Question
Determine whether the function yequalsnegative 5x cubed is increasing or decreasing for the following conditions.
(a) xless than0
(b) xgreater than0
Question content area bottom
Part 1
(a) Is the function increasing or decreasing for xless than0?
(a) xless than0
(b) xgreater than0
Question content area bottom
Part 1
(a) Is the function increasing or decreasing for xless than0?
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine whether the function \( y = -5x^3 \) is increasing or decreasing for \( x < 0 \), we need to analyze the derivative of the function.
1. **Find the derivative**:
\[
y = -5x^3
\]
The derivative is given by:
\[
y' = \frac{dy}{dx} = -15x^2
\]
2. **Analyze the sign of the derivative for \( x < 0 \)**:
- Since \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)) and multiplying by -15 makes it non-positive, \( y' = -15x^2 \) is:
- **Zero** at \( x = 0 \)
- **Negative** for all \( x < 0 \)
Since the derivative \( y' \) is negative for \( x < 0 \), this means that the function \( y = -5x^3 \) is **decreasing** for \( x < 0\).
### Conclusion:
(a) The function is **decreasing** for \( x < 0 \).
1. **Find the derivative**:
\[
y = -5x^3
\]
The derivative is given by:
\[
y' = \frac{dy}{dx} = -15x^2
\]
2. **Analyze the sign of the derivative for \( x < 0 \)**:
- Since \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)) and multiplying by -15 makes it non-positive, \( y' = -15x^2 \) is:
- **Zero** at \( x = 0 \)
- **Negative** for all \( x < 0 \)
Since the derivative \( y' \) is negative for \( x < 0 \), this means that the function \( y = -5x^3 \) is **decreasing** for \( x < 0\).
### Conclusion:
(a) The function is **decreasing** for \( x < 0 \).
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