C. Polynomial and Rational Functions · Polynomial Functions and Equations

Polynomial Inequalities: Reading Solutions from Graphs

C4.1 C4.2 C4.3About 70 minutes

Learning goals

I can explain the difference between the solution of a polynomial equation and the solution of the related inequality.
I can write solutions using inequalities, interval notation, set notation and number lines.
I can solve linear inequalities, reversing the sign when multiplying or dividing by a negative number.
I can solve polynomial inequalities from a graph, including inequalities that compare two functions.

Words to know

inequality
A statement using <, >, ≤ or ≥. Its solution is usually a whole set of numbers, not just a few.

interval notation
A way to write a set of numbers: (a, b) means a < x < b; [a, b] means a ≤ x ≤ b. A round bracket excludes the endpoint and a square bracket includes it. ∞ always gets a round bracket.

union (∪)
The symbol joining two intervals: (−2, 1) ∪ (3, ∞) means x is in either interval.

set notation
A way to write a set using braces, such as {x ∈ ℝ | x > 3}: "all real x such that x > 3".

From points to intervals

An equation such as (x + 2)(x − 1)(x − 3) = 0 asks: where exactly is the graph on the x-axis? The answer is a few numbers: x = −2, 1, 3.

An inequality such as (x + 2)(x − 1)(x − 3) > 0 asks: where is the graph above the x-axis? The answer is whole stretches of the number line: −2 < x < 1 or x > 3.

So the solutions of the equation are the boundary points of the solutions of the inequality. Whether those boundary points are included depends on the inequality symbol: they are included for ≤ and ≥, and excluded for < and >.

f(x) = (x + 2)(x − 1)(x − 3) is above the x-axis (shaded strips) for −2 < x < 1 and for x > 3.

Three ways to write the answer

For the solution of f(x) > 0 above:

Inequalities: −2 < x < 1 or x > 3
Interval notation: (−2, 1) ∪ (3, ∞)
Set notation: {x ∈ ℝ | −2 < x < 1 or x > 3}

And on a number line: solid (filled-in) dots for included endpoints, open dots for excluded ones (drawn here as a ring with a lightly shaded centre, never filled solid), and a thick line or arrow for the intervals.

Equation versus inequalities for f(x) = (x + 2)(x − 1)(x − 3). The equation has three solutions (points). The inequalities have intervals as solutions. Solid (filled-in) dots are included. Open dots, drawn as a ring with a lightly shaded centre, are not included: for f(x) > 0 the values x = −2, 1 and 3 make f(x) = 0, which is not greater than 0.

Inequality	Interval notation	In words
−2 < x < 1	(−2, 1)	between −2 and 1, endpoints excluded
1 ≤ x ≤ 3	[1, 3]	between 1 and 3, endpoints included
x > 3	(3, ∞)	greater than 3
x ≤ −2	(−∞, −2]	−2 or less
all real numbers	(−∞, ∞)	every x

Translating between inequalities and interval notation.

Linear inequalities: a quick review

Solve a linear inequality like a linear equation, with one new rule: when you multiply or divide both sides by a negative number, reverse the inequality sign. (Example: 2 < 5, but −2 > −5.)

Worked example

Example 1: Linear inequalities

Solve (a) 3 − 2x ≤ 7 (b) −1 < 2x + 3 ≤ 9. Write each answer in interval notation.

(a) Subtract 3: −2x ≤ 4. Divide by −2 and reverse the sign: x ≥ −2.
(b) Subtract 3 from all three parts: −4 < 2x ≤ 6. Divide by 2 (positive, so no reversal): −2 < x ≤ 3.

(a) x ≥ −2, [−2, ∞) (b) −2 < x ≤ 3, (−2, 3]

Worked example

Example 2: Equation versus inequality

For f(x) = (x + 2)(x − 1)(x − 3), use the graph to solve (a) f(x) = 0 (b) f(x) > 0 (c) f(x) ≤ 0.

(a) The x-intercepts: x = −2, 1, 3.
(b) Above the x-axis, not including the intercepts: (−2, 1) ∪ (3, ∞).
(c) On or below the x-axis, including the intercepts: (−∞, −2] ∪ [1, 3].

The equation has 3 solutions; each inequality has infinitely many.

Worked example

Example 3: A double root

Solve g(x) ≥ 0 and g(x) > 0 for g(x) = (x + 1)²(x − 2).

Zeros: −1 (order 2, touches) and 2 (crosses). Degree 3 with a > 0: Q3 to Q1.
Signs: negative for x < −1, still negative for −1 < x < 2 (no change at the double zero), positive for x > 2.
g(x) ≥ 0 includes every zero: x = −1 and x ≥ 2.
g(x) > 0 excludes the zeros: x > 2 only.

g(x) ≥ 0: {−1} ∪ [2, ∞). g(x) > 0: (2, ∞). The single point x = −1 is easy to miss.

Worked example

Example 4: Comparing two functions

Solve x³ ≥ 4x graphically.

Graph y = x³ and y = 4x. They intersect where x³ = 4x, that is x(x − 2)(x + 2) = 0: at x = −2, 0, 2.
Read where the curve y = x³ is on or above the line y = 4x: between −2 and 0, and from 2 onward.
(Equivalently, graph y = x³ − 4x and find where it is ≥ 0.)

[−2, 0] ∪ [2, ∞)

y = x³ (solid) is on or above y = 4x (dashed) for −2 ≤ x ≤ 0 and x ≥ 2 (shaded).

With graphing technology

Graph the polynomial (or the difference of the two sides) and use the calculator's zero or intersection feature to find the boundary points. Then read which intervals lie above or below. When a boundary is irrational (such as 2.09 for x³ − 2x − 5), round it sensibly and say so: for example, x³ − 2x − 5 > 0 for x > 2.09 (approximately).

Just for fun

An equation books a few exact seats in the theatre. An inequality books whole rows, sometimes all the way to the back wall at infinity.

Hands-on activity

Read a graph you made

You need: grid paper, a pencil, a ruler and a calculator.

Make a table of V(x) = x(27.9 − 2x)(21.6 − 2x) (the letter-paper box from lesson 1) for x = 0, 1, 2, ..., 10, and plot it. Join the points smoothly.
Draw the horizontal line V = 900.
Read from your graph the interval of cut sizes x for which the box holds at least 900 cm³. Write it in interval notation.
Write the inequality you just solved. How are its boundary points related to the equation V(x) = 900?

What to look for: V(2) ≈ 841, V(3) ≈ 1025, V(4) ≈ 1083, V(5) ≈ 1038, V(6) ≈ 916, V(7) ≈ 739. So V(x) ≥ 900 for x from a bit above 2 to a bit above 6, roughly [2.2, 6.1] (graphing technology gives about [2.25, 6.10]). The boundary points are the solutions of the equation V(x) = 900.

Your turn

Solve 5 − 3x > 11. Give interval notation.

Solve −4 ≤ 2x − 6 < 8.

Write "x ≤ 3 or x > 5" in interval notation.

For f(x) = (x + 2)(x − 1)(x − 3), solve f(x) < 0.

For f(x) = x(x − 4), compare the solutions of f(x) = 0 and f(x) ≥ 0.

For g(x) = (x − 1)²(x + 3), solve g(x) > 0.

Use the graphs of y = x³ and y = 4x to solve x³ < 4x.

More practice: warm-up

Does [2, 5) include 2? Does it include 5?

Which bracket always goes with ∞?

Solve −3x < 12.

Write −1 < x ≤ 4 in interval notation.

Write x > 2 in set notation.

More practice: on your own

Solve (2 − x)/3 ≥ x + 2.

For f(x) = −(x + 1)(x − 2)², solve f(x) ≥ 0.

The equation x³ + x − 3 = 0 has one real root, about 1.21. Solve x³ + x − 3 < 0.

Compare the solutions of x⁴ − 5x² + 4 = 0 and x⁴ − 5x² + 4 ≤ 0.

For the 30 cm by 20 cm box, V(x) = 1000 at x = 10 − 5√2 and x = 5 (lesson 14). For which cut sizes is V(x) ≥ 1000?

Challenge

Solve 2x³ − 3x² + 8x − 12 > 0.

Create a polynomial inequality whose solution is [−3, 1] ∪ [4, ∞).

Explain why the inequality (x − 2)² > 0 has almost every real number as a solution, but (x − 2)² < 0 has none.

Check your understanding

Quiz

How do the solutions of P(x) = 0 relate to the solutions of P(x) > 0?

What is the solution of −2x + 5 ≥ 11?

Which interval notation means 4 < x ≤ 9?

f(x) = (x + 3)(x − 2) has zeros −3 and 2. What is the solution of f(x) < 0?

g(x) = x²(x − 5). What is the solution of g(x) ≤ 0?

The graphs of y = f(x) and y = g(x) intersect at x = −1 and x = 4, and f is above g only between them. What is the solution of f(x) > g(x)?

For parents and teachers

Big idea (C4.1, C4.2, C4.3): an equation has a few solutions (where the graph meets the x-axis); an inequality has intervals of solutions (where the graph is above or below). The equation's roots are the boundaries of the intervals.

How to help: Draw a quick graph and ask, "Where is it above the x-axis? Below? Should the endpoints be included?" Have your teen say the answer in words, then write it in interval notation.

Common difficulties:

Forgetting to reverse the sign when dividing by a negative.
Using square brackets with ∞.
Missing an isolated point, such as a double zero in a ≥ or ≤ inequality.
Giving x-intercepts as the answer to an inequality.