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Math Help - Concavity

  1. #1
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    Concavity

    The graph of f(x)= integral from 0 to x (15(t^2) - 2(t^3) + 24)dt is concave up on (a,b). Find b-a.

    Okay so I know from the curve sketching lesson in AB calculus that to find concavity you take the second derivative and then set that to 0. Do I take the integral and then take the first derivative, and then the second?
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  2. #2
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    Hello, abclarinetuvwxyz!

    The graph of: . f(x)\;= \;\int^x_0(15t^2 - 2t^3 + 24)\,dt . is concave up on (a,b).]
    Find b-a

    The first derivative is: . f'(x) \:=\:15x^2 - 2x^3 + 24

    The second derivative is: . f''(x) \:=\:30x - 6x^2

    Now where is the graph concave up?
    . . Where f''(x) is positive.

    So we have: . 30x - 6x^2 \:>\:0\quad\Rightarrow\quad 6x(5 - x) \:>\:0

    This is true when: .  0 \,<\,x\,<\,5

    Therefore, the interval is: . (0,5) \quad\Rightarrow\quad b - a \:=\:5-0 \:=\:\boxed{5}

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  4. #4
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    Thanks very much!
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