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Thread: Find zero/Find positive intervals

  1. #1
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    Find zero/Find positive intervals

    I am studying for a test and I cannot figure out how to do these two questions. Any help would be greatly appreciated. Thanks!

    1. Find the intervals on which (x-4)/(x-1) is positive

    2. Find all of x for which (x-x)(2x+1) - (x+x)(2x-1) is zero
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  2. #2
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    Hello, cam_ddrt!

    1. Find the intervals on which $\displaystyle f(x) \:=\:\frac{x^2-4}{x-1}$ is positive.
    We have: .$\displaystyle f(x) \:=\:\frac{(x-2)(x+2)}{x-1}$

    The function is zero or undefined at $\displaystyle x \:=\:-2, 1, 2$
    These are the only values where $\displaystyle f(x)$ changes signs.

    These three values divide the number line into four intervals.

    . . $\displaystyle \underbrace{\;- - - -}\;[\text{-}2]\;\underbrace{ - - - }\;[1]\;\underbrace{ - - - }\;[2]\;\underbrace{ - - -\;}$


    Test a value in each interval.

    . . $\displaystyle f(\text{-}3) \:=\:\frac{(\text{-}5)(\text{-}1)}{\text{-}3-1}\:=\:\frac{5}{\text{-}4}\;\text{ neg.}$

    . . $\displaystyle f(0) \:=\:\frac{0-4}{0-1} \:=\:4\;\text{ pos.}$

    . . $\displaystyle f(\tfrac{3}{2}) \:=\:\frac{(-\frac{1}{2})(\frac{7}{2})}{\frac{1}{2}} \:=\: \text{-}\frac{7}{2}\;\text{ neg.}$

    . . $\displaystyle f(3) \:=\:\frac{(1)(5)}{(2} \:=\:\frac{5}{2}\;\text{ pos.}$


    We have:

    . . $\displaystyle \underbrace{\;- - - -}_{{\color{red}-}}\;[\text{-}2]\;\underbrace{ - - - }_{{\color{red}+}}\;[1]\;\underbrace{ - - - }_{{\color{red}-}}\;[2]\;\underbrace{ - - -\;}_{{\color{red}+}}$


    Therefore, $\displaystyle f(x)$ is positive on: .$\displaystyle (\text{-}2,1) \cup (2,\infty)$




    2. Find all $\displaystyle x$ for which $\displaystyle f(x) \:=\:(x^2-x)(2x+1)-(x^2+x)(2x-1)$ is zero.
    We have: .$\displaystyle f(x) \:=\:0$

    Simplify: .$\displaystyle (x^2-x)(2x+1) - (x^2+x)(2x-1) \:=\:0$

    . . . . . . . . . $\displaystyle (2x^3 - x^2 - x) - (2x^3 + x^2 - x) \:=\:0$

    . . . . . . . . . . . $\displaystyle 2x^3 - x^2 - x - 2x^3 - x^2 + x \:=\:0$

    . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle -2x^2 \:=\:0$

    . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x \:=\:0$

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