To determine whether the function \( y = \sqrt{3x - 1} \) is increasing or decreasing, we can find its derivative.
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Differentiate the function: \[ y = \sqrt{3x - 1} \] Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{2}(3x - 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x - 1}} \]
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Analyze the derivative: The expression \( \sqrt{3x - 1} \) is positive for \( x > \frac{1}{3} \), since it is only defined for \( 3x - 1 \geq 0 \). Therefore, the derivative \( \frac{dy}{dx} \) will also be positive for all \( x > \frac{1}{3} \).
Since the derivative is positive for \( x > \frac{1}{3} \), the function is increasing in that interval.
Thus, the correct answer is:
Option 1: increasing.