precalculus

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solve the inequality express your solution using interval notation

x^4-7x^3+12x^2<0

  • precalculus -

    y = x^2 * (x^2 - 7x + 12)
    = x^2 (x-3)(x-4)

    x^2 is always positive, so (x-3)(x-4) must be negative. That is true only for

    3 < x < 4 or x in (3,4)

  • precalculus -

    x^2(x^2 - 7x + 12) < 0
    x^2(x-3)(x-4) < 0

    critical values are x = 0, 3, 4
    test for x< 0 , say x = -1
    all we care about is the sign, so
    (+)(-)(-) < 0 false
    test for x between 0 and 3, says x = 1
    (+)(-)(-) < 0 false
    test for x between 3 and 4, say x = 3.5
    (+)(+)(-) < 0 ---> TRUE
    test for x > 4 , say x = 5
    (+)(+)(+) < 0 false

    so the solution is
    3 < x < 4

    the graph shown here, confirms this
    http://www.wolframalpha.com/input/?i=x%5E4-7x%5E3%2B12x%5E2





    http://www.wolframalpha.com/input/?i=x%5E4-7x%5E3%2B12x%5E2

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