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4.1 Maximum and Minimum Values

4.1 Maximum and Minimum Values. Maximum Values Local Maximum Absolute Maximum |c2|c2 |c1|c1 I

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Page 1: 4.1 Maximum and Minimum Values. Maximum Values Local Maximum Absolute Maximum |c2|c2 |c1|c1 I

4.1 Maximum and Minimum Values

Page 2: 4.1 Maximum and Minimum Values. Maximum Values Local Maximum Absolute Maximum |c2|c2 |c1|c1 I

Maximum Values

Local Maximum

Absolute Maximum

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Minimum Values

Local Minimum

Absolute Minimum

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Collectively, maximum and minimum values are called extreme values.

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DefinitionsA 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 a local (relative) maximum at c if f (c) ≥ f (x) where x is in a small open interval about c.

A function f has a local (relative) minimum at c if f (c) ≤ f (x) where x is in a small open interval about c.

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The Extreme Value Theorem

If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f (c) and an absolute minimum f (d) at some numbers c and d in [a, b].

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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)?

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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

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Finding Absolute Extrema on a Closed Interval [a,b]

1. Find the critical numbers of f on (a, b).

2. Compute the value of f at each of • the critical numbers on (a,b) • the endpoints a and b.

3. The largest of these values is the absolute maximum. The smallest is the absolute minimum.

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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]

4 3 2( ) 4 4 1 [0,2]f t t t t on 8 2

3 3( ) 16 3 [ 3,1]h x x x on