Ex: 3x 4 – 16x 3 + 18x 2 for -1 < x < 4 3.1 – Maximums and Minimums Global vs. Local Global = highest / lowest point in the domain or interval… Local =

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Presentation transcript:

ex: 3x 4 – 16x x 2 for -1 < x < – Maximums and Minimums Global vs. Local Global = highest / lowest point in the domain or interval… Local = high / low points relative to points NEAR a particular x value… Endpoints of an interval cannot be local… Local Global

Extreme Value Theorem “If f is continuous on a closed interval [a, b] then f has a global maximum and a global minimum at some numbers in that interval.” Extreme values can occur more than once. Absence of continuity or closed interval requirements negates the theorem. (An extreme value might not exist). Fermat’s Theorem “If f has a local maximum or minimum at c, and if f ’(c) exits, then f ’(c) = 0” The converse of this is false in general… f’(c) = 0 does NOT automatically guarantee a local max or min… ex: f(x) = x 3, f(x) = |x| We can extend this theorem to the idea of ‘critical numbers’. A critical number of a function f is a number c in the domain of f such that f ’(c) = 0 OR f ’(c) does not exist.

ex: Find the critical numbers of: Product Rule: Find values where the derivative = 0 or is undefined: f’(x) = 0 where the top = 0 f’(x) = undefined where the bottom = 0 Critical numbers are 0, 3/2

Re-wording of Fermat: “If f has a local maximum or minimum at c, then c is a critical number of f …” The Closed Interval Method Continuous functions on closed intervals always have a max and a min. (EVT) Global maximums or minimums can be local or at endpoints of an interval. If they are local, they are at critical numbers (Fermat) So to find global max/min we test any critical numbers and both endpoints of an interval in a function and compare the results. To find the global max/min of a continuous function f on a closed interval [a,b]: 1. Evaluate f at any critical numbers in [a,b]. 2. Evaluate f at the endpoints of the interval. 3. Compare the results – the largest value is the global max, the smallest value is the global min.

ex: Use calculus to find the exact minimum and maximum values of the following function: 1. Find critical values… * (the derivative of f is continuous on the interval, thus there are no critical numbers where the derivative is undefined…)

2. Evaluate critical values… 3. Evaluate end points… 4. Compare MAX MIN