Maximum & Minimum Values; Mean Value Theorem

Front 1

Fermats Theorem

Back 1

Suppose (c, f(c)) is a local extreme value for f, & assume f is differentiable at c.
Then f '(c) = 0

Remark: Just because f '(c) = 0 doesn't mean that (c, f(c)) is a local extreme value

Front 2

Extreme Value Theorem

Back 2

Suppose f is continous on [a, b] then f attains an absolute max & absolute min on [a, b]

Front 3

critical number

Back 3

A critical number of a function f is a number x in the domain of f such that either f '(c) = 0 or
f '(c) does not exist.

note: if f has a local maximum or minimum at c, then c is a critical number of f.

Front 4

The Closed Interval Method - Finding the absolute maximum or minimum of a continuous function f on a closed interval [a,b]

Back 4

1. Find the values of f at the critical numbers of f in (a,b).
2. Find the values of f at the endpoints of the interval.
3. The largest of the values from Steps 1 and 2 is the absolute maximum; the smallest of these values is the absolute minimum value.

Front 5

Rolle's Theorem

Back 5

Let f be a function that satisfies the following three hypotheses
1. f is continuous on the interval [a,b]
2. f is differentiable on the open interval (a,b)
3. f(a) = f(b)
Then there is a number x in (a,b) such that f '(c) = 0

Front 6

The Mean Value Theorem

Back 6

Let f be a function that satisfies the following hypotheses:
1. f is continuous on the interval [a,b]
2. f is differentiable on the open interval (a,b)
Then there is a number c in (a,b) such that:
f '(c) = [f (b) - f (a)] / [b - a]