To determine whether the function \( y = 3^{\sqrt[3]{x}} - 1 \) is increasing or decreasing, we can analyze its derivative.
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First, let’s find the derivative of the function:
- The function can also be written as \( y = 3^{\sqrt[3]{x}} - 1 \).
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To differentiate this function, we can use the chain rule:
- Let \( u = \sqrt[3]{x} \), so \( y = 3^u - 1 \).
- The derivative \( dy/du = 3^u \ln(3) \).
- Now, differentiate \( u = x^{1/3} \):
- The derivative \( du/dx = \frac{1}{3}x^{-2/3} \).
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Now, applying the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3^{\sqrt[3]{x}} \ln(3) \cdot \frac{1}{3}x^{-2/3} = \frac{3^{\sqrt[3]{x}} \ln(3)}{3 \cdot x^{2/3}} \]
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Now, we can analyze the sign of \( \frac{dy}{dx} \):
- The term \( 3^{\sqrt[3]{x}} \) is always positive for all \( x \geq 0 \).
- The term \( \ln(3) \) is also positive.
- The term \( 3 \cdot x^{2/3} \) is positive for all \( x > 0 \).
Since \( \frac{dy}{dx} \) is positive for all \( x > 0 \), it indicates that the function is increasing on the interval where \( x > 0 \).
Thus, the correct answer is Option 1: increasing.