Math Problem Statement
Solution
To identify the point where the local maximum occurs for the function
we need to follow these steps:
-
Find the First Derivative: Calculate to find the critical points by setting .
-
Set the Derivative Equal to Zero: Solve for .
This is a quadratic equation in , which can be solved using the quadratic formula:
where , , and .
-
Calculate the Values of :
Simplifying inside the square root:
This gives two solutions:
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Determine Which is a Local Maximum: Use the second derivative test or plug these values into the original function to see which gives a maximum.
-
Evaluate the Function at Each Critical Point: Calculate and , then identify the local maximum.
Let me complete these calculations to confirm the exact values and find the local maximum point.The critical points are approximately:
- with
- with
Since is greater, the local maximum occurs at the point:
Let me know if you'd like further details on this solution or have any questions!
Here are some related questions for further practice:
- How do you find local minima using derivatives?
- What is the significance of the second derivative in determining concavity?
- Can a cubic function have more than one local maximum or minimum?
- How would the result change if the coefficients of the function were altered?
- What happens to the maximum point if we apply a vertical shift to the function?
Tip: When solving for critical points, always check the function values to distinguish between maximum and minimum points.
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivative
Critical Points
Local Maximum
Formulas
First derivative of the function y' = d/dx (3x^3 + 7x^2 - 5x - 4)
Quadratic formula for finding roots: x = (-b ± sqrt(b^2 - 4ac)) / 2a
Theorems
First Derivative Test
Second Derivative Test
Suitable Grade Level
Grades 11-12
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