Higher Engineering Mathematics, Sixth Edition

304 Higher Engineering Mathematics

It is possible to have a turning point, the gradient on

either side of which is the same. Such a point is given

the special name of apoint of inflexion, and examples

are shown in Fig. 28.5.

0

y

x

Maximum

point

Minimum point

Maximum

point

Points of

inflexion

Figure 28.5

Maximum and minimum points and points of

inflexion are given the general term ofstationary

points.

Procedure for finding and distinguishing between

stationary points:

(i) Giveny=f(x), determine

dy

dx

(i.e. f′(x))

(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 correspondingy-

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 signof thegradient ofthe curve just

before and just after the stationary points. If the

sign change for the gradient of the curve is:

(a) positivetonegative—thepointisamaximum

one,

(b) negative to positive—thepoint is a minimum

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+5then

dy

dx

= 3 x^2 − 3

For a maximum or minimum value

dy

dx

= 0

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