Math Problem Statement
Solution
We are given the function:
We need to find the local minimum and local maximum. Let's start by finding the critical points where the derivative .
Step 1: Find the first derivative of
Step 2: Solve to find critical points
Dividing the equation by 6: Now, factor the quadratic equation: So, the critical points are and .
Step 3: Determine whether these points are minima or maxima
We use the second derivative test.
Find the second derivative:
Now evaluate at the critical points:
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For : Since , there is a local maximum at .
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For : Since , there is a local minimum at .
Step 4: Find the values of the function at these points
To find the actual values at the local minimum and maximum, substitute and into the original function .
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For : So, the local maximum value is .
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For : So, the local minimum value is .
Final Answer:
- The local minimum is at with value .
- The local maximum is at with value .
Would you like a more detailed explanation or further clarification on any steps?
Here are 5 related questions you could explore:
- What is the graphical interpretation of local maxima and minima?
- How do the critical points relate to the shape of the graph of the function?
- What happens if both the first and second derivatives at a critical point are zero?
- How does one determine global maxima and minima on a closed interval?
- How can we apply this method to functions with more complex terms?
Tip: Always remember that a second derivative greater than zero indicates concavity upwards (local minimum), while less than zero indicates concavity downwards (local maximum).
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Math Problem Analysis
Mathematical Concepts
Calculus
Critical Points
Maxima and Minima
Formulas
f'(x) = 6x^2 - 54x + 48 (First Derivative)
f''(x) = 12x - 54 (Second Derivative)
Quadratic formula: ax^2 + bx + c = 0
Theorems
First derivative test to find critical points
Second derivative test to determine concavity (local minima/maxima)
Suitable Grade Level
Grades 10-12 or early college level
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