2.7 Polynomial and
Rational Inequalies
Sample Inequalities
Polynomial Inequality
ex.
3x  4x  7  0
3
Rational Inequality
ex.
2x  1
0
7x  10
Think of the “related equation” and solve.
Review: Find the zeros of
f(x) = x³ + x² - 6x.
The values of 0, -3, and 2 are considered to be
boundary points.
Boundary points are values of x for which P(x) is
undefined or equal to zero.
Polynomial Inequalities
• A quadratic inequality can be written in the
form ax2 + bx + c > 0, (where the symbol
could be replaced with either <, <, or >.)
• A quadratic inequality is one type of
polynomial inequality.
Examples of Other Types:
3
4
2
2 x  x  2  4
x60
4
8x  2 x  6 x  5
3
2
Given the following graph of f(x), give
interval notation for x-values such that
f(x)>0.
a ) ( 3, 1)  (0, 2)  (4, )
b) ( , 3)  ( 1, 0)  (2, 4)
c) ( , )
d ) ( 3, 4)
Steps for Solving Polynomial
Inequalities
1. Rewrite the inequality so that there is a zero
on the right side.
2. Find the x-intercepts. (Solve the polynomial
equation.)
3. The x-intercepts divide the x-axis into
intervals. Select test values in each interval
and determine the sign of the polynomial on
that interval.
4. The solution will be those intervals in which
the function has the correct signs satisfying
the inequality.
Start up Example – Watch & listen.
Solve: 4x3  7x2  15x.
Example continued 4x3  7x2  15x
• The zeros divide the x-axis into four intervals.
We need to determine the sign of the
polynomial in each interval.
Interval
Test Value
Sign of f(x)
Since we are solving 4x3  7x2  15x  0, the
solution set consists of only two of the four
intervals, those in which the sign of f(x) is negative.
Solve and graph the solution set: x2 – x > 20.
Interval
Test Value
Sign of f(x)
Solve and graph the solution set on a real
number line: x3 + 3x2 < x +3
Interval
Test Value
Sign of f(x)
Rational Inequalities
• What does the word “rational” mean? In
other words, what can you expect in these
problems? Answer: A FRACTION!
• A rational inequality is any inequality that
can be put in the form
f(x) < 0, f(x) > 0, f(x) < 0, or f(x) > 0.
x3
• Solve 2
0 .
x 1
• First, find the domain and
determine if f(x) is not defined for
any values of x.
Example continued
• Next, solve f(x) = 0.
x3
0
2
x 1
Example continued
• The boundary points are 3, 1, and 1.
• These values divide the x-axis into four
intervals. We use a test value to determine
the sign of f(x) in each interval.
Interval
Test Value
Sign of f(x)
(, 3)
f(4) =1/15

(3, 1)
f(2) = 1/3
+
(1, 1)
f(0) = 3
(1, )

f(2) = 5/3
+
Example continued
x3
0
2
x 1
• Function values are positive in the
intervals (3, 1) and (1, ).
• Since neither 1 or 3 is in the domain of
f, they cannot be part of the solution set.
Therefore, the solution set is
(3, 1)  (1, ).
Steps for Solving Rational Inequalities
1.
Rewrite the inequality so that there is a zero on the
right side.
2.
Find the values for which the rational function is
undefined and equal to zero. (Values that make
zero in the denominator and zero in the numerator.)
These are the boundary points
3.
The boundary points divide the x-axis into intervals.
Select test values in each interval and determine the
sign of the polynomial on that interval.
4.
The solution will be those intervals in which the
function has the correct signs satisfying the
inequality. Boundary points that made the function
undefined (denominator = 0)will never be included
in the solution set.
• Solve and graph the solution set:
Interval
Test Value
2x
1
x 1
Sign of f(x)