f(x) and g(x)

• Apr 11th 2013, 04:40 PM
Mr Rayon
f(x) and g(x)
Let $f(x)$ and $g(x)$ be defined by:

$f(x)=x+3$

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

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

Any help will be appreciated!
• Apr 11th 2013, 07:10 PM
Soroban
Re: f(x) and g(x)
Hello, Mr Rayon!

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

Quote:

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

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

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

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