If `f(x)=sqrt(3x-2)`, what is the smallest possible value of f(x)?

Options

A. `0`
B. `(2)/(3)`
C. `1`
D. `2`

Answer

To find the smallest possible value of the function \( f(x) = \sqrt{3x - 2} \), we will follow these steps:

### Step 1: Identify the domain of the function
The expression under the square root, \( 3x - 2 \), must be non-negative for \( f(x) \) to be defined. Therefore, we need to solve the inequality:
\[ 3x - 2 \geq 0 \]

### Step 2: Solve the inequality
To solve the inequality, we can isolate \( x \):
\[ 3x \geq 2 \]
\[ x \geq \frac{2}{3} \]
This means that the function \( f(x) \) is defined for \( x \) values greater than or equal to \( \frac{2}{3} \).

### Step 3: Find the minimum value of \( f(x) \)
The smallest value of \( x \) in the domain is \( x = \frac{2}{3} \). We will now substitute this value into the function to find the corresponding value of \( f(x) \):
\[ f\left(\frac{2}{3}\right) = \sqrt{3\left(\frac{2}{3}\right) - 2} \]
Calculating inside the square root:
\[ = \sqrt{2 - 2} = \sqrt{0} = 0 \]

### Step 4: Conclusion
The smallest possible value of \( f(x) \) is \( 0 \).

### Final Answer
The smallest possible value of \( f(x) \) is \( 0 \).

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