Presentation on theme: "Concavity and the Second Derivative Test"— Presentation transcript:
1Concavity and the Second Derivative Test Determine the intervals on which the graphs of functions are concave upward or concave downward.Find the points of inflection of the graphs of functions.Use the Second Derivative Test to find the relative extrema of functions.Standard 4.5a
2Concavity – the property of curving upward or downward Concave upwardf’ is increasing
4Definition of Concavity Let f be differentiable on an open interval I. The graph of f isConcave upward on I if f’ is increasing on the interval.Concave downward on I if f’ is decreasing on the interval.
5Test for ConcavityLet f be a function whose second derivative exists on an open interval I.If f ´´(x) > 0 for all x in I, then f is concave upward on I.If f ´´(x) < 0 for all x in I, then f is concave downward on I.
6Determine the intervals on which the graph is concave upward or concave downward. 1. Locate the x-values at which f ´´(x) = 0 or f ´´(x) is undefined.
7Use these x-values to determine the test intervals. Test the signs of f ´´(x) in each test interval.Interval(-∞, -√3)(-√3, √3)(√3, ∞)Test Valuesx = -2x = 0x = 2Sign of f ´´(x)f ´´(-2) > 0f ´´(0) < 0f ´´(2) > 0ConclusionConcave upwardConcave downward
9Defintion of Point of Inflection If the graph of a continuous function has a tangent line at point where the concavity changes from upward to downward (or vice versa) then the point is a point of inflection.
10Property of Points of Inflection If (c, f(c)) is a point of inflection of the graph of f, then either f ´´(c) = 0 or f ´´(c) is undefined at c.
11Find the points of inflection of the graph. Possible inflection point
12Inflection point (4, 16) (-∞, 4) (4, ∞) x = 0 x = 5 f ´´(0) < 0 Concave downConcave upInflection point (4, 16)
13It is possible for the second derivative to be zero at a point that is not a point of inflection. * You must test to be certain that the concavity actually changes.
14Find the points of inflection and discuss the concavity of the graph of the function.
15Possible points of inflections: x = 0, x = 3 (0, 3)(3, ∞)x = 1x = 4f ´´(1) > 0f ´´(4) < 0Concave upwardConcave downwardInflection Point:
16If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum Concave Upwardf ´´(c) > 0cIf f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
17If f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum Concave downwardf ´´(c) < 0cIf f ´(c) = 0 and f ´´(c) > 0, f (c) is a relative minimum
18Second – Derivative Test Let f ´(c) = 0 and let f ´´exist on an open interval containing c.If f ´´(c) > 0, then f(c) is a relative minimum.If f ´´(c) < 0, then f(c) is a relative maximum.If f ´´(c) = 0 then the test fails. Use the First Derivative Test.
19Find the relative extrema using the Second-Derivative Test 1. Find the critical numbers.Critical Numbers
20(0,0) is neither a relative max or a relative min 2. Find the second derivative.3. Plug the critical numbers into the second derivative to determine relative extrema.Relative minimum (-1, 2)Test failsRelative maximum (1, 2)(-1, 0)(0, 1)f ´(-1/2) >0f ´(1/2) >0Increasing(0,0) is neither a relative max or a relative min
21(0,1) is neither a relative max or a relative min Find the relative extrema using the Second-Derivative Test.Test failsRelative min (3,-26 )(-∞, 0)(0, 3)f ´(-1) < 0f ´(1) < 0Decreasing(0,1) is neither a relative max or a relative min