Posted by Greg on Thursday, October 18, 2012 at 9:21am.
Find the critical points of f(x)=x^4e^x.
You would use the product rule.
Then what?

Calculus  Reiny, Thursday, October 18, 2012 at 9:40am
set the derivative equal to zero and solve for x
plug those x values back into the original equation to find the critical points

Calculus  Greg, Thursday, October 18, 2012 at 10:15am
Would the derivative be e^x (x^44x^3)?

Calculus  Steve, Thursday, October 18, 2012 at 10:25am
almost.
d/dx(e^x) = e^x
so,
df/dx = (4x^3)(e^x) + (x^4)(e^x)
= e^x (4x^3  x^4)
= x^3 e^x (4x)
set that to zero, which occurs at x=4
luckily, in this case, you would still have gotten the correct answer, since the only mistake was the sign of df/dx.

Calculus  Reiny, Thursday, October 18, 2012 at 10:30am
dy/dx= x^4(e^x)(1) + e^x (4x^3)
= e^x( x^4  4x^3)
(looks like you forgot the derivative of the exponent x )
to solve:
e^x = 0 or x^4  4x^3 = 0
e^x = 0 has no solution
x^4 + 4x^3 = 0
x^3(x+4) = 0
x = 0 or x = 4
now sub back in

Calculus  Greg, Thursday, October 18, 2012 at 10:46am
Sub 0 and 4 into f(x). or f'(x)?

Calculus  Greg, Thursday, October 18, 2012 at 10:57am
f(x)=x^4e^x
f(0)=0^4e^0 = 0
f(4)=4^4e^4 = 13977.13
I feel like i didn't do that right. It seems like a weird number for a critcal point.

Calculus  Steve, Thursday, October 18, 2012 at 11:01am
must be because the value for x is 4, not 4

Calculus  Greg, Thursday, October 18, 2012 at 11:05am
f(4)= 4^4 e^4 = 4.6888
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