Define and plot
`(i) y=|x-2|+3|x-3|` `(ii) y=||x-2|-3|+|x|`
`(iii)y=|x-1|+|x-4|-2|x+1|`

Answer

To solve the given problems step by step, we will define and plot the following functions:

### (i) \( y = |x - 2| + 3|x - 3| \)

**Step 1: Identify critical points.**
The critical points occur where the expressions inside the absolute values equal zero:
- \( x - 2 = 0 \) gives \( x = 2 \)
- \( x - 3 = 0 \) gives \( x = 3 \)

**Step 2: Determine intervals.**
The critical points divide the number line into intervals:
1. \( (-\infty, 2) \)
2. \( [2, 3) \)
3. \( [3, \infty) \)

**Step 3: Evaluate the function in each interval.**

- **Interval 1: \( x < 2 \)**
\[ y = -(x - 2) + 3(- (x - 3)) = -x + 2 - 3x + 9 = -4x + 11 \]

- **Interval 2: \( 2 \leq x < 3 \)**
\[ y = (x - 2) + 3(- (x - 3)) = x - 2 - 3x + 9 = -2x + 7 \]

- **Interval 3: \( x \geq 3 \)**
\[ y = (x - 2) + 3(x - 3) = x - 2 + 3x - 9 = 4x - 11 \]

**Step 4: Plot the function.**
- For \( x < 2 \): Use \( y = -4x + 11 \)
- For \( 2 \leq x < 3 \): Use \( y = -2x + 7 \)
- For \( x \geq 3 \): Use \( y = 4x - 11 \)

### (ii) \( y = ||x - 2| - 3| + |x| \)

**Step 1: Identify critical points.**
The critical points occur where the expressions inside the absolute values equal zero:
1. \( |x - 2| - 3 = 0 \) gives \( x - 2 = 3 \) or \( x - 2 = -3 \) leading to \( x = 5 \) and \( x = -1 \)
2. \( |x| = 0 \) gives \( x = 0 \)

**Step 2: Determine intervals.**
The critical points divide the number line into intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 0) \)
3. \( [0, 2) \)
4. \( [2, 5) \)
5. \( [5, \infty) \)

**Step 3: Evaluate the function in each interval.**

- **Interval 1: \( x < -1 \)**
\[ y = -(|x - 2| - 3) + (-x) = -(-x + 2 - 3) - x = x - 1 - x = -1 \]

- **Interval 2: \( -1 \leq x < 0 \)**
\[ y = -(|x - 2| - 3) + (-x) = -(-x + 2 - 3) - x = x - 1 - x = -1 \]

- **Interval 3: \( 0 \leq x < 2 \)**
\[ y = -(|x - 2| - 3) + x = -(-x + 2 - 3) + x = x + 1 \]

- **Interval 4: \( 2 \leq x < 5 \)**
\[ y = -(|x - 2| - 3) + x = -((x - 2) - 3) + x = -x + 5 + x = 5 \]

- **Interval 5: \( x \geq 5 \)**
\[ y = -(|x - 2| - 3) + x = -((x - 2) - 3) + x = -x + 5 + x = 5 \]

**Step 4: Plot the function.**
- For \( x < -1 \): \( y = -1 \)
- For \( -1 \leq x < 0 \): \( y = -1 \)
- For \( 0 \leq x < 2 \): \( y = x + 1 \)
- For \( 2 \leq x < 5 \): \( y = 5 \)
- For \( x \geq 5 \): \( y = 5 \)

### (iii) \( y = |x - 1| + |x - 4| - 2|x + 1| \)

**Step 1: Identify critical points.**
The critical points occur where the expressions inside the absolute values equal zero:
1. \( x - 1 = 0 \) gives \( x = 1 \)
2. \( x - 4 = 0 \) gives \( x = 4 \)
3. \( x + 1 = 0 \) gives \( x = -1 \)

**Step 2: Determine intervals.**
The critical points divide the number line into intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 1) \)
3. \( [1, 4) \)
4. \( [4, \infty) \)

**Step 3: Evaluate the function in each interval.**

- **Interval 1: \( x < -1 \)**
\[ y = -(x - 1) - (x - 4) - 2(-x - 1) = -x + 1 - x + 4 + 2x + 2 = 3 \]

- **Interval 2: \( -1 \leq x < 1 \)**
\[ y = -(x - 1) - (x - 4) - 2(-x - 1) = -x + 1 - x + 4 + 2x + 2 = 7 - 2x \]

- **Interval 3: \( 1 \leq x < 4 \)**
\[ y = (x - 1) - (x - 4) - 2(-x - 1) = x - 1 - x + 4 + 2x + 2 = 5 + 2x \]

- **Interval 4: \( x \geq 4 \)**
\[ y = (x - 1) + (x - 4) - 2(x + 1) = x - 1 + x - 4 - 2x - 2 = -7 \]

**Step 4: Plot the function.**
- For \( x < -1 \): \( y = 3 \)
- For \( -1 \leq x < 1 \): \( y = 7 - 2x \)
- For \( 1 \leq x < 4 \): \( y = 5 + 2x \)
- For \( x \geq 4 \): \( y = -7 \)