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  1. #1
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    Math Problem Help

    * Solve for x, xER


    A) (x+1) (x-2) (x-4)^2 >
    0


    B) -x^3+25<0


    Thanks
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  2. #2
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    Quote Originally Posted by agent2421 View Post
    * Solve for x, xER


    A) (x+1) (x-2) (x-4)^2 > 0


    B) -x^3+25<0


    Thanks
    A) Answer this question by first drawing a graph of y = (x+1) (x-2) (x-4)^2 (the x-intercepts are all you'll need to show on the graph).

    B) -x^3 < -25 => x^3 > 25 and the last step should be obvious.
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  3. #3
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    A) Here is solution for - Wolfram|Alpha

    1. Solve (x+1)(x-2)(x-4)^2 = 0

    As you can see we have 3 roots here: x_1=-1, x_2=2, x_3=4. It means function y=(x+1)(x-2)(x-4)^2 changes it's sign in this points.

    2. Find sign of y in following intervals by selecting random value and evaluating y: (-inf, -1), (-1, 2), (2, 4) and (4,+inf).

    3. Select intervals where y > 0
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  4. #4
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    Helolo, agent2421!

    A) Solve for x\!:\;\;(x+1) (x-2) (x-4)^2 \:\geq\:0
    Note that (x-4)^2 is never negative.

    So we are concerned with (x+1) and (x-2) only.

    The product is greater than or equal to zero in two cases:
    . . (1) Both factors are positive
    . . (2) Both factors are negative


    Case 1: . \begin{array}{ccccccc}x+1 & \geq & 0 & \Rightarrow & x & \geq & \text{-}1 \\ x-2 &\geq & 0 & \Rightarrow & x & \geq & 2 \end{array} \quad\Rightarrow\quad x \:\geq\:2

    Case 2: . \begin{array}{ccccccc}x + 1 & \leq & 0 & \Rightarrow & x & \leq & \text{-}1 \\ x-2 & \leq & 0 & \Rightarrow & x & \leq & 2 \end{array}\quad\Rightarrow\quad x \:\leq \:\text{-}1


    Therefore: . (x \:\leq\: \text{-}1) \:\vee \:(x \:\geq\: 2)\quad \hdots \text{or: }\;(-\infty,\:\text{-}1] \:\cup\:[2,\,\infty) <br />

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  5. #5
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    so my final answer would be... (x< 1) (x>2) ?

    Like if I put that on my answer sheet, it is the correct one?
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  6. #6
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    I think you did typo.

    The answer is x<=-1 and x>=2
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