Math Problem Statement
Solution
The problem is to find the sum of the minimum and maximum values of the function over the interval .
Step 1: Analyze the function
The given function is a quadratic function:
The coefficient of is negative (), so the parabola opens downward. This means the function has a maximum value at its vertex, and the minimum values will occur at the endpoints of the interval.
Step 2: Find the critical point
The critical point is where the derivative of is zero. Compute the derivative:
Set :
The critical point is . Check if this point lies in the interval . Since is within , it is a valid candidate for the maximum or minimum value.
Step 3: Evaluate at the critical point and endpoints
Now calculate at , , and :
-
At :
-
At :
-
At :
Step 4: Identify the maximum and minimum values
From the calculations:
- Maximum value:
- Minimum value:
Step 5: Compute the sum
The sum of the maximum and minimum values is:
Final Answer:
The sum of the maximum and minimum values of over the interval is:
Additional Related Questions:
- How do we determine whether a quadratic function has a maximum or minimum value?
- What does the vertex of a parabola represent in terms of optimization?
- How does the interval of evaluation impact the extrema of a function?
- Why is it necessary to evaluate endpoints in addition to critical points for optimization?
- Can we use a different method (e.g., completing the square) to find the maximum value of ?
Tip: For optimization problems on a closed interval, always evaluate the function at critical points and the endpoints, as the extrema could occur at any of these points.
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Optimization
Calculus
Formulas
Derivative: f'(x) = -4x - 4
Vertex formula for a parabola: x = -b/(2a)
Theorems
Extreme Value Theorem (ensures maximum and minimum values on a closed interval)
Suitable Grade Level
Grades 10-12
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