# absolute functions help

• Sep 19th 2010, 09:55 AM
skatefallen15
absolute functions help
This problem was proposed to me recently and the absolute values are throwing me off.

Given the functions f(x) = |-2x-4| and g(x) = |x+3|, considering appropriate domain restrictions, state the defining equations for the three regions of the function (f+g)(x).

Obviously the solution will be a piecewise function, but I'm having trouble getting the equations for the three regions of the function(f+g)(x). Absolute signs can be tricky!
• Sep 19th 2010, 10:20 AM
skeeter
Quote:

Originally Posted by skatefallen15
This problem was proposed to me recently and the absolute values are throwing me off.

Given the functions f(x) = |-2x-4| and g(x) = |x+3|, considering appropriate domain restrictions, state the defining equations for the three regions of the function (f+g)(x).

Obviously the solution will be a piecewise function, but I'm having trouble getting the equations for the three regions of the function(f+g)(x). Absolute signs can be tricky!

case 1 ...

$\displaystyle (-2x-4) \ge 0$ and $\displaystyle (x+3) \ge 0$

$\displaystyle -2(x+2) \ge 0$ ... $\displaystyle x \ge -3$

$\displaystyle x+2 \le 0$

$\displaystyle x \le -2$

intersection is the interval $\displaystyle [-3,-2]$

case 2 ...

$\displaystyle (-2x-4) \ge 0$ and $\displaystyle (x+3) < 0$

$\displaystyle x \le -2$ ... $\displaystyle x < -3$

intersection is $\displaystyle (-\infty,-3)$

case 3 ...

$\displaystyle (-2x-4) < 0$ and $\displaystyle (x+3) \ge 0$

$\displaystyle -2(x+2) < 0$ ... $\displaystyle x \ge -3$

$\displaystyle x+2 > 0$

$\displaystyle x > -2$

intersection is $\displaystyle (-2,\infty)$

case 4 ...

$\displaystyle (-2x-4) < 0$ and $\displaystyle (x+3) < 0$

$\displaystyle x > -2$ ... $\displaystyle x < -3$

no intersection.

so ... the three regions are $\displaystyle (-\infty,-3)$ , $\displaystyle [-3,-2]$ , and $\displaystyle (-2,\infty)$

I'll leave you to define $\displaystyle (f+g)(x)$ for each region.
• Sep 19th 2010, 10:31 AM
Plato
It is interesting to notice that $\displaystyle |-2x-4|=|2x+4|$.