# Thread: solving critical point problems??

1. ## 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??

2. Critical points are found by makin' $\displaystyle f'(x)=0,$ do it.

3. i dont know how f! when theres a e????

4. can any one else tell me how to slove this? Please

5. 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.

6. 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??

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

Use product rule

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

The derivative of e is

$\displaystyle f(x)=e^x$

$\displaystyle f'(x)=e^xdx$

7. thanks for your help i understand how to get the dervi , now how do i use that -1 critical point?

8. 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