If `y=" min "(x^(2)+2, 6-3x)`, then the greatest value of y for `x gt0`

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the greatest value of \( y \) for the function \( y = \min(x^2 + 2, 6 - 3x) \) when \( x > 0 \), we will follow these steps:

### Step 1: Identify the Functions
We have two functions:
1. \( f_1(x) = x^2 + 2 \)
2. \( f_2(x) = 6 - 3x \)

### Step 2: Find the Intersection Point
To find the point where these two functions intersect, we set them equal to each other:
\[ x^2 + 2 = 6 - 3x \]
Rearranging gives:
\[ x^2 + 3x - 4 = 0 \]

### Step 3: Solve the Quadratic Equation
We can factor the quadratic equation:
\[ (x + 4)(x - 1) = 0 \]
This gives us the solutions:
\[ x = -4 \quad \text{or} \quad x = 1 \]
Since we are only interested in \( x > 0 \), we take \( x = 1 \).

### Step 4: Calculate the Value of \( y \) at the Intersection Point
Now, we substitute \( x = 1 \) back into either function to find \( y \):
\[ y = f_1(1) = 1^2 + 2 = 3 \]
or
\[ y = f_2(1) = 6 - 3(1) = 3 \]
Both functions yield \( y = 3 \).

### Step 5: Determine the Behavior of the Functions
Next, we need to analyze the behavior of both functions for \( x > 0 \):
- \( f_1(x) = x^2 + 2 \) is a parabola that opens upwards and increases as \( x \) increases.
- \( f_2(x) = 6 - 3x \) is a linear function that decreases as \( x \) increases.

### Step 6: Find the Maximum Value of \( y \)
Since \( f_2(x) \) intersects \( f_1(x) \) at \( x = 1 \) and \( f_2(x) \) is decreasing while \( f_1(x) \) is increasing, the minimum value of \( y \) will occur at the intersection point. For \( x > 1 \), \( f_1(x) \) will be greater than \( f_2(x) \), meaning \( y \) will be determined by \( f_2(x) \) until \( x = 1 \).

Thus, the greatest value of \( y \) for \( x > 0 \) is:
\[ \boxed{3} \]