Higher Engineering Mathematics

SOME APPLICATIONS OF DIFFERENTIATION 303

G

Figure 28.5

(ii) Let

dy

dx

=0 and solve for the values ofx.

(iii) Substitute the values of x into the original
equation, y=f(x), to find the correspond-
ing y-ordinate values. This establishes the
co-ordinates of the stationary points.

To determine the nature of the stationary points:

Either

(iv) Find

d^2 y

dx^2

and substitute into it the values ofx

found in (ii).

If the result is:

(a) positive—the point is a minimum one,

(b) negative—the point is a maximum one,

(c) zero—the point is a point of inflexion

or

(v) Determine the sign of the gradient of the curve

just before and just after the stationary points.

If the sign change for the gradient of the

curve is:

(a) positive to negative—the point is a max-

imum one

(b) negative to positive—the point is a min-

imum one

(c) positive to positive or negative to negative—

the point is a point of inflexion

Problem 10. Locate the turning point on the

curvey= 3 x^2 − 6 xand determine its nature by

examining the sign of the gradient on either side.

Following the above procedure:

(i) Sincey= 3 x^2 − 6 x,

dy

dx

= 6 x−6.

(ii) At a turning point,

dy

dx

=0. Hence 6x− 6 =0,

from which,x=1.

(iii) Whenx=1,y=3(1)^2 −6(1)=−3.

Hence the co-ordinates of the turning point

are (1,−3).

(iv) Ifxis slightly less than 1, say, 0.9, then

dy

dx

=6(0.9)− 6 =− 0 .6,

i.e. negative.

Ifxis slightly greater than 1, say, 1.1, then

dy

dx

=6(1.1)− 6 = 0 .6,

i.e. positive.

Since the gradient of the curve is negative just

before the turning point and positive just after

(i.e.−∨+), (1,− 3 )is a minimum point.

Problem 11. Find the maximum and minimum

values of the curvey=x^3 − 3 x+5by

(a) examining the gradient on either side of the

turning points, and

(b) determining the sign of the second

derivative.

Sincey=x^3 − 3 x+5 then

dy

dx

= 3 x^2 − 3

For a maximum or minimum value

dy

dx

= 0

Hence 3x^2 − 3 =0, from which, 3x^2 =3 andx=± 1

Whenx=1,y=(1)^3 −3(1)+ 5 = 3

Whenx=−1,y=(−1)^3 −3(−1)+ 5 = 7

Hence (1, 3) and (−1, 7) are the co-ordinates of the

turning points.

(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.