The minimum value of the fuction `f(x)=2|x-2|+5|x-3|` for all ` x in R , is `

Options

A. 3
B. 2
C. 5
D. 7

Explanation

To find the minimum value of the function \( f(x) = 2|x-2| + 5|x-3| \) for all \( x \in \mathbb{R} \), we will analyze the function by considering the critical points where the expressions inside the absolute values change sign, which are at \( x = 2 \) and \( x = 3 \).

### Step 1: Identify intervals based on critical points
The critical points divide the real line into three intervals:
1. \( x < 2 \)
2. \( 2 \leq x < 3 \)
3. \( x \geq 3 \)

### Step 2: Define the function in each interval

**Interval 1: \( x < 2 \)**
- Here, both \( |x-2| \) and \( |x-3| \) will be negative:
\[ f(x) = 2(2 - x) + 5(3 - x) = 4 - 2x + 15 - 5x = -7x + 19 \]

**Interval 2: \( 2 \leq x < 3 \)**
- In this interval, \( |x-2| \) is positive and \( |x-3| \) is negative:
\[ f(x) = 2(x - 2) + 5(3 - x) = 2x - 4 + 15 - 5x = -3x + 11 \]

**Interval 3: \( x \geq 3 \)**
- Here, both \( |x-2| \) and \( |x-3| \) are positive:
\[ f(x) = 2(x - 2) + 5(x - 3) = 2x - 4 + 5x - 15 = 7x - 19 \]

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

Now we have three piecewise functions:
1. For \( x < 2 \): \( f(x) = -7x + 19 \)
2. For \( 2 \leq x < 3 \): \( f(x) = -3x + 11 \)
3. For \( x \geq 3 \): \( f(x) = 7x - 19 \)

### Step 4: Find the minimum value in each interval

**Interval 1: \( x < 2 \)**
- As \( x \) approaches \( 2 \), \( f(x) \) approaches \( f(2) = -7(2) + 19 = 5 \).

**Interval 2: \( 2 \leq x < 3 \)**
- At \( x = 2 \): \( f(2) = -3(2) + 11 = 5 \)
- At \( x = 3 \): \( f(3) = -3(3) + 11 = 2 \)

**Interval 3: \( x \geq 3 \)**
- As \( x \) increases beyond \( 3 \), \( f(x) \) increases since the slope is positive. Thus, the minimum in this interval is at \( x = 3 \): \( f(3) = 2 \).

### Step 5: Compare the minimum values
- From interval 1: \( f(x) \) approaches \( 5 \) as \( x \) approaches \( 2 \).
- From interval 2: The minimum value is \( 2 \) at \( x = 3 \).
- From interval 3: The function increases from \( 2 \).

### Conclusion
The overall minimum value of the function \( f(x) = 2|x-2| + 5|x-3| \) is:
\[ \boxed{2} \]