Posted by Greg on .
Find the critical points of f(x)=x^4e^x.
You would use the product rule.
Then what?

Calculus 
Reiny,
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,
Would the derivative be e^x (x^44x^3)?

Calculus 
Steve,
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,
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,
Sub 0 and 4 into f(x). or f'(x)?

Calculus 
Greg,
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,
must be because the value for x is 4, not 4

Calculus 
Greg,
f(4)= 4^4 e^4 = 4.6888