To determine whether the function \( y = \sqrt{3x - 1} \) is increasing or decreasing, we can analyze the derivative of the function.
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The function \( y = \sqrt{3x - 1} \) is defined for values of \( x \) such that \( 3x - 1 \geq 0 \) (i.e., \( x \geq \frac{1}{3} \)).
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Now, we can find the derivative of the function:
\[ y = (3x - 1)^{1/2} \]
Using the chain rule, the derivative is:
\[ \frac{dy}{dx} = \frac{1}{2}(3x - 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x - 1}} \]
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The numerator \( 3 \) is positive, and the denominator \( 2\sqrt{3x - 1} \) is also positive for \( x \geq \frac{1}{3} \). This means that \( \frac{dy}{dx} > 0 \) for all \( x \) in the domain of the function.
Since the derivative is positive, we can conclude that the function is increasing.
The correct answer is option 1: increasing.