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March 26, 2017

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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

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