Modulus Functions - Sketching Graphs (AQA A-Level Mathematics): Revision Notes

Example 1: Solving $|x| \leq 3$

The inequality $|x| \leq 3$ means that the absolute value of $x$ is less than or equal to 3. This can be rewritten as:

$-3 \leq x \leq 3$

So, the solution is all $x$ -values between -3 and 3, inclusive.

Example 2: Solving $|x| > 2$

The inequality $|x| > 2$ means that the absolute value of $x$ is greater than 2. This can be rewritten as:

$x < -2 \quad \text{or} \quad x > 2$

In this case, the solution is any $x$ -value outside the range between -2 and 2.

Example 3: Solving $|x - 1| \leq 4$

To solve this inequality, follow these steps:

1. Rewrite the inequality without the modulus: $-4 \leq x - 1 \leq 4$

2. Solve for $x$: $-4 + 1 \leq x \leq 4 + 1$

$-3 \leq x \leq 5$

So, the solution is -3 ≤ x ≤ 5 .

Example 4: Solving $|2x + 3| > 7$

1. First, split it into two cases:

• $2x + 3 > 7$
• $2x + 3 < -7$

2. Solve each case separately:

• For $2x + 3 > 7$ :

• $2x > 4 \quad \Rightarrow \quad x > 2$

• For $2x + 3 < -7$ :

• $2x < -10 \quad \Rightarrow \quad x < -5$ So, the solution is x < -5 or x > 2 .

Key Steps for Solving Modulus Inequalities

1. Understand the modulus: Write the modulus inequality as two separate inequalities.
2. Solve the inequalities: Treat each part of the inequality as a regular linear inequality.
3. Combine the solutions: Depending on the sign of the inequality (≤, ≥, <, >), you either combine or exclude parts of the number line.

Practice Problem

Solve the inequality: $|x + 2| \geq 5$ .

4. Split into two inequalities:

• $x + 2 \geq 5$
• $x + 2 \leq -5$

5. Solve each:

• $x \geq 3$
• $x \leq -7$ Final solution: x ≤ -7 or x ≥ 3.

e.g. Solve $|2x + 3| \leq |x - 1|$ (graphical method recommended)

6. Find the points of intersection using the graphical method:

1. Decide from the graph which region(s) satisfy the inequality: We can see from $|2x+3| \leq |x-1|$ that we require the regions where the graph of $y = |2x+3|$ is under $y = |x-1|$.

• Note: The squaring both sides method does work but you must check the answer.

Q1 (June 2007, Q2) Solve the inequality $|4x - 3| < |2x + 1|$.

Steps to Solve: 7. Finding the Points of Intersection:

• Setting the two equations equal:

• Setting the negative of one side equal to the positive of the other:

8. Solution:

• The points of intersection are $x = \frac{1}{3}$ and $x = 2$.
• The inequality is satisfied between these points. Hence, the solution is:

Examples:

• $|4| = 4$
• $|-2| = 2$
• $\begin{vmatrix} 2\\ 3 \end{vmatrix}$
• $| \sqrt{2^2 + 3^2} | = \sqrt{13} \quad \text {(an example with vectors)}$

Example 1: Draw $y = 2x$ and $y = |2x|$ on the same set of axes.

• For $y = 2x$:

• If $x = -1, y = 2(-1) = -2$

• For $y = |2x|$:

• If $x = -1, y = |2(-1)| = |-2| = 2$ In the graph:

• The line $y = 2x$ is shown in blue.

• The line $y = |2x|$ is shown in red and reflects the negative part above the x-axis.

Example 2: Draw $y=|x-3|$.