The textbook asks questions on domain and range that require you to be able to solve some inequalities eventhough the topic of inequalities doesn't come up until later in the course.  At that time we will go into more depth, but for now I want to introduce just enough of the topic to be able to solve the types of domain problems that are the most common.

Type 1: For some problems, all you need to do is solve an equation such as: (exprss1)¹0, where (exprss1) stands for any algebraic expression. This is the same as solving (exprss1)=0, except your answer will be everything except the answer you get to the second equation.
Example 1: Solve 5x2 + 6x + 1 ¹ 0 and write your answer in interval notation.
Note: This would be used to solve a problem like, "Find the domain of ", since the bottom can't be zero.
(5x + 1)(x + 1) ¹ 0
5x = 1 ¹ 0 and x = 1 ¹ 0
x ¹ -1/5 and x ¹ -1
Solution: - {-1/5, -1} = (-¥, -1)È(-1, -1/5)È(-1/5, ¥)
Type 2: For other problems you will need to solve things such as express ³ 0 or express > 0.
Example 2: Solve 5x + 3 ³ 0
Note: This example would be used to solve a problem such as: Let . Find the domain of g.
5x ³ -3
x ³ -3/5   so the solution is: [-3/5, ¥

Example 3: Solve -4x + 12 > 0
Note: Example 3 would be used to solve a problem such as: Find the domain of .
-4x > -12
-4x/(-4) < -12/(-4)  Recall from Intermediate or High School Algebra that whenever you multiply or divide by a negative you must flip the inequality.)
x < 3   so the solution is (-¥, 3).

Example 4: 6x2 + 5x - 1 ³ 0.  First solve 6x2 + 5x - 1 ³ = 0
Factor and get: (6x - 1)(x + 1) = 0
Set each factor equal to zero: 6x - 1 = 0 x + 1 = 0
Solve: x = 1/6 x = -1
You can figure out which of the intervals: (-¥, -1] and/or [1, 1/6] and/or [1/6, ¥) result in the desired greater than zero graphically. Graph y = 6x2 + 5x - 1.

So as you can see from the graph the solution is (-¥, -1]U[1/6, ¥)
Type 3: The messiest example that you might want to solve at this point in the course is one that combines both of the above types of problems.
Example 5: Solve x + 3 ³ and x2 - 4 ¹ 0
This example would be used to solve a problem such as . Find the domain of g.
x ³ -3 and x2 ¹ 4, so x ¹ ±2
Thus the solution is [-3, ¥) - {-2, 2} = [-3, 2)U(2, ¥)
If you run across problems more complicated than these at this point in the course, you can just solve them graphically.