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Math Help - Function composition problem

  1. #1
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    Function composition problem

    I have two functions f,g:\Re \rightarrow \Re
    f(x)=\left\{\begin{array}{cc}1 - x^2,&\mbox{ if }<br />
x\geq 0\\5x + 1, & \mbox{ if } x<0\end{array}\right.
    g(x)=\left\{\begin{array}{cc}x^2,&\mbox{ if }<br />
x\leq -2\\2x - 1, & \mbox{ if } x>-2\end{array}\right.

    and I have to find function h(x)=f \circ g
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  2. #2
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    Hello, cristian!

    I have two functions f,g:\Re \rightarrow \Re

    f(x)=\left\{\begin{array}{cc}5x+1 & \text{ if }<br />
x < 0 \\ 1-x^2 & \text{ if } x\ge 0\end{array}\right.

    g(x)=\left\{\begin{array}{cc}x^2 & \text{ if }<br />
x\leq -2 \\ 2x - 1 & \text{ if } x>-2\end{array}\right.

    and I have to find: . h(x)\:=\: f\circ g

    We are concerned with three intervals: . (\text{-}\infty, \text{-}2),\;(\text{-}2,0),\;(0,\infty)


    \text{On }(\text{-}\infty,\text{-}2): \;\begin{Bmatrix}f(x) &=& 5x-1 \\ g(x) &=& x^2\end{Bmatrix}

    . . Hence: . f\circ g \;=\;5(x^2)-1 \;=\;5x^2-1


    \text{On }(\text{-}2,0):\; \begin{Bmatrix}f(x) &=& 5x+1 \\ g(x) &=& 2x-1 \end{Bmatrix}

    . . Hence: . f\circ g \;=\;5(2x-1) + 1 \;=\;10x - 4


    \text{On }(0,\infty):\;\begin{Bmatrix}f(x) &=& 1-x^2 \\ g(x) &=& 2x-1 \end{Bmatrix}

    . . Hence: . f\circ g \;=\;1 - (2x-1)^2 \;=\;4x - 4x^2


    Therefore: . \left\{ \begin{array}{cc}<br />
5x^2 + 1 & \text{ if }x \leq \text{-}2 \\ 10x-4 & \text{ if }\text{-}2 < x < 0 \\<br />
4x - 4x^2 & \text{ if }x \ge 0 \end{array}\right.

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  3. #3
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    I tried that too, but the book says the correct answer is:
    f(x) = \left\{ \begin{array}{cc}1 - x^4 & \text{ if }x \leq \text{-}2 \\ 4x(1 - x) & \text{ if } x \geq \frac{1}{2} \\2(5x - 2) & \text{ if } -2<x<\frac{1}{2} \end{array}\right.
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  4. #4
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    Quote Originally Posted by cristian View Post
    I tried that too, but the book says the correct answer is:
    f(x) = \left\{ \begin{array}{cc}1 - x^4 & \text{ if }x \leq \text{-}2 \\ 4x(1 - x) & \text{ if } x \geq \frac{1}{2} \\2(5x - 2) & \text{ if } -2<x<\frac{1}{2} \end{array}\right.
    The textbbok is correct.
    Notice that g(x)<0 on \left(-2,\frac{1}{2}\right).
    Otherwise it is not negative.
    So use the corresponding definitions of f on those three intervals.
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  5. #5
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    Thank you.
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