Higher Engineering Mathematics

304 DIFFERENTIAL CALCULUS

Since the gradient changes from negative to

positive,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

negative,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 be seen that the second 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−6 then
d^2 y
dx^2
= 2 x−1.
Whenx=−2,
d^2 y
dx^2
=2(−2)− 1 =−5,
which is negative.
Hence (−2, 9) is a maximum point.
Whenx=3,
d^2 y
dx^2
=2(3)− 1 =5,
which is positive.