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Analyzing f(x) and f’(x) /
Graphical Representations Among f(x) and f’(x)
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12. Finding Relative Extrema
The lowest or highest point on f(x) Set f’(x) equal to 0 Solve for x Create a sign chart for f’(x) to determine min or max ex: f(x) = x^2 f’(x) = 2x - + 2x = 0 X = 0 Max - Min
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13. Finding Absolute Extrema on a Closed Interval
The lowest or highest point on f(x) including endpoints Find the relative extrema Compare those points to your endpoints Ex: find the absolute min for x^2 on the interval [-2,2] f(x) = x^2 f’(x) = 2x X 2x = 0 X = 0 Y f(x) has an absolute min at x = 0. -2 2 4
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14. Intervals of Increasing/Decreasing
f(x) is Increasing if: f’(x) is positive f(x) is Decreasing if: f’(x) is negative
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15. Determining Concavity and Points of Inflection
f(x) is Concave Up if: f’’(x) is positive f(x) is Concave Down if : f’’(x) is negative f(x) has a point of inflection when: f’’(x) = 0 and changes signs. Ex: + + 2 - + 2
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16. Second Derivative Test
Substitute each critical point into the second derivative If: f’’>0 then there is a relative minimum f’’<0 then there is a relative maximum f’’=0 then the test fails
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Justifications pt.1 Justification Chart for increasing/decreasing intervals of functions Situation Justification f(x) is increasing on the interval from (a,b) f(x) is increasing on the interval from (a,b) because f’(x) > 0 f(x) is decreasing on the interval from (a,b) f(x) is decreasing on the interval from (a,b) because f’(x) < 0
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Justifications pt.2 Justifications of relative maximums/minimums and points of inflections Situation Justification Relative minimum f(x) has a relative minimum at x=a because f’(x) change points from negative to positive when x=a Relative maximum f(x) has a relative maximum at x=a because f’(x) changes signs from positive to negative when x=a Point of inflection f(x) has a point of inflection at x=a because f’’(x) changes signs when x=a
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Homework Analyzing F and the Derivative MC: 1, 6, 9-12 FRQ: 13-15
Graphical Representation Among F and the Derivative MC: 2, 3, 6-9,12 FRQ: 14-16
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