# Calculus

posted by on .

Find the critical points of f(x)=x^4e^-x.

You would use the product rule.

Then what?

• Calculus - ,

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 - ,

Would the derivative be e^-x (x^4-4x^3)?

• Calculus - ,

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 (4-x)

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 - ,

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 - ,

Sub 0 and -4 into f(x). or f'(x)?

• Calculus - ,

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 - ,

must be because the value for x is 4, not -4

• Calculus - ,

f(4)= 4^4 e^-4 = 4.6888