Math Problem Statement
4x−x2
Solution
Given the quadratic function , we can find the position of the maximum or minimum using calculus, following a similar procedure as before.
Step 1: Express the quadratic function
The quadratic function is:
This is in the standard form , where:
Step 2: Find the first derivative
To find the critical points, take the first derivative of with respect to :
Applying the power rule:
Step 3: Set the first derivative to zero
Set the first derivative equal to zero and solve for :
Step 4: Identify the nature of the critical point
To determine whether this critical point is a maximum or minimum, find the second derivative:
Since the second derivative is negative (), the function is concave down, indicating that the critical point is a maximum.
Step 5: Find the corresponding -coordinate
Substitute back into the original function to find the maximum value:
Conclusion
The maximum value of the function occurs at , and the maximum value is .
Would you like further details or have any questions?
Here are some related questions you might find interesting:
- How would you graph the function ?
- What is the significance of the vertex in the context of this problem?
- How does the quadratic function relate to physical problems like projectile motion?
- What are the intercepts of the function ?
- Can you express the function in vertex form?
- What happens to the maximum value if the coefficient of changes?
- How would the graph change if the function were ?
- Can the function have any other critical points?
Tip: When dealing with quadratic functions, the vertex is always a critical point, and it gives the maximum or minimum value depending on the sign of the leading coefficient.
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Math Problem Analysis
Mathematical Concepts
Quadratic Equations
Calculus
Critical Points
Maximum and Minimum
Formulas
Quadratic function standard form: y = ax^2 + bx + c
First derivative for critical points
Second derivative test for concavity
Theorems
Second Derivative Test
Suitable Grade Level
Grades 11-12