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

408 Chapter 12 Applications of differentiation
Example12.1 Determinethepositionandnatureofallmaximumandminimumpointsofthefollowing
functions:
(a)y=x 2
(b)y=−t 2 +t+1
(c)y= x3
3 + x2
2 −2x+1
(d) y = |t|
Solution (a) Ify =x 2 , then by differentiation dy
dx = 2x.
Recall thatatmaximum and minimum points either
(i)
(ii)
dy
does notexist, or
dx
dy
= 0.We mustcheck both of these conditions.
dx
The function 2x exists for all values of x, and so we move to examine any points
where dy = 0.So, wehave
dx
dy
dx =2x=0
Theequation2x = 0hasonesolution,x = 0.Weconcludethataturningpointexists
atx = 0. Furthermore, from the given functiony =x 2 , we see that whenx = 0 the
valueofyisalso0,soaturningpointexistsatthepointwithcoordinates (0,0).To
determine whether this point is a maximum or minimum we use the first-derivative
test and examine the sign of dy on either side ofx = 0. To the left ofx = 0,xis
dx
clearly negative and so 2x is also negative. To the right ofx = 0,xis positive and
so2xisalsopositive.Henceyhasaminimumatx = 0.Agraphofy =x 2 showing
thisminimum isgiven inFigure 12.2.
(b) Ify=−t 2 +t + 1, theny ′ = −2t + 1 and this function exists for all values oft.
Solvingy ′ = 0 we have
−2t+1=0 andsot= 1 2
Weconcludethatthereisaturningpointatt = 1 2 .Theycoordinatehereis − ( 1
2) 2
+
1
2 +1=11 4 . We now inspect the sign ofy′ to the left and to the right oft = 1 2 . A
little to the left, say att = 0, we see thaty ′ = −2(0) +1 = 1 which is positive. A
littletotheright,sayatt = 1,weseethaty ( ′ = −2(1) +1 = −1whichisnegative.
1
Hence thereisamaximum atthe point
2 4)
,11 .
Agraph ofthe function isshown inFigure 12.3.