Originally Posted by
earboth Use the definition of the absolute value:
$\displaystyle |x|=\left\{ \begin{array}{l}x,\ if \ x \geq 0 \\ -x,\ if \ x<0 \end{array} \right.$
With your example you have:
$\displaystyle |x+4|=\left\{ \begin{array}{l}x+4,\ if \ x \geq -4 \\ -(x+4),\ if \ x<-4 \end{array} \right.$
$\displaystyle |x-1|=\left\{ \begin{array}{l}x-1,\ if \ x \geq 1 \\ -(x-1),\ if \ x<1 \end{array} \right.$
You now have 3 intervals with different inequalities:
$\displaystyle \begin{array}{ll}x\geq 1 :& x+4+x-1>5 \\ -4\leq x<1 :& x+4+(-(x-1)) >5 \\ x<-4: &-(x+4)+(-(x-1))>5 \end{array}$
Solve for x.
Draw the graphs of
$\displaystyle f(x)=|x+4|+|x-1|$ and y = 5
Determine those points on the graph of f where the y-coordinate is greater than 5.