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calculus-derivatives

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Hello, please help me. I needed to find the derivative of f(x)= x^4 -2x^3 + x +1. I got f'(x)= 4x^3 -6x^2 +1. I would like to figure out the critical numbers so I can figure out where it increases and decreases, but I have no idea how to factor it... Is it even possible?

Thank you for your time.

  • calculus-derivatives - ,

    tried x = ± 1 , no good
    tried ± 1/2 and ahhh! x = 1/2 is a root
    so (2x-1) is a factor
    long division:
    4x^3 -6x^2 +1 = (2x-1)((2x^2 - 2x-1)

    2x^2 - 2x-1 = 0 gave me
    x = (1 ± √3)/2

    so there are 3 critical points to worry about.
    take it from here.

  • calculus-derivatives - ,

    this factors into

    2(2x-1)(x^2 - 2x - 1)
    x = 1/2 or (1±√3)/2

    Having that, you can find where f'<0 or f'>0

  • calculus-derivatives - ,

    Oooooh! Thank you two so much :) I'm more confident in regards to my test now, thank you

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