Q.Redefine the function $f(x) = |x - 2| + |2 + x|$, $-3 \le x \le 3$

Concept: Piecewise Definition of Absolute Value Functions

The function involves two absolute value terms. We remove the absolute values by considering where each expression inside changes sign.

Step 1: Identify critical points where expressions inside absolute values are zero:

• $x - 2 = 0 \implies x = 2$
• $2 + x = 0 \implies x = -2$

These divide $[-3, 3]$ into three intervals: $[-3, -2)$, $[-2, 2]$, and $(2, 3]$.

Step 2: Rewrite $f(x)$ on each interval by determining the sign of each expression:

• For $x \in [-3, -2)$: both $(x-2) < 0$ and $(2+x) \le 0$, so

  $f(x) = -(x-2) + (-(2+x)) = -x + 2 - 2 - x = -2x$

• For $x \in [-2, 2]$: $(x-2) \le 0$ and $(2+x) \ge 0$, so …

The key to understanding any function built from absolute values is recognising that $|u|$ is really a piecewise definition: it equals $u$ when $u \ge 0$ and $-u$ when $u < 0$. The function switches its formula at the zeros of the expressions inside the absolute values.

Here we have two absolute values: $|x - 2|$ changes at $x = 2$, and $|2 + x| = |x + 2|$ changes at $x = -2$. These critical points divide our domain $[-3, 3]$ into three intervals, and on each interval both expressions have constant sign, so we can drop the absolute value bars.

Finding the piecewise formula

1. Region I: $-3 \le x < -2$

   When $x < -2$, we have $x - 2 < 0$ (so $|x - 2| = -(x - 2) = 2 - x$) and $x + 2 < 0$ (so $|x + 2| = -(x + 2) = -x - 2$).

2. Region II: $-2 \le x \le 2$

   When $-2 \le x \le 2$, we have $x - 2 \le 0$ (so $|x - 2| = 2 - x$) and $x + 2 \ge 0$ (so $|x + 2| = x + 2$).

3. Region III: $2 < x \le 3$

   When $x > 2$, both $x - 2 > 0$ (so $|x - 2| = x - 2$) and $x + 2 > 0$ (so $|x + 2| = x + 2$).

Behaviour and graph …