Example 9

For what values of $x$ is $| 3x +4 | \leq | x - 2 |$?

(1) $ [-3,\dfrac{-1}{2}]$

(2) $ (-∞, -3] \cup [ 2, ∞)$

(3) $ (-∞, \dfrac{-1}{2}]$

(4) $ [\dfrac{-4}{3}, \dfrac{-1}{2}]$

Solution

The deflection points for linear modulus functions can be ascertained by equating them to 0.

When $3x + 4 = 0$ ⇒ $ \bm{x = - \dfrac{4}{3}}$ and when $ x - 2 =0 $ ⇒ $\bm{ x = 2}$

These are deflection points for these lines.

Condition	Expansion of modulus	Acceptable Range
$ x \geq 2$	$ 3x +4 \leq x - 2 $ ⇒ $\bm{x \leq - 3}$	None
$\dfrac{-4}{3} \leq x \lt 2 $	$ 3x +4 \leq -(x - 2) $ ⇒ $4x \leq - 2$ ⇒ $\bm{x \leq \dfrac{-1}{2}}$	$\dfrac{-4}{3} \leq x \leq \dfrac{-1}{2}$
$x \lt \dfrac{-4}{3} $	$ -(3x +4) \leq -(x - 2)$ ⇒ $-2x \leq - 6$ ⇒ $ \bm{x \geq - 3}$	$ -3 \leq x \leq \dfrac{-4}{3}$

Combining the two acceptable ranges we get the final acceptable range of $ \bm{- 3 \leq x \leq \dfrac{-1}{2}}$

Answer: $ (1) \left[- 3, \dfrac{-1}{2}\right]$

Example 10

How many values of $x$ satisfy $| 2x - 1| + | x - 3| = 8$?

Solution

The deflection points to determine the sign when the modulus is removed are ascertained by equating each linear function in the modulus to 0.
The deflection points for $| 2x - 1| + | x - 3| = 8$ are at $x = \dfrac{1}{2}$ and $ x = 3$

Condition	Expansion of modulus	Acceptable Range
$ x \geq 3$	$ 2x -1 + x - 3 = 8$ ⇒ $\bm{x =4}$	Accepted
$\dfrac{1}{2} \leq x \lt 3 $	$ 2x -1 - (x - 3)=8$ ⇒ $\bm{ x = 6}$	Rejected (not in range)
$x \lt \dfrac{1}{2} $	$ -(2x - 1) - (x - 3) = 8$ ⇒ $ \bm{x = \dfrac{-4}{3}} $	Accepted

∴ This equation has 2 solutions.

Answer: $2$

Example 11

For what values of $x$ is $| x + 1 | + | 2 - x | + | x + 3 | \leq x + 8$?

Solution

We first rewrite $| 2 - x|$ as $| x - 2|$ so that we get a positive result for $x \gt 2$.
⇒ $| x + 1| + | x - 2 | + | x + 3 | \leq x + 8 $

Deflection points are $x = -3, -1$ and $2$. Rewriting the terms in the ascending order of thier deflection values,

⇒ $| x + 3| + | x + 1 | + | x - 2 | \leq x + 8 $

Condition	Expansion of modulus	Acceptable Range
$ x \geq 2$	$ x + 3 + x + 1 + x - 2 \leq x+8 $ ⇒ $2x \leq 6$ ⇒ $\bm{x \leq 3}$	$2 \leq x \leq 3$
$- 1 \leq x \lt 2 $	$ x + 3 +x + 1 - (x - 2)\leq x + 8$ ⇒ $\bm{6 \leq 8}$ This is always true	$-1 \leq x \lt 2$
$- 3 \leq x \lt - 1$	$ x + 3 - (x + 1) - (x - 2) \leq x+8 $ ⇒ $-2x \leq 4$ ⇒ $\bm{x \geq -2}$	$-2 \leq x \lt -1$
$x \lt - 3$	$ -(x + 3) - (x + 1) - (x - 2) \leq x + 8$ ⇒ $- 4x \lt 10 $ ⇒ $\bm{x \gt - 2.5}$	None

Combining the acceptable ranges we get $ - 2 \leq x \leq 3 $

Answer: $ - 2 \leq x \leq 3$

Example 12

What is the minimum value of $f(x)$, where$ f(x) $= $| x - 5 | + | x + 6 | + | x - 3 | $?

Solution

As this is a sum of modulus functions with linear expression, the minimum value will be at one of the deflection points, i.e., when one of the modulus function equals 0.

The deflection points are $ x = 5, -6$ and $3$.

$f(5)$ $=$ $ 0 + 11 +2 = 13$

$f(-6)$ $=$ $ 11 + 0 + 9 = 20$

$f(3)$ $=$ $ 2 + 9 + 0 = 11$

∴ Minimum value of $f(x) = 11$

Answer: $11$

CAT 2025 Lesson : Inequalities - Modulus on a Number Line