TC3 → Stan Brown → Calculus → Finding Relative Extrema
revised Jun 8, 2005

Finding Relative Extrema
(Lecture Notes for L-H-E Calculus 7e sections 3.3–3.4)

Copyright © 2002–2008 by Stan Brown, Oak Road Systems

Summary:  To find relative maxima and minima, first find the critical points (where f′ is 0 or doesn’t exist). Then examine each critical point. It is a relative maximum if f′ changes from positive to negative or f″ is negative. It is a relative minimum if f′ changes from negative to positive or f″ is positive.

3.3 / The First Derivative Test

page 
174definitions: increasing, decreasing, and constant functions
174Theorem 3.5: sign of f′ (negative, zero, positive) on open interval tells you direction of original function (decreasing, constant, increasing) on closed interval.
175to find all open intervals where function is increasing or decreasing, see Example 1 and “Guidelines” box: find critical numbers, divide interval into sub-intervals, test each for sign of f′
175definition: monotonic functions
176Theorem 3.6, the First Derivative Test, shows how to find relative max or min of a continuous and differentiable function:
  • Where f′ changes positive to negative, the graph has reached the top of the hill and started back down; therefore you have a relative maximum.
  • Where f′ changes negative to positive, the graph has reached the bottom of the valley and started back up; therefore you have a relative minimum.
  • If the function is continuous at a point but not differentiable at that single point, you can still apply the First Derivative Test.
How to remember this? Think shapes of graphs, and realize which change of sign of f′ is a hilltop and which is a valley.

3.4 / The Second Derivative Test

page 
184definitions: concave up holds water, concave down spills water
185Theorem 3.7: sign of f″ (positive, negative) on an open interval tells you concavity of f (up, down) on that interval.
186definition: point of inflection = where concavity changes, where graph crosses its tangent.
187Theorem 3.8: find possible points of inflection by looking where f″=0 or f′ does not exist. But you have an actual point of inflection only if the concavity changes there.
188Theorem 3.9, the Second Derivative Test, helps you weed through critical points (possible extrema) to find actual extrema based on sign of f″:
  • If f′(x)=0 and f″ is negative, the graph is concave down and you have a relative maximum.
  • If f′(x)=0 and f″ is positive, the graph is concave up and you have a relative minimum.
  • If f′(x)=f″(x)=0, you must use the First Derivative Test: look at sign changes in f′.

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