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

410 Chapter 12 Applications of differentiation
y
dy
—
dx > 0
dy
—
dx = 0
Figure12.5
The derivative dy decreases on
dx
passingthrough amaximum point.
dy
—
dx < 0
x
y
dy
—
dx < 0
dy
—
dx = 0
dy
—
dx > 0
Figure12.6
Thederivative dy increases onpassing
dx
through aminimum point.
x
Rather than examine the sign of y ′ on both sides of the point, a second-derivative
test may be used. On passing through a maximum pointy ′ changes from positive to 0
to negative, as shown in Figure 12.5. Hence,y ′ is decreasing. Ify ′′ is negative then this
indicates y ′ is decreasing and the point is therefore a maximum point. Conversely, on
passingthroughaminimumpoint,y ′ increases,goingfromnegativeto0topositive(see
Figure 12.6). Ify ′′ ispositive theny ′ isincreasing and this indicates a minimum point.
So,havinglocatedthepointswherey ′ = 0,welookatthesecondderivative,y ′′ .Thus
y ′′ > 0 implies a minimum point;y ′′ < 0 implies a maximum point. Ify ′′ = 0, then we
must return to the earlier, more basic test of examiningy ′ on both sides of the point. In
summary:
Thesecond-derivative test todistinguish maximafrom minima:
Ify ′ = 0 andy ′′ < 0 atapoint,thenthisindicates thatthe pointisamaximum
turningpoint.
Ify ′ = 0 andy ′′ > 0 atapoint,thenthisindicates thatthe pointisaminimum
turningpoint.
Ify ′ = 0 andy ′′ = 0 atapoint,the second-derivative testfailsand you must
use the first-derivative test.
Example12.2 Usethesecond-derivativetesttofindallmaximumandminimumpointsofthefunctions
inExample 12.1.
Solution (a) Given y = x 2 then y ′ = 2x and y ′′ = 2. We locate the position of maximum and
minimum points by solvingy ′ = 0 and so such a point exists atx = 0. Evaluating
y ′′ atthispointweseethaty ′′ (0) = 2whichispositive.Usingthesecond-derivative
testweconcludethatthe pointisaminimum.
(b) Giveny = −t 2 +t + 1 theny ′ = −2t + 1 andy ′′ = −2. Solvingy ′ = 0 we find
t = 1 ( ) 1
2 . Evaluatingy′′ at this point we findy ′′ = −2 which is negative. Using
2
the second-derivative testwe conclude thatt = 1 isamaximum point.
2
(c) Given y = x3
3 + x2
2 −2x+1,theny′ =x 2 +x−2andy ′′ =2x+1.y ′ =0
at x = 1 and x = −2. At x = 1, y ′′ = 3 which is positive and so the point