Find global min and max of quadratic functions on a closed interval
f(x) = ax^2 + bx + c on interval [L, R]
f(x)=x^2 +x +, interval [,]
Result
Absolute Min
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Absolute Max
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Derivation
Absolute Extrema Method
1. Find critical points: f(x)=0 inside [L,R]
2. Evaluate f at all critical points
3. Evaluate f at endpoints x=L and x=R
4. Largest = max, Smallest = min
Absolute extrema are the highest and lowest function values on a given interval. The Extreme Value Theorem guarantees their existence for continuous functions on closed intervals. Compare candidate values from critical points and endpoints.
⚠Always check endpoints! The absolute extremum may occur at an endpoint rather than a critical point inside the interval.
What Are Absolute Extrema?
The absolute maximum is the largest y-value of the function on the interval. The absolute minimum is the smallest. To find them: evaluate the function at critical points (where f=0) and endpoints, then compare all values.
Critical Points
Where f(x)=0. For quadratics: f=2ax+b, critical point at x=-b/(2a). Candidates for extrema.
Endpoints
Evaluate f at both ends of the interval. The extremum can occur at boundaries, especially for monotonic functions.
Comparison
List all candidate y-values. The largest is absolute max. The smallest is absolute min. Both exist by EVT.
Extreme Value Theorem
Continuous function on [a,b] always attains global max and min. This theorem justifies the candidate comparison method.
Teaching Example: f(x)=x^2-4x+3 on [0,5]. f=2x-4=0 at x=2. f(0)=3, f(2)=-1, f(5)=8. Absolute min = -1 at x=2. Absolute max = 8 at x=5.
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