Math Problem Statement
Solution
To solve this problem, let’s go through each part step by step.
Problem Analysis
Given:
- A function defined by:
- Points to examine for extremal (maximum or minimum) behavior: .
Part (a): Determine Which Points Are Local Extrema
To determine which of these points are local extrema, we need to:
- Find the derivative of and set it to zero to find the critical points.
- Evaluate the second derivative at each critical point or use the first derivative test to classify them as local minima, maxima, or neither.
Step 1: Find the Derivative of
The first derivative is: Setting to find the critical points: Thus, is a critical point of .
Step 2: Evaluate the Second Derivative
The second derivative of is: Since for all , the function is concave up everywhere, indicating that is a local minimum.
Conclusion for Part (a)
Among the points , only is a local extremum (specifically, a local minimum).
Part (b): Find the Local Extremum Value
Since we have identified as a local minimum, we evaluate at to find the extremal value:
Summary
- Part (a): The point is a local extremum.
- Part (b): The local minimum value at is .
Would you like further details on the solution steps?
Here are 5 related questions for deeper understanding:
- How would you find local extrema for more complex functions?
- What changes if had a different leading coefficient, e.g., ?
- How does the second derivative test help determine concavity?
- What if we had to find global extrema on a closed interval?
- How can the first derivative test be used as an alternative to the second derivative test?
Tip: Always check the concavity using the second derivative, especially if you need to classify the nature of critical points accurately.
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivative
Critical Points
Second Derivative Test
Formulas
f(x) = x^2 + 2x
First derivative: f'(x) = 2x + 2
Setting f'(x) = 0 to find critical points
Second derivative: f''(x) = 2
Theorems
Second Derivative Test
Suitable Grade Level
Grades 10-12
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