Posted by Greg on Thursday, October 18, 2012 at 9:21am.
set the derivative equal to zero and solve for x
plug those x values back into the original equation to find the critical points
Would the derivative be e^-x (x^4-4x^3)?
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.
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
Sub 0 and -4 into f(x). or f'(x)?
f(x)=x^4e^-x
f(0)=0^4e^0 = 0
f(-4)=-4^4e^4 = 13977.13
I feel like i didnt do that right. It seems like a weird number for a critcal point.
must be because the value for x is 4, not -4
f(4)= 4^4 e^-4 = 4.6888
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