The range of `f(x) = sqrt(3x^(2) - 10 x + 12)` is

Options

A. `[sqrt(11/3), infty)`
B. `[-sqrt(11)/3, infty)`
C. `[sqrt(3), infty)`
D. `(-sqrt(3), infty) `

Answer

To find the range of the function \( f(x) = \sqrt{3x^2 - 10x + 12} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
Let \( y = f(x) \).
\[ y = \sqrt{3x^2 - 10x + 12} \]

### Step 2: Square both sides
To eliminate the square root, we square both sides:
\[ y^2 = 3x^2 - 10x + 12 \]

### Step 3: Rearrange the equation
Rearranging gives us a quadratic equation in \( x \):
\[ 3x^2 - 10x + (12 - y^2) = 0 \]

### Step 4: Determine the condition for real \( x \)
For \( x \) to have real solutions, the discriminant of this quadratic must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = 3 \), \( b = -10 \), and \( c = 12 - y^2 \). Thus, the discriminant is:
\[ D = (-10)^2 - 4(3)(12 - y^2) \geq 0 \]
Simplifying gives:
\[ 100 - 12 + 4y^2 \geq 0 \]
\[ 4y^2 + 88 \geq 0 \]

### Step 5: Solve the inequality
Since \( 4y^2 + 88 \) is always positive (as \( 4y^2 \) is non-negative and 88 is positive), we need to find when:
\[ 12 - y^2 \geq 0 \]
This simplifies to:
\[ y^2 \leq 12 \]

### Step 6: Find the bounds for \( y \)
Taking the square root of both sides gives:
\[ -\sqrt{12} \leq y \leq \sqrt{12} \]
Since \( y \) represents \( f(x) \) which is always non-negative, we discard the negative part:
\[ 0 \leq y \leq \sqrt{12} \]

### Step 7: Express the range
Thus, the range of \( f(x) \) is:
\[ [0, \sqrt{12}] \]
Since \( \sqrt{12} = 2\sqrt{3} \), we can express the range as:
\[ [0, 2\sqrt{3}] \]

### Final Answer
The range of \( f(x) = \sqrt{3x^2 - 10x + 12} \) is:
\[ [0, 2\sqrt{3}] \]

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