Question

Does the graph of the function y=3^3√x−1−−−−− show that the function is increasing or decreasing?

Option 1: increasing

Option 2: decreasing

(1 point)
The correct answer is option
.

Answer 1

To determine whether the function \( y = 3^{\sqrt[3]{x}} - 1 \) is increasing or decreasing, we can analyze its derivative.

1. First, let’s find the derivative of the function:
- The function can also be written as \( y = 3^{\sqrt[3]{x}} - 1 \).

2. 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 \( du/dx = \frac{1}{3}x^{-2/3} \).

3. Now, applying the chain rule:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3^{\sqrt[3]{x}} \ln(3) \cdot \frac{1}{3}x^{-2/3} = \frac{3^{\sqrt[3]{x}} \ln(3)}{3 \cdot x^{2/3}}
\]

4. 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.