082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

12.3 Points of inflexion 417
y ′ = 0 atx = 0. Alsoy ′′ = 0 atx = 0 and so the second-derivative test is of no help
in determining the position of maximum and minimum points. We return to examiney ′
on both sides ofx = 0. To the left ofx = 0,y ′ < 0; to the righty ′ > 0 and sox = 0 is
a minimum point. Figure 12.12 illustrates this. Note that at the pointx = 0, the second
derivativey ′′ is zero. However,y ′′ is positive both to the left and to the right ofx = 0;
thusx = 0 isnot a point of inflexion.
Example12.5 Find any maximum points,minimum points and points of inflexion ofy =x 3 +2x 2 .
Solution Giveny =x 3 +2x 2 theny ′ = 3x 2 +4xandy ′′ = 6x +4.Letusfirstfindanymaximum
andminimumpoints.Thefirstderivativey ′ iszerowhen3x 2 +4x =x(3x+4) = 0,that
iswhenx=0orx=− 4 3 . Using the second-derivative test we findy′′ (0) = 4 which
(
corresponds to a minimum point. Similarly,y ′′ − 4 )
= −4 which corresponds to a
3
maximum point.
We seek points of inflexion by looking for points wherey ′′ = 0 and then examining
the concavity on either side.y ′′ = 0 whenx = − 2 3 .
Sincey ′′ is negative whenx < − 2 3 , theny′ is decreasing there, that is the function is
concavedown.Also,y ′′ ispositivewhenx > − 2 3 andsoy′ isthenincreasing,thatisthe
functionisconcave up. Hence thereisapointofinflexion whenx = − 2 .Thegraph of
3
y =x 3 +2x 2 isshown inFigure 12.13.
y
Figure12.13
– 4–
3
– 2–
3
x
There isamaximum atx = − 4 ,aminimum at
3
x = 0 and apoint ofinflexion atx = − 2 3 .
FromExamples12.4 and 12.5 wenotethat:
(1) The conditiony ′′ = 0 is not sufficient to ensure a point is a point of inflexion.
The concavity of the function on either side of the point wherey ′′ = 0 must be
considered.
(2) At a pointofinflexion itis notnecessarytohavey ′ = 0.
(3) At a pointofinflexiony ′′ = 0 ory ′′ doesnotexist.