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

416 Chapter 12 Applications of differentiation
y
y
A
A
B
(a)
x
Figure12.10
(a) There isapoint ofinflexion atA; (b)there are pointsofinflexion at A
and B.
(b)
y
x
y = x 4 x
y
y = x 3
y'' > 0
y'' = 0
x
y' < 0, y'' > 0 y' > 0, y'' > 0
y'' < 0
Figure12.11
The second derivative,y ′′ ,changes
sign atx = 0.
Figure12.12
Thederivative,y ′ ,changessignatx = 0,but
y ′′ remainspositive.
Example12.3 Locateany points ofinflexion ofthe curvey =x 3 .
Solution Given y = x 3 , then y ′ = 3x 2 and y ′′ = 6x. Points of inflexion can only occur where
y ′′ = 0 or does not exist. Clearlyy ′′ exists for allxand is zero whenx = 0. It is possible
that a point of inflexion occurs whenx = 0 but we must examine the concavity of
the curve on either side. To the left ofx = 0,xis negative and soy ′′ is negative. Hence
to the left, the curve is concave down. To the right of x = 0, x is positive and so y ′′
is positive. Hence to the right, the curve is concave up. Thus the concavity changes at
x = 0.Weconcludethatx = 0isapointofinflexion.AgraphisshowninFigure12.11.
Note thatatthis pointofinflexiony ′ = 0 too.
Acommonerroristostatethatify ′ =y ′′ = 0thenthereisapointofinflexion.This
isnotalways true;considerthe next example.
Example12.4 Locateall maximumpoints,minimum points and points ofinflexion ofy =x 4 .
Solution y ′ = 4x 3 y ′′ = 12x 2