90 likes | 218 Views
4.1 – Extrema on an Interval. Maximum and Minimum Values. Absolute Maximum f ( c 1 ) f ( x ) for all x. Local Maximum f ( c 2 ) f ( x ) for all x in I . •. •. | c 2. | c 1. I. Maximum and Minimum Values. Absolute Minimum ( f ( c 1 ) f ( x ) for all x ). c 2 |.
E N D
Maximum and Minimum Values Absolute Maximum f (c1) f (x) for all x Local Maximum f (c2) f (x) for all x in I • • |c2 |c1 I
Maximum and Minimum Values Absolute Minimum ( f (c1) f(x) for all x) c2| |c1 • I I • Collectively, maximum and minimum values are called extreme values. Local Minimum ( f (c2) f(x) for all x in I )
Definitions A function f has an absolute (global) maximum at c if f (c) ≥ f (x) for all x in the domain. A function f has an absolute (global) minimum at c if f (c) ≤ f (x) for all x in the domain. The maximum and minimum values are called extreme values. A function f has an local (relative) maximum at c if f (c) ≥ f (x) where x is in a small open interval about c. A function f has an local (relative) minimum at c if f (c) ≤ f (x) where x is in a small open interval about c.
[ ] ● ● The Extreme Value Theorem If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f (c) and absolute minimumf (d) at some numbers c and d in [a, b]
Fermat’s Theorem If f has a local maximum or minimum at c, and if f'(c) exists, then f'(c) = 0. Question If a (local) maximum or minimum occur at c, then what is the value of f '(c)?
Critical Number or Value A critical number of a function f is a number c in the domain of f such that either f '(c) = 0 or f '(c) does not exist ( f (x) is not differentiable) Fact An absolute extremum occurs at two places: • Critical points • End points
Finding Absolute Extrema on a Closed Interval [a,b] 1. Find the critical numbers of f on (a, b). • Compute the value of f at each of • the critical numbers on (a,b) • the endpointsa and b. 3. The largest of these values is the absolute maximum. The smallest is the absolute minimum.
Examples Locate the absolute extrema of the function on the closed interval f(x) = x3 – 12x on [-3,4] g(x) = 4x / (x2+1) on [0,3] h(t) = 2 sec(t) - tan(t) on [0,π/4]