To determine whether the function \( y = 3^{\sqrt{x} - 1} \) is increasing or decreasing, we can consider its derivative.
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First, rewrite the function for clarity: \[ y = 3^{\sqrt{x} - 1} \]
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Take the derivative of the function using the chain and power rules: To differentiate \( y \), we first recognize that \( y = 3^{u} \) where \( u = \sqrt{x} - 1 \). The derivative of \( 3^{u} \) is given by: \[ \frac{dy}{du} = 3^{u} \ln(3) \] Next, we differentiate \( u \): \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \]
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Now apply the chain rule: Thus, the derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3^{\sqrt{x} - 1} \ln(3) \cdot \frac{1}{2\sqrt{x}} \]
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Analyze the derivative:
- \( 3^{\sqrt{x} - 1} \) is always positive for \( x \geq 0 \).
- \( \ln(3) \) is a positive constant.
- \( \frac{1}{2\sqrt{x}} \) is positive for \( x > 0 \).
Since the derivative \( \frac{dy}{dx} \) is positive for \( x > 0 \), it indicates that the function \( y = 3^{\sqrt{x} - 1} \) is increasing for \( x > 0 \).
Conclusion:
The function \( y = 3^{\sqrt{x} - 1} \) is increasing.