• Oct 28th 2008, 02:46 PM
ab32
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 PM
11rdc11
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

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

Now just solve for a