Extremum & Inflection Finding and Confirming the Points of Extremum & Inflection.

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Presentation transcript:

Extremum & Inflection

Finding and Confirming the Points of Extremum & Inflection

InflectionExtremum Candidate Points for Inflection Points at which the second derivative is equal 0 or does not exist. Critical Points Points at which the first derivative is equal 0 or does not exist. I The second Derivative Test for inflection Testing the sign of the second derivative about the point The First Derivative Test for Local Extremum Testing the sign of the first derivative about the point II The Third Derivative Test for Inflection Testing the sign of the third derivative at the point The Second Derivative Test for Local Extremum Testing the sign of the second derivative at the point III

Extremum Critical Points We find critical points, which include any point ( x 0, f((x 0 ) ) of f at which either the derivative f’ (x 0 ) equal 0 or does not exist. The First Derivative Test for Local Extremum For each critical point ( x 0, f((x 0 ) ) we examine the sign of the first derivative f’ (x) on the immediate left and the immediate right of this point x 0. If there is a change of sign at x 0, then the point ( x 0, f((x 0 ) ) is a point of local extremum, with the extremum being: (1) A local maximum if the sign of f’ (x) is positive on the immediate left and negative on the immediate right of the critical point x 0 (2) A local minimum if the sign of f’ (x) is negative on the immediate left and positive on the immediate right of the critical point I II.

Extremum The Second Derivative Test for Local Extremum Only for critical points ( x 0, f((x 0 ) ) for which the first derivative f’ (x 0 ) = 0 (1) If f’’ (x 0 ) is negative, then the point ( x 0, f(x 0 ) ) is a point of local maximum (2) If f’’ (x 0 ) is positive, then the point ( x 0, f(x 0 ) ) is a point of local minimum III.

Inflection Candidate Points We find candidate points, which include any point (x 0,y 0 ) of f at which either the second derivative f’’ (x 0 ) equal 0 or does not exist The Second Derivative Test for Inflection For each candidate point ( x 0, f((x 0 ) ) we examine the sign of the second derivative f’’ (x) on the immediate left and the immediate right of this point x 0. If there is a change of sign at x 0, then the point ( x 0, f((x 0 ) ) is a point of inflection, at which: (1) The graph of f changes from concave upward to concave downward if the sign of second derivative f’’ (x) changes from positive on the immediate left to negative on the immediate right of the candidate point x 0 (2) The graph of f changes from concave downward to concave upward if the sign of second derivative f’’ (x) changes from negative on the immediate left to positive on the immediate right of the candidate point x 0 I II.

Inflection The Third Derivative Test for Inflection Only for candidate points ( x 0, f( ( x 0 ) ) for which the second derivative f’’ (x 0 ) = 0 (1) If f’’’ (x 0 ) is negative, then the point ( x 0, f( ( x 0 ) ) is a point of inflection at which the graph of f changes from concave upward to concave downward (2) If f’’ (x 0 ) is positive,, then the point ( x 0, f( ( x 0 ) ) is a point of inflection at which the graph of f changes from concave downward to concave upward III.

One Page Summary Word Doc : PDF:

Example Let: f(x) = 2x 3 – 9x 2 +12x + 1 Determine all points of extremum and inflection of the function f and use this information and other information to sketch its graph.

Solution

Graphing f

Graph of f

f’f’

Graphing g=f ’

Graph of g=f ’

Question? Home Quiz (1) Can you graph g without using the derivatives?

f ’’

f ’’’

The relation between f and f ’

The relation between f and f ’’

The relation between f and f ’’’

The Four Graphs graphs

Homework (1)

Homework (2) For each of the functions f of the previous homework (1) a. Determine the intervals on which f is increasing or decreasing. b. Determine the intervals on which f is concave upward or concave downward.