Unit 4: Functions and Graphs – Exercise 4.2

(i) \( f(x) = |x - 2| \)
Vertex: \( x - 2 = 0 \Rightarrow x = 2 \), vertex \((2,0)\). Symmetric about \(x=2\).
\[ \begin{array}{c|c|c|c|c|c} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x)=|x-2| & 2 & 1 & 0 & 1 & 2 \end{array} \]

(ii) \( f(x) = 3|x+3| - 4 \)
Vertex at \(x=-3\) ⇒ \((-3,-4)\). Table:
\[ \begin{array}{c|c|c|c|c|c} x & -5 & -4 & -3 & -2 & -1 \\ \hline f(x)=3|x+3|-4 & 2 & -1 & -4 & -1 & 2 \end{array} \]

(iii) \( f(x) = 5|x| \)
Vertex at \((0,0)\). Table: \(x=-2,10\); \(x=-1,5\); \(x=0,0\); \(x=1,5\); \(x=2,10\).

(iv) \( f(x) = |x+2| + 3 \)
Vertex: \(x=-2\) → \((-2,3)\). Symmetry about \(x=-2\).
\[ \begin{array}{c|c|c|c|c|c} x & -4 & -3 & -2 & -1 & 0 \\ \hline f(x) & 5 & 4 & 3 & 4 & 5 \end{array} \]

(v) \( f(x) = 4|x+4| - 3 \)
Vertex: \(x=-4\) → \((-4,-3)\). Values: \(x=-6 \to 1\), \(-5 \to -1\), \(-4 \to -3\), \(-3 \to -1\), \(-2 \to 1\).

(vi) \( f(x) = 2|x+1| - 6 \)
Vertex: \(x=-1\) ⇒ \((-1,-6)\).
\[ \begin{array}{c|c|c|c|c|c} x & -3 & -2 & -1 & 0 & 1 \\ \hline f(x) & -2 & -4 & -6 & -4 & -2 \end{array} \]

(i) \( |x-2| = 6 \)
\[ \begin{aligned} |x-2| &= 6 \\ \pm (x-2) &= 6 \;\Longrightarrow\; x-2 = \pm 6 \\[4pt] \text{Case 1:}&\quad x-2 = 6 \;\Rightarrow\; x = 8 \\ \text{Case 2:}&\quad x-2 = -6 \;\Rightarrow\; x = -4 \end{aligned} \] Solution set: \( \{-4,\; 8\} \).

(ii) \( |2x+1| = 7 \)
\[ \begin{aligned} |2x+1| &= 7 \;\Rightarrow\; 2x+1 = \pm 7 \\ 2x+1 &= 7 \;\Rightarrow\; 2x = 6 \;\Rightarrow\; x = 3 \\ 2x+1 &= -7 \;\Rightarrow\; 2x = -8 \;\Rightarrow\; x = -4 \end{aligned} \] \( S.S = \{-4,\;3\} \)

(iii) \( |4x-9| = 3 \)
\[ \begin{aligned} 4x-9 &= \pm 3 \\ 4x-9 = 3 &\;\Rightarrow\; 4x = 12 \;\Rightarrow\; x = 3 \\ 4x-9 = -3 &\;\Rightarrow\; 4x = 6 \;\Rightarrow\; x = \frac{3}{2} = 1\frac{1}{2} \end{aligned} \] \( S.S = \left\{1\frac{1}{2},\; 3\right\} \)

(iv) \( |7-2x| = 1 \)
\[ \begin{aligned} 7-2x &= \pm 1 \\ 7-2x = 1 &\;\Rightarrow\; -2x = -6 \;\Rightarrow\; x = 3 \\ 7-2x = -1 &\;\Rightarrow\; -2x = -8 \;\Rightarrow\; x = 4 \end{aligned} \] \( S.S = \{3,\;4\} \)

(i) \( |5x+8| \le 3 \)
\[ \begin{aligned} |5x+8| &\le 3 \;\Longrightarrow\; -3 \le 5x+8 \le 3 \\ \text{Left:}&\quad -3 \le 5x+8 \;\Rightarrow\; -11 \le 5x \;\Rightarrow\; x \ge -\frac{11}{5} = -2\frac{1}{5} \\ \text{Right:}&\quad 5x+8 \le 3 \;\Rightarrow\; 5x \le -5 \;\Rightarrow\; x \le -1 \end{aligned} \] Solution: \(\left[-2\frac{1}{5},\; -1\right]\)

(ii) \( |4x-12| \le 0 \)
\[ \begin{aligned} |4x-12| &\le 0 \;\Longrightarrow\; 4x-12 = 0 \;\Rightarrow\; x = 3 \end{aligned} \] \( S.S = \{3\} \) (single point).

(iii) \( |3-4x| > 0 \)
\[ \begin{aligned} |3-4x| > 0 &\;\Longrightarrow\; 3-4x \neq 0 \\ &\;\Longrightarrow\; x \neq \frac{3}{4} \end{aligned} \] Solution: \( (-\infty,\frac{3}{4}) \cup (\frac{3}{4},\infty) \)

(iv) \( 1 - 2\left|\frac{3}{2}x - 5\right| > -3 \)
\[ \begin{aligned} 1 - 2\left|\frac{3}{2}x - 5\right| &> -3 \\ - 2\left|\frac{3}{2}x - 5\right| &> -4 \\ \left|\frac{3}{2}x - 5\right| &< 2 \\ -2 < \frac{3}{2}x - 5 &< 2 \\ 3 < \frac{3}{2}x &< 7 \\ 2 < x &< \frac{14}{3} = 4\frac{2}{3} \end{aligned} \] \( S.S = \left(2,\; 4\frac{2}{3}\right) \)

(v) \( |3x-2| \le 12 \)
\[ \begin{aligned} |3x-2| &\le 12 \;\Longrightarrow\; -12 \le 3x-2 \le 12 \\ -10 &\le 3x \le 14 \\ -\frac{10}{3} &\le x \le \frac{14}{3} \end{aligned} \] Solution: \(\left[-3\frac{1}{3},\; 4\frac{2}{3}\right]\)

(vi) \( |1-2x| > 5 \)
\[ \begin{aligned} |1-2x| > 5 &\;\Longrightarrow\; 1-2x < -5 \quad \text{or} \quad 1-2x > 5 \\ &\text{Case 1: } 1-2x < -5 \;\Rightarrow\; -2x < -6 \;\Rightarrow\; x > 3 \\ &\text{Case 2: } 1-2x > 5 \;\Rightarrow\; -2x > 4 \;\Rightarrow\; x < -2 \end{aligned} \] Solution: \( (-\infty,-2) \cup (3,\infty) \)