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- Fermat's theorem
- If a function ƒ(x) is defined on the interval (a, b), and there exists a point
- on the interval for which x = c such that a < c < b, then if the point
- (c, ƒ(c)) is a local extremum and ƒ′(c) exists, then (c, ƒ(c)) must be a
- critical point of ƒ(x) such that ƒ′(c) = 0.
- +------------------------------------------------------------------------------+
- | FIRST DERIVATIVE TEST |
- | |
- +------------------------------------------------------------------------------+
- | |
- | Suppose that c is a point such that the first derivative is 0, f '(c) = 0 |
- | |
- | If f' changes from positive to negative at c, then c is a local maximum. |
- | |
- | If f' changes from negative to positive at c, then c is a local minimum. |
- | If f' does not change at c, no minimum/maximum exists at c. |
- | |
- +------------------------------------------------------------------------------+
- | SECOND DERIVATIVE TEST |
- | |
- +------------------------------------------------------------------------------+
- | |
- | Let f '' is continuous near c. |
- | |
- | If f '' (c) > 0, there is a local minimum at c. |
- | |
- | If f '' (c) < 0, there is a local maximum at c. |
- | |
- +------------------------------------------------------------------------------+
- To find critical points, set the first derivative equal to zero and solve for
- the zeros.
- To find the extrema for the function f over the closed interval [a, b]:
- 1. Find the critical numbers of f in (a, b).
- 2. Evaluate f at each critical number found in Step 1 over (a, b).
- 3. Evaluate f at each end point of the interval [a, b].
- 4. The least of these values is the minimum and the greatest is the maximum.
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