The minimum value of the function `y = |2x+1| + 2|x-2|`, is

Options

A. `4x - 3`
B. `3x - 1`
C. 5
D. 1

Answer

To find the minimum value of the function \( y = |2x + 1| + 2|x - 2| \), we will analyze the function by considering the points where the expressions inside the absolute values change sign.

### Step 1: Identify critical points
The expressions inside the absolute values are \( 2x + 1 \) and \( x - 2 \). We need to find when these expressions equal zero:
- For \( 2x + 1 = 0 \):
\[ 2x = -1 \implies x = -\frac{1}{2} \]
- For \( x - 2 = 0 \):
\[ x = 2 \]

Thus, the critical points are \( x = -\frac{1}{2} \) and \( x = 2 \).

### Step 2: Analyze intervals
We will analyze the function in the following intervals:
1. \( (-\infty, -\frac{1}{2}) \)
2. \( [-\frac{1}{2}, 2] \)
3. \( (2, \infty) \)

### Step 3: Evaluate the function in each interval

**Interval 1: \( x < -\frac{1}{2} \)**
- Here, both expressions are negative:
\[ y = -(2x + 1) + 2(-x + 2) = -2x - 1 - 2x + 4 = -4x + 3 \]

**Interval 2: \( -\frac{1}{2} \leq x \leq 2 \)**
- In this interval, \( 2x + 1 \) is non-negative and \( x - 2 \) is negative:
\[ y = (2x + 1) + 2(-x + 2) = 2x + 1 - 2x + 4 = 5 \]

**Interval 3: \( x > 2 \)**
- Here, both expressions are positive:
\[ y = (2x + 1) + 2(x - 2) = 2x + 1 + 2x - 4 = 4x - 3 \]

### Step 4: Find minimum values at critical points
Now we will evaluate the function at the critical points:
- At \( x = -\frac{1}{2} \):
\[ y = |2(-\frac{1}{2}) + 1| + 2|-\frac{1}{2} - 2| = |0| + 2|-\frac{5}{2}| = 0 + 5 = 5 \]
- At \( x = 2 \):
\[ y = |2(2) + 1| + 2|2 - 2| = |5| + 2|0| = 5 + 0 = 5 \]

### Step 5: Determine the minimum value
From our evaluations:
- In the interval \( (-\infty, -\frac{1}{2}) \), \( y = -4x + 3 \) approaches infinity as \( x \) approaches \(-\infty\).
- In the interval \( [-\frac{1}{2}, 2] \), \( y = 5 \).
- In the interval \( (2, \infty) \), \( y = 4x - 3 \) also approaches infinity as \( x \) approaches \(\infty\).

Thus, the minimum value of \( y \) occurs at both critical points \( x = -\frac{1}{2} \) and \( x = 2 \), and the minimum value is:
\[ \boxed{5} \]