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

12.2 Maximum points and minimum points 409
y
y = x 2
y
y = –t 2 + t + 1
1–
2
t
x
Figure12.2
Thefunctionyhas a minimumatx = 0.
Figure12.3
Thefunctionyhas amaximum att = 1 2 .
(c) Ify= x3
3 + x2
2 −2x +1,theny′ =x 2 +x−2andthisfunctionexistsforallvalues
ofx. Solvingy ′ = 0 wefind
x 2 +x−2=0
(x−1)(x+2)=0
x=1,−2
Therearethereforetwoturningpoints,oneatx = 1andoneatx = −2.Weconsider
each inturn.
Atx = 1,weexaminethesignofy ′ totheleftandtotherightofx = 1.Alittleway
to the left, say atx = 0, we see thaty ′ = −2 which is negative. A little to the right,
sayatx = 2,weseethaty ′ = 2 2 +2−2 = 4whichispositive.Sothepointwhere
x = 1 isaminimum.
Atx = −2, we examine the sign ofy ′ to the left and to the right ofx = −2. A little
way to the left, say atx = −3, we see thaty ′ = (−3) 2 + (−3) − 2 = 4 which is
positive.Alittletotheright,sayatx = −1,weseethaty ′ = (−1) 2 +(−1)−2 = −2
which isnegative. So the point wherex = −2 isamaximum.
Agraph of the function is shown inFigure 12.4.
(d) Recall that the modulus functiony = |t| isdefined as follows:
{ −t t0
y=|t|=
t t>0
A graph of this function was given in Figure 10.13(a) and this should be looked at
before continuing. Note that dy = −1 fort negative, and dy = +1 fort positive.
dt
dt
Thederivativeisnotdefinedatt = 0becauseofthecornerthere.Therearenopoints
when dy = 0. Because the derivative is not defined att = 0 this point requires
dt
further scrutiny. To the left oft = 0, dy dy
< 0; to the right,
dt dt >0andsot=0isa
minimum point.
y
y = —
x 3
+ —
x 2
– 2x + 1
3 2
–2
1
x
Figure12.4
Thefunctionyhas a maximum atx = −2
and aminimum atx = 1.