Higher Engineering Mathematics, Sixth Edition

Some applications of differentiation 305

(a) Considering the point (1, 3):

Ifxis slightly less than 1, say 0.9, then

dy

dx

= 3 ( 0. 9 )^2 − 3 ,

which is negative.

Ifxis slightly more than 1, say 1.1, then

dy

dx

= 3 ( 1. 1 )^2 − 3 ,

which is positive.

Since the gradient changes from negative to posi-

tive,the point (1, 3) is a minimum point.

Considering the point (− 1 ,7):

Ifxis slightly less than−1, say−1.1, then

dy

dx

= 3 (− 1. 1 )^2 − 3 ,

which is positive.

Ifxis slightly more than−1, say−0.9, then

dy

dx

= 3 (− 0. 9 )^2 − 3 ,

which is negative.

Since the gradient changes from positive to nega-

tive,the point (−1, 7) is a maximum point.

(b) Since

dy

dx

= 3 x^2 −3, then

d^2 y

dx^2

= 6 x

Whenx=1,

d^2 y

dx^2

is positive, hence (1, 3) is a

minimum value.

Whenx=−1,

d^2 y

dx^2

is negative, hence (−1, 7) is

amaximum value.

Thus the maximum value is 7 and the min-

imum value is 3.

It can beseen that thesecond differential method of

determining the nature of the turning points is, in

this case, quicker than investigating the gradient.

Problem 12. Locate the turning point on the

following curve and determine whether it is a

maximum or minimum point:y= 4 θ+e−θ.

Since y= 4 θ+e−θ

then

dy

dθ

= 4 −e−θ= 0

for a maximum or minimum value.

Hence 4=e−θ,^14 =eθ,givingθ=ln^14 =− 1 .3863 (see

Chapter 4).

Whenθ=− 1 .3863,y= 4 (− 1. 3863 )+e−(−^1.^3863 )

= 5. 5452 + 4. 0000 =− 1. 5452

Thus (−1.3863,−1.5452) are the co-ordinates of the

turning point.

d^2 y

dθ^2

=e−θ.

Whenθ=− 1 .3863,

d^2 y

dθ^2

=e+^1.^3863 = 4. 0 ,

which is positive, hence (−1.3863,−1.5452) is a

minimum point.

Problem 13. Determine the co-ordinates of the

maximum and minimum values of the graph

y=

x^3

3

−

x^2

2

− 6 x+

5

3

and distinguish between

them. Sketch the graph.

Following the given procedure:

(i) Sincey=

x^3

3

−

x^2

2

− 6 x+

5

3

then

dy

dx

=x^2 −x− 6

(ii) At a turning point,

dy

dx

=0. Hence

x^2 −x− 6 =0, i.e.(x+ 2 )(x− 3 )=0,

from whichx=−2orx=3.

(iii) Whenx=−2,

y=

(− 2 )^3

3

−

(− 2 )^2

2

− 6 (− 2 )+

5

3

= 9

Whenx=3,

y=

( 3 )^3

3

−

( 3 )^2

2

− 6 ( 3 )+

5

3

=− 11

5

6

Thus the co-ordinates of the turning points are

(−2, 9) and

(

3,− (^1156)
)
.
(iv) Since
dy
dx
=x^2 −x−6then
d^2 y
dx^2
= 2 x−1.
Whenx=−2,
d^2 y
dx^2
= 2 (− 2 )− 1 =− 5 ,
which is negative.