Math Problem Statement
Solution
To solve this problem, let’s go through each part carefully.
Given function:
Part (a): Find the intervals on which the function is increasing and decreasing.
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Find the derivative :
Using the product rule for differentiation where : Simplifying each term:
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Set to find critical points:
Solving for :
So, we have a critical point at .
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Determine the sign of on the intervals around the critical point :
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For : Choose a point like .
- , and since , , so is decreasing on this interval.
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For : Choose a point like .
- , so , meaning is increasing on this interval.
Therefore, the function is increasing on the interval .
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Answer for Part (a):
The correct answer is A: The function is increasing on the open interval .
Part (b): Identify local extreme values, if any.
Since changes from decreasing to increasing at , there is a local minimum at this point.
- Local Minimum Value: Substitute into :
Therefore, there is a local minimum of at .
Summary:
- Intervals of Increase:
- Local Minimum: at
Do you need further clarification on any step?
Here are some related questions for deeper understanding:
- How would you find intervals of concavity for the function ?
- Can you determine if the local minimum at is also a global minimum?
- How would the behavior of change if we considered negative values of ?
- What happens to the function as approaches 0 from the right?
- How would you calculate the derivative of a similar function, such as ?
Tip: Always check the domain of the function before starting any calculus operations to ensure your intervals are valid.
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Math Problem Analysis
Mathematical Concepts
Calculus
Differentiation
Critical Points
Intervals of Increase and Decrease
Formulas
f(x) = x * ln(x)
Derivative: f'(x) = ln(x) + 1
Theorems
Product Rule
Critical Points in Calculus
Suitable Grade Level
College Level Calculus
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