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Math Help - Range of Composite Functions

  1. #1
    Senior Member Stroodle's Avatar
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    Range of Composite Functions

    Hi.

    If:
    f:R\rightarrow R,\;f(x)=x^2-4
    and
    g:R+\cup\left\{0\right\}\rightarrow R,\;g(x)=\sqrt x

    Then what is the range of f\circ g and why?

    And what's the latex command for the little + symbol that normally goes next to R in the domain?

    Thanks for your help.
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  2. #2
    Senior Member pankaj's Avatar
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     <br />
fog=(g(x))^2-4=(\sqrt x)^2-4\geq 0-4\geq -4<br />

    Thus range =(-4,\infty]
    Last edited by pankaj; June 17th 2009 at 09:01 AM.
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  3. #3
    Senior Member Stroodle's Avatar
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    Thanks heaps for your reply, but I don't get that, and my text says the answer is [-4,\infty)
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  4. #4
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    Quote Originally Posted by Stroodle View Post
    Hi.

    If:
    f:R\rightarrow R,\;f(x)=x^2-4
    and
    g:R+\cup\left\{0\right\}\rightarrow R,\;g(x)=\sqrt x

    Then what is the range of f\circ g and why?
    g(x) maps the non-negative real numbers to the set of non-negative real numbers: given any y\ge 0, take x= y^2 so y= g(x).
    Now apply f(x) to such a number. If x\ge 0, then x^2\ge 0 so x^2- 4\ge -4. The range of f\circ g(x) is all x\ge -4.

    And what's the latex command for the little + symbol that normally goes next to R in the domain?
    R^+ is the set of positive real numbers. Adding 0 to that set gives the set of "non-negative" real numbers. You need that for the domain of g because you cannot take the square root of a negative number.

    Thanks for your help.
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  5. #5
    Senior Member Stroodle's Avatar
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    Awesome. Thankyou
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  6. #6
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Stroodle View Post
    Thanks heaps for your reply, but I don't get that, and my text says the answer is [-4,\infty)
    Yor text is right.I have edited my post.
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