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January 30, 2015

January 30, 2015

Posted by **Greg** on Thursday, October 18, 2012 at 9:21am.

You would use the product rule.

Then what?

- Calculus -
**Reiny**, Thursday, October 18, 2012 at 9:40amset 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 10:15amWould the derivative be e^-x (x^4-4x^3)?

- Calculus -
**Steve**, Thursday, October 18, 2012 at 10:25amalmost.

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 -
**Reiny**, Thursday, October 18, 2012 at 10:30amdy/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 10:46amSub 0 and -4 into f(x). or f'(x)?

- Calculus -
**Greg**, Thursday, October 18, 2012 at 10:57amf(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.

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

- Calculus -
**Greg**, Thursday, October 18, 2012 at 11:05amf(4)= 4^4 e^-4 = 4.6888

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