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calculus!!

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f(x)= x^3/x^2-4

defined on the interval [-18,20]

f(x) is concave up on the region ? to ? and ? to ?

the inflection points are ?, ? and ?

  • calculus!! - ,

    I assume you mean f(x) = x^3/(x^2-4)
    First find the points where f'(x) = 0

    f'(x) = [(x^2-4)(3x^2)-2x^4]/(x^2 -4)^2
    (x^2-4)(3x^2)-2x^4 = 0 when f'(x) = 0
    3x^4 -12 x^2 -2x^4 = 0
    x^2 (x^2-12)
    x = 0, +2sqrt3 and -2sqrt3

    Now evaluate the second derivative at those three critical points. The function is concave up if f''(x) > 0 and is an inflection point if f''(x) = 0

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