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Thread: Find the domain and range

  1. #1
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    Find the domain and range

    g(x) = sqrt.(25 - x^2)
    r(x) = |2x - 3|

    plz help me thanks!
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  2. #2
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    Hello, NeedHelp18!

    Find the domain and range.

    $\displaystyle 1)\;\;g(x) \:= \:\sqrt{25 - x^2}$

    We can see that $\displaystyle (25-x^2)$ must not be negative.

    We have: .$\displaystyle 25-x^2 \:\geq \:0$

    Then we have: .$\displaystyle -x^2 \:\geq \:-25$

    Multiply by -1: .$\displaystyle x^2 \:\leq \:25$

    And we have: .$\displaystyle |x| \:\leq \:5$

    . . $\displaystyle \boxed{\text{Domain: }\;-5 \:\leq\:x\:\leq\:5}$


    We see that $\displaystyle g(x)$ is never negative: .$\displaystyle g(x) \:\geq\:0$


    When $\displaystyle x=0,\;g(0) \:=\:\sqrt{25} \:=\:5$ (maximum value)

    . . $\displaystyle \boxed{\text{Range: }\;0 \:\leq \:g(x) \:\leq \:5}$




    $\displaystyle 2)\;\;r(x) \:= \:|2x - 3|$

    We can use any value for $\displaystyle x.$

    . . $\displaystyle \boxed{\text{Domain: }\:(-\infty,\:\infty)}$


    An absolute value is never negative: .$\displaystyle r(x) \:\geq\:0$

    As $\displaystyle x \to \pm\infty,\;r(x) \to \infty$

    . . $\displaystyle \boxed{\text{Range: }\;r(x) \:\geq 0}$

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