Functions

Furthermore

The modulus (or absolute value) of a number represents its distance from zero on the number line, regardless of its sign. When solving equations and inequalities involving modulus, it's essential to consider both the positive and negative cases. This often results in a system of equations or inequalities to solve.

Key Formulae:

$$ |a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases} $$

For any real numbers \( a \) and \( b \):

1) \( |a| = b \) is equivalent to \( a = b \) or \( a = -b \)
2) \( |a| > b \) is equivalent to \( a > b \) or \( a < -b \)
3) \( |a| < b \) is equivalent to \( -b < a < b \)

Example 1:

Solve the equation \( |x - 3| = 5 \).

Using the key formulae, this equation can be broken down into two possible equations:

$$ x - 3 = 5 \quad \text{or} \quad x - 3 = -5 $$

Solving each equation, we get:

\( x = 8 \) or \( x = -2 \)

Thus, the solutions to the equation \( |x - 3| = 5 \) are \( x = 8 \) and \( x = -2 \).

Example 2:

Solve the inequality \( |3x \arccos(x)| > 1 \) for \(-1 \le x \le 1 \).

Using the key formulae for modulus inequalities, this inequality can be broken down into two possible inequalities:

$$ 3x \arccos(x) > 1 \quad \text{or} \quad 3x \arccos(x) < -1 $$

$$ x \arccos(x) > \frac{1}{3} $$

$$ x \arccos(x) < -\frac{1}{3} $$

Using the GDC graph plotting function the following solution is obtained:

$$ -1 \le x \lt -0.189 \text{ and } 0.254 \lt x \lt 0.937 $$