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Thread: f(x) and g(x)

  1. #1
    Member Mr Rayon's Avatar
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    f(x) and g(x)

    Let $\displaystyle f(x)$ and $\displaystyle g(x)$ be defined by:

    $\displaystyle f(x)=x+3$

    $\displaystyle g(x)=\left\{\begin{matrix}
    2-x &x< 0 \\x^2-2x-1
    & x\geq 0
    \end{matrix}\right.$

    (a) Evaluate $\displaystyle (f+g)(-2)$, $\displaystyle (f-g)(-1)$, $\displaystyle (fg)(1)$ and $\displaystyle (f/g)(2)$.
    (b) Find all $\displaystyle x$ for which $\displaystyle f(x)=g(x)$. You must give the exact answer and show all working.

    Any help will be appreciated!
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  2. #2
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    Re: f(x) and g(x)

    Hello, Mr Rayon!

    $\displaystyle \text{Let }f(x)\text{ and }g(x)\text{ e defined by:}$

    . . $\displaystyle f(x)\:=\:x+3$

    . . $\displaystyle g(x) \:=\:\left \begin{Bmatrix}2-x && x< 0 \\ x^2-2x-1 && x\geq 0\end{array}$

    $\displaystyle \text{(a) Evaluate: }\:(f+g)(\text{-}2),\;(f-g)(\text{-}1),\;(fg)(1),\; \left(\trac{f}{g}\right)(2)$

    $\displaystyle \text{For }x < 0:\;f+g \:=\:(x+3) + (2-x) \:=\:5$

    . . $\displaystyle \text{Therefore: }\:(f+g)(\text{-}2) \:=\:5$


    $\displaystyle \text{For }x<0:\;f-g \:=\:(x+3) - (2-x) \:=\:2x+3$

    . . $\displaystyle \text{Therefore: }\:(f-g)(\text{-}1) \:=\:2(\text{-}1) + 3 \:=\:1$


    $\displaystyle \text{For }x > 0:\:fg \:=\:(x+3)(x^2-2x-1)$

    . . $\displaystyle \text{Therefore: }\:(fg)(1) \:=\:(1+3)(1^2-2[1] - 1) \:=\;(4)(-2) \;=\;-8$


    $\displaystyle \text{For }x > 0:\:\frac{f}{g} \:=\:\frac{x+3}{x^2-2x-1}$

    . . $\displaystyle \text{Therefore: }\:\left(\frac{f}{g}\right)(2) \;=\;\frac{2+3}{2^2-2(2) - 1} \;=\;\frac{5}{-1} \;=\;-5$




    $\displaystyle \text{(b) Find all }x\text{ for which }f(x)\,=\,g(x)$

    $\displaystyle \text{If }x < 0:\;x+3 \:=\<2-x \quad\Rightarrow\quad 2x \,=\, \text{-}1 \quad\Rightarrow\quad \boxed{x \,=\,\text{-}\tfrac{1}{2}}$

    $\displaystyle \text{If }x > 0: \; x+3 \:=\:x^2-2x-1 \quad\Rightarrow\quad x^2 - 3x - 4 \:=\:0$

    . . . . $\displaystyle (x-4)(x+1) \:=\:0 \quad\Rightarrow\quad \boxed{x = 4}\;\;{\color{red}\rlap{/////}}x = \text{-}1$
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