Question

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

decreasing

Answer 1

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

1. **First, rewrite the function for clarity:**
\[
y = 3^{\sqrt{x} - 1}
\]

2. **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 \( u \):
\[
\frac{du}{dx} = \frac{1}{2\sqrt{x}}
\]

3. **Now apply the chain rule:**
Thus, the derivative of \( y \) with respect to \( x \) is:
\[
\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3^{\sqrt{x} - 1} \ln(3) \cdot \frac{1}{2\sqrt{x}}
\]

4. **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**.