3.4 Concavity and the Second Derivative Test

In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator or a computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs. First derivative: is positive Curve is rising. is negative Curve is falling. is zero Possible local maximum or minimum.

Concavity of a Function As you look at the graph of a function … … if the function CURVES UP, like a cup, we say the function is _______________. …if the function CURVES DOWN, like a frown, we say the function is _______________. CONCAVE UP CONCAVE DOWN

++–– What do their eyes mean???

Second derivative: is positive Curve is concave up. is negative Curve is concave down. is zero Possible inflection point (where concavity changes). ++–– is positive is negative

Example 1 Graph There are roots at and. Set First derivative test: negative positive We can use a chart to organize our thoughts. Possible local extrema at x = 0, 2.

Example 1 Graph There are roots at and. Set First derivative test: maximum at minimum at Possible local extrema at x = 0, 2.

Local Maximum at x = 0 Local Minimum at x = 2 f ’’(x) = 6x – 6 = 6(x – 1) f ’’(2) = 6(2 – 1)= 6> 0 f ’’(0) = 6(0 – 1) = –6<0

Theorem 3.7 Test for Concavity

Example 2 Graph First derivative test: NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation! There is a local maximum at (0,4) because for all x in and for all x in (0,2). There is a local minimum at (2,0) because for all x in (0,2) and for all x in.

Because the second derivative at x = 0 is negative, the graph is concave down and therefore (0,4) is a local maximum. Example 2 Graph There are roots at and. Or you could use the second derivative test: Because the second derivative at x = 2 is positive, the graph is concave up and therefore (2,0) is a local minimum. Possible local extrema at x = 0, 2.

Example 3 Graph There are one zero ( x = 0 ) for f ’ ( x ) = 0, and there are no the zeros for f ’’ (x) = 0, but f ( x ) is not continuous at x = ±2. Interval –∞ < x <–2–2< x < 22 < x < +∞ Test Value x = –3x = 0x = 3 Sign of f ’’ ( x ) f ’’( – 3)> 0f ’’(0) < 0f ’’(3)> 0 Conclusion Concave UpConcave downConcave Up

do not exist.

Definition of Point of Inflection

Theorem 3.8 Points of Inflection Can you give an example (or draw a sketch of a graph) for why the point of inflection could occur where f ’’(c) does not exist?

inflection point at There is an inflection point at x = 1 because the second derivative changes from negative to positive. Example 4 Graph We then look for inflection points by setting the second derivative equal to zero. Possible inflection point at. negative positive

Make a summary table: rising, concave down local max falling, inflection point local min rising, concave up

Example 5 Determine the points of inflection and discuss the concavity of the graph of Solution Taking the 1 st and 2 nd derivative: Setting f ’’(x) = 0 to find the zeros of f ’’(x) is x = 0 and x = 2: Interval –∞ < x < 00< x < 22 < x < +∞ Test Value x = –1x = 1x = 3 Sign of f’ ’ ( x ) f ’’(–1)> 0 f ’’(1) < 0f ’’(3)> 0 Conclusion Concave UpConcave downConcave Up Points of inflection (x^4-4x^3)/15

Theorem 3.9 Second Derivative Test The second derivative can be used to perform a simple test for the relative min. and max. The test is based on f ’(c) = 0.

Example 6 Find the relative extrema for Solution Taking the 1 st and 2 nd derivative. The zeros of 1 st derivative are:

Example 6 Point (–2, 0)(1, –81/8)(4, 0) Sign of f ’’ ( x ) f ’’(–2)=–9 < 0f ’’(1) = 9/2 > 0f ’’(3)=–9 < 0 Conclusion Relative MaximumRelative MinimumRelative Maximum Relative Minimum

Setting f ’’(x) = 0 to find the zeros of f ’’(x): Example 6 Find the points of inflection Points of Inflection Interval –∞< x < 1– 3 1/2 1– 3 1/2 < x < / /2 < x < +∞ Sign of f ’’ ( x ) f ’’(x) < 0f ’’(x) > 0f ’’(x) < 0 Conclusion Concave DownConcave UpConcave Down

Homework Pg odds, 29, 35, 37, 48, odds, 69