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Exercise 1.5 · Q4

Q.If f(x)=∣x−2∣+∣x+2∣, x∈Rf(x)=|x-2|+|x+2|,\ x\in R, then

(1) f(x)={−2xif x∈(−∞,−2]4if x∈(−2,2]2xif x∈(2,∞)f(x)=\begin{cases}-2x & \text{if } x\in(-\infty,-2]\\ 4 & \text{if } x\in(-2,2]\\ 2x & \text{if } x\in(2,\infty)\end{cases}
(2) f(x)={2xif x∈(−∞,−2]4xif x∈(−2,2]−2xif x∈(2,∞)f(x)=\begin{cases}2x & \text{if } x\in(-\infty,-2]\\ 4x & \text{if } x\in(-2,2]\\ -2x & \text{if } x\in(2,\infty)\end{cases}
(3) f(x)={−2xif x∈(−∞,−2]−4xif x∈(−2,2]2xif x∈(2,∞)f(x)=\begin{cases}-2x & \text{if } x\in(-\infty,-2]\\ -4x & \text{if } x\in(-2,2]\\ 2x & \text{if } x\in(2,\infty)\end{cases}
(4) f(x)={−2xif x∈(−∞,−2]2xif x∈(−2,2]2xif x∈(2,∞)f(x)=\begin{cases}-2x & \text{if } x\in(-\infty,-2]\\ 2x & \text{if } x\in(-2,2]\\ 2x & \text{if } x\in(2,\infty)\end{cases}
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Step 1. For x≤−2x\le-2: x−2<0⇒∣x−2∣=2−xx-2<0\Rightarrow|x-2|=2-x; x+2≤0⇒∣x+2∣=−(x+2)=−x−2x+2\le0\Rightarrow|x+2|=-(x+2)=-x-2. Sum: (2−x)+(−x−2)=−2x(2-x)+(-x-2)=-2x.

Step 2. For −2<x≤2-2<x\le2: x−2≤0⇒∣x−2∣=2−xx-2\le0\Rightarrow|x-2|=2-x; x+2>0⇒∣x+2∣=x+2x+2>0\Rightarrow|x+2|=x+2. Sum: (2−x)+(x+2)=4(2-x)+(x+2)=4.

Step 3. For x>2x>2: x−2>0⇒∣x−2∣=x−2x-2>0\Rightarrow|x-2|=x-2; x+2>0⇒∣x+2∣=x+2x+2>0\Rightarrow|x+2|=x+2. Sum: (x−2)+(x+2)=2x(x-2)+(x+2)=2x. …

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