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Math Help - I am to graph the function but i cant find where f''(x) equals zero

  1. #1
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    I am to graph the function but i cant find where f''(x) equals zero

    f(x)= x^4 - 4x^3 - 8x^2 + 48x
    f'(x)= 4 (x-2)(x+2)(x-3)
    f''(x)= 4(3x^2 -6x- 4)


    my problem is i am unable to get the concavity.
    what would be my intervals?
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  2. #2
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    Re: I am to graph the function but i cant find where f''(x) equals zero

    Have you tried factoring the 3x^2-6x-4 and then setting those to 0
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  3. #3
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    Re: I am to graph the function but i cant find where f''(x) equals zero

    Hello, Alioyo!

    f(x)\:=\: x^4 - 4x^3 - 8x^2 + 48x
    f'(x)\:=\: 4x^3-12x^2 - 16x + 48 \:=\: 4 (x-2)(x+2)(x-3)
    f''(x)\:=\: 4(3x^2 -6x- 4)

    My problem is i am unable to get the concavity. .Why not?
    What would be my intervals?

    You're taking Calculus and you don't know how to solve a quadratic equation?


    You have: . 3x^2-6x-4 \:=\:0

    Quadratic Formula: . x \:=\:\frac{6 \pm\sqrt{36+48}}{6} \:=\:\frac{6\pm\sqrt{84}}{6} \:=\:\frac{6\pm2\sqrt{21}}{6} \:=\:\frac{3\pm\sqrt{21}}{3}

    The intervals are: . \left(\text{-}\infty,\:\tfrac {3 -\sqrt{21}}{3} \right),\;\left(\tfrac{3-\sqrt{21}}{3},\:\tfrac{3+\sqrt{21}}{3}\right),\; \left(\tfrac{3+\sqrt{21}}{3},\:\infty\right)
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