Question
What is the mathematical operation used to determine if the function y=3^(3√x)−1 is increasing or decreasing?
Answers
GPT-4o mini
To determine whether the function \( y = 3^{\sqrt[3]{x}} - 1 \) is increasing or decreasing, you can use the first derivative test. The basic idea is to find the derivative of the function and analyze its sign.
1. **Find the derivative**:
The first step is to compute the derivative \( \frac{dy}{dx} \).
The function is:
\[
y = 3^{\sqrt[3]{x}} - 1
\]
To find the derivative, we can use the chain rule. The derivative of \( 3^u \) with respect to \( u \) is \( 3^u \ln(3) \), where \( u = \sqrt[3]{x} \).
Therefore, using the chain rule:
\[
\frac{d}{dx}(3^{\sqrt[3]{x}}) = 3^{\sqrt[3]{x}} \ln(3) \cdot \frac{d}{dx}(\sqrt[3]{x})
\]
The derivative of \( \sqrt[3]{x} \) is \( \frac{1}{3} x^{-\frac{2}{3}} \).
Putting it all together:
\[
\frac{dy}{dx} = 3^{\sqrt[3]{x}} \ln(3) \cdot \left(\frac{1}{3} x^{-\frac{2}{3}}\right)
\]
Simplifying this gives:
\[
\frac{dy}{dx} = \frac{\ln(3)}{3} \cdot 3^{\sqrt[3]{x}} \cdot x^{-\frac{2}{3}}
\]
2. **Analyze the sign of the derivative**:
To determine where the function is increasing or decreasing, you need to check the sign of \( \frac{dy}{dx} \).
- Since \( \ln(3) > 0 \) and \( 3^{\sqrt[3]{x}} > 0 \) for all \( x \) (as the exponential function is always positive),
- The term \( x^{-\frac{2}{3}} \) will be positive for \( x > 0 \) and undefined for \( x < 0 \) (it grows very large as \( x \) approaches 0 from the right).
Thus, for \( x > 0 \), \( \frac{dy}{dx} > 0 \), indicating that the function is **increasing** for \( x > 0 \).
3. **Conclusion**:
- For \( x > 0 \), \( y = 3^{\sqrt[3]{x}} - 1 \) is increasing.
- For \( x < 0 \), the function is not defined because the cube root is real but we need to evaluate closer to determine behavior in that region. However, the focus here is primarily on the domain where \( y \) is considered increasing.
You can conclude that \( y = 3^{\sqrt[3]{x}} - 1 \) is an increasing function for \( x > 0 \).
1. **Find the derivative**:
The first step is to compute the derivative \( \frac{dy}{dx} \).
The function is:
\[
y = 3^{\sqrt[3]{x}} - 1
\]
To find the derivative, we can use the chain rule. The derivative of \( 3^u \) with respect to \( u \) is \( 3^u \ln(3) \), where \( u = \sqrt[3]{x} \).
Therefore, using the chain rule:
\[
\frac{d}{dx}(3^{\sqrt[3]{x}}) = 3^{\sqrt[3]{x}} \ln(3) \cdot \frac{d}{dx}(\sqrt[3]{x})
\]
The derivative of \( \sqrt[3]{x} \) is \( \frac{1}{3} x^{-\frac{2}{3}} \).
Putting it all together:
\[
\frac{dy}{dx} = 3^{\sqrt[3]{x}} \ln(3) \cdot \left(\frac{1}{3} x^{-\frac{2}{3}}\right)
\]
Simplifying this gives:
\[
\frac{dy}{dx} = \frac{\ln(3)}{3} \cdot 3^{\sqrt[3]{x}} \cdot x^{-\frac{2}{3}}
\]
2. **Analyze the sign of the derivative**:
To determine where the function is increasing or decreasing, you need to check the sign of \( \frac{dy}{dx} \).
- Since \( \ln(3) > 0 \) and \( 3^{\sqrt[3]{x}} > 0 \) for all \( x \) (as the exponential function is always positive),
- The term \( x^{-\frac{2}{3}} \) will be positive for \( x > 0 \) and undefined for \( x < 0 \) (it grows very large as \( x \) approaches 0 from the right).
Thus, for \( x > 0 \), \( \frac{dy}{dx} > 0 \), indicating that the function is **increasing** for \( x > 0 \).
3. **Conclusion**:
- For \( x > 0 \), \( y = 3^{\sqrt[3]{x}} - 1 \) is increasing.
- For \( x < 0 \), the function is not defined because the cube root is real but we need to evaluate closer to determine behavior in that region. However, the focus here is primarily on the domain where \( y \) is considered increasing.
You can conclude that \( y = 3^{\sqrt[3]{x}} - 1 \) is an increasing function for \( x > 0 \).