Math Problem Statement
Solution
The given function is . We found the critical points by solving the first derivative and determining the concavity with the second derivative. Here are the results:
- The critical points are and .
- The second derivative test shows that at , the second derivative is negative, indicating a local maximum. At , the second derivative is positive, indicating a local minimum.
- Evaluating the function at these points gives approximate values of:
- (local maximum)
- (local minimum)
Therefore, the local maximum occurs at the point , which corresponds to option (D).
Would you like more details or have any questions?
Here are 5 related questions to expand on this:
- How do you find critical points of a function?
- What does the second derivative test indicate?
- What is the significance of concavity in determining local maxima or minima?
- How would you graph a cubic function like this one?
- How can you confirm whether a point is a local maximum or minimum graphically?
Tip: Always check the second derivative to confirm whether a critical point is a maximum or minimum!
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Math Problem Analysis
Mathematical Concepts
Calculus
Critical Points
Second Derivative Test
Polynomial Functions
Formulas
First derivative: dy/dx = 3x^2 - 4x - 5
Second derivative: d²y/dx² = 6x - 4
Theorems
Second Derivative Test
Suitable Grade Level
Grades 11-12 (Calculus)
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