3.6 Polynomial and Rational Inequalities

Introduce infant mortality module.

Definition of a Polynomial Inequality

A polynomial inequality is any inequality that can be put in one of the forms

anxn an-1x

n-1 … a2x2 a1x a0 < 0 anx

n an-1x

n-1 … a2x2 a1x a0 > 0

anxn an-1x

n-1 … a2x2 a1x a0 < 0 anx

n an-1x

n-1 … a2x2 a1x a0 > 0

where the coefficients are real numbers and the degree is 2 or higher.

Procedure for Solving Polynomial Inequalities• Express the inequality in the standard form (leading coefficient positive

set against zero.)

ex: anxn an-1x

n-1 … a2x2 a1x a0 < 0

• Solve the equation anxn an-1x

n-1 … a2x2 a1x a0=0. The real

solutions are the boundary points.• Locate these boundary points on a number line, thereby dividing the

number line into test intervals.• Choose one representative number within each test interval. If substituting

that value into the original inequality produces a true statement, then all real numbers in the test interval belong to the solution set. If substituting that value into the original inequality produces a false statement, then no real numbers in the test interval belong to the solution set.

• Write the solution set; the interval(s) that produced a true statement.

Solve and graph the solution set on a real number line: 2x2 – 3x > 2. (Don’t look at notes, no need to write.)

Example

Solution

Step 1 Write the inequality in standard form. We can write by subtracting 2 from both sides to get zero on the right.

2x2 – 3x – 2 > 2 – 2

Step 2 Solve the related quadratic (polynomial) equation. Replace the inequality sign with an equal sign. Thus, we will solve.

2x2 – 3x – 2 = 0 This is the related quadratic equation.

(2x + 1)(x – 2) = 0 Factor.

2x + 1 = 0 or x – 2 = 0 Set each factor equal to 0.

x = -1/2 or x = 2 Solve for x.

The boundary points are –1/2 and 2.

So far, we have taken similar steps to solve quadratic (polynomial) EQUATIONS.

2x2 – 3x – 2 > 0

Solve and graph the solution set on a real number line: 2x2 – 3x > 2.

Solution

Step 3 Locate the boundary points on a number line. The number line with the boundary points is shown as follows:

The boundary points divide the number line into ________ test intervals. Including the boundary points (because of the given greater than or equal to sign), the intervals are:

-1/2

-5 -4 -3 -2 -1 0 1 2 3 4 5

2

x

Example cont.

Solve and graph the solution set on a real number line: 2x2 – 3x > 2.

Solution

Step 4 Take one representative number within each test interval and substitute that number into the original inequality (belongs or not to the solution set?).

[2, ºº) _________

to the solution set.

[2, ºº)

[-1/2, 2] ________

to the solution set.

[-1/2, 2]

(-ºº, -1/2] ______

to the solution set.

(-ºº, -1/2]

Conclusion

Substitute into

2x2 – 3x > 2 Representative NumberTest Interval

Example cont.

Solve and graph the solution set on a real number line: 2x2 – 3x > 2.

Solution

Step 5 The solution set are the intervals that produced a true statement. Our analysis shows that the solution set is

(-ºº, -1/2] or [2, ºº).

The graph of the solution set on a number line is shown as follows:

-1/2

-5 -4 -3 -2 -1 0 1 2 3 4 5

2

) ( x

Example cont.

Find the graphing error(s).

When you get tired of this method, let me know, there is a more conceptual method that is much quicker (using “signs”.)

Ex: Graph the solutions of 3 24 8 2x x x

Ex:Do p 378 # 44.

Ex: Finally find the equation of a graph that has no solutions. (Use your calculator to check your results.)

Text Example(Don’t look at notes, not necessary to write. If

struggling, do p 378#48 first.)

Solution

Step 1 Express the inequality so that one side is zero and the other side is a single quotient. We subtract 2 from both sides to obtain zero on the right.

Solve and graph the solution set:

x 1

x 32 This is the given inequality.

x 1

x 3 2 0 Subtract 2 from both sides, obtaining 0 on the right.

x 1

x 3

2(x 3)

x 30 The least common denominator is x + 3. Express 2

in terms of this denominator.

x 1 2(x 3)

x 30 Subtract rational expressions.

x 1 2x 6

x 30 Apply the distributive property.

x 5

x 30 Simplify.

x 1

x 32

IMPORTANT: We MUST set = 0 first!

Solution

Step 2 Find boundary points by setting the numerator and the denominator equal to zero.

Solve and graph the solution set:

The boundary points are -5 and -3. Because equality is included in the given less-than-or-equal-to symbol, we include the value of x that causes the quotient to be zero. Thus, -5 is included in the solution set. By contrast, we do not include 1 in the solution set because -3 makes the denominator ________.

41xx

Set the numerator and denominator equal to 0. These are the values that make the previous quotient zero or undefined.

0=x + 30=-x - 5

Solve for x.-3=xx = -5

Text Example cont.x 1

x 32

-x-5 < 0

x + 3

Step 4 Take one representative number within each test interval and substitute that number into the original equality. (Fill in your notes.)

(-3, ºº) ___________

belong to the solution set.

(-3, ºº)

[-5,-3) _____________

belong to the solution set.[-5,-3)

(-ºº, -5] ____________

belong to the solution set.(-ºº, -5]

ConclusionSubstitute into Representative Number

Test Interval

Text Example cont. Solve and graph the solution set: x 1

x 32

x 1

x 32

Step 3 Locate boundary points on a number line.

(In this case, -5 and –3. How many testing intervals now? Name them.)

Text Example cont.Solve and graph the solution set: x 1

x 32

Step 5 The solution set are the intervals that produced a true

statement. Our analysis shows that the solution set is (-ºº, -5]

or (-3, ºº)

The Position Formula for a Free-Falling Object Near Earth’s Surface

An object that is falling or vertically projected into the air has its height in feet above the ground given by

s = -16 t 2 + v0 t + s0

where s is the height in feet, v0 is the original velocity (initial velocity) of the object in feet per second, t is the time that the object is in motion in seconds, and s0 is the original height (initial height) of the object in feet.

Q: What shape would this have if we were to graph it? What variables would go on each axis? What does this mean physically?

Examples = -16 t 2 + v0 t + s0

Solution

An object is propelled straight up from ground level with an initial velocity of 80 fps. Its height at time t is described by __________________ where the height, s, is measured in feet and the time, t, is measured in seconds. In which time interval will the object be more than 64 feet above the ground? (What is s0 ?) How does this graph look? Put it in your calculator to check your guess.

Q: This is the inequality implied by the problem’s question. We must find t.

Now what? Work in pairs to solve this step by step. Be able to verbalize it.

.

t = Solve for t. The boundary points are ___ and ___.

Find the intervals.

Test. Draw conclusions to answer this real world problem.

t =