Transcript Sullivan College Algebra Section 4.4

Sullivan Algebra and
Trigonometry: Section 4.5
Solving Polynomial and
Rational Inequalities
Objectives
• Solve Polynomial Inequalities
• Solve Rational Inequalities
Steps for Solving Polynomial and
Rational Inequalities
1. Write the inequality so that the polynomial or rational
expression is on the left and zero is on the right.
2. Determine when the expression on the left is equal to
zero or, with a rational expression, is undefined.
3. Use the numbers found in step 2 to separate the real
number line into intervals.
4. Select a number in each interval and determine if the
function on the left is positive (> 0) or negative (< 0).
If the inequality isn’t strict, include the solutions to
f(x) = 0 in the solution set.
Example: Solve the inequality x(x+1) > 20.
x( x  1)  20
x 2  x  20  0
 x  5  x  4   0
So, divide the number line at x = – 5 and at x = 4
For the region   x  5
Choose x = – 6, then – 6(– 6 +1) > 20 is TRUE.
So, this region is included in the solution set.
For the region  5  x  4
Choose x = 0, then 0(0+1) > 20 is FALSE.
So, this region is not included in the solution set.
For the region 4  x  
Choose x = 5, then 5(5+1) > 20 is TRUE.
So, this region is included in the solution set.
Since this is a strict inequality, we do not include the
endpoints in the solution set.
The solution set is x | x  5  x | x  4
Example: Solve the inequality
2x  3
x4
1
2x  3
1 0
x  4
2x  3
x4

 0
x4
x4
2x  3  x  4
x4
0
x7
0
x4
So, divide the number line at x = – 7 and at x = 4.
Note that 4 is NOT in the solution set.
For the region    x   7
Choose x = – 8, then 2(–8) + 3 > 1
(–8) – 4
So, this region is included in the solution set.
For the region  7  x  4
Choose x = 0, then 2(0) + 3 > 1
(0) – 4
So, this region is not included in the solution set.
For the region 4  x  
Choose x = 5, then 2(5) + 3 > 1
(5) – 4
So, this region is included in the solution set.
Since this is not a strict inequality but involves an
equal sign, we do include the endpoint x = – 7 in the
solution set, since the left hand side is zero at x = – 7.
The other endpoint, x = 4, cannot be included since it
makes the denominator equal to zero.
The solution set is
x | x  7 x | x  4