Exercise 34 Page 120

We have been given an inequality. g(x) > |x+1| if x≤ -4 - |x| if - 4 < x < 2 |x-4| if x ≥ 2 We will first graph each case of the inequality separately. Then we will combine them on the same coordinate plane. Let's start with the first case. We will begin by determining it boundary line. To do that we replace the inequality symbol with the equals sign. &Case I &&Boundary Line & g(x) > |x+1| && g(x) = |x+1|The graph of the boundary line is the graph of parent function f(x)=|x| translated 1 unit to the left. Since the inequality is strict, the boundary line will be dashed. Let's graph it!

For the first case, the domain is the x-values less than or equal to -4, so the boundary line will be bound by x=-4 on the right. We will indicate its right endpoint with a closed point because the inequality that represents the restriction is non-strict.

Now we will test a point to decide which region we should shade. Let's test the point (-5,5)!

The point satisfied the inequality, so we will shade the region that contains point.

Applying the same steps, other cases can be graphed.

Case	Boundary Line	Domain	Transformation of f(x)=|x|	Graph	Testing a Point	Inequality
g(x) > |x+1|	g(x) = |x+1|	x ≤ -4	Translation 1 unit to the left		5 ? > | -5+1| ⇓ 5> 4
g(x) > - |x|	g(x) = - |x|	-4 < x < 2	Reflection in the x-axis		2 ? > -| 0| ⇓ 2> 0
g(x) > |x-4|	g(x) = |x-4|	x ≥ 2	Translation 4 units to the right		4 ? > | 4-4| ⇓ 4> 0

Finally, we can combine the each case.