Basic Engineering Mathematics, Fifth Edition

314 Basic Engineering Mathematics

3. Givenf(x)= 3 x^3 + 2 x^2 − 3 x+2, prove that
   f( 1 )=

1

7

f( 2 ).

34.3 The gradient of a curve

If a tangent is drawn at a pointPon a curve, the gradient

of this tangent is said to be thegradient of the curve

atP. In Figure 34.1, the gradient of the curve atPis

equal to the gradient of the tangentPQ.

0 x

Q

P

f(x)

Figure 34.1

For the curve shown in Figure 34.2, let the points

A and B have co-ordinates (x 1 ,y 1 )and(x 2 ,y 2 ),

respectively. In functional notation,y 1 =f(x 1 )and

y 2 =f(x 2 ),asshown.

0

B

A

E D

C f(x^2 )

f(x 1 )

f(x)

x 1 x 2 x

Figure 34.2

The gradient of the chordAB

=

BC

AC

=

BD−CD

ED

=

f(x 2 )−f(x 1 )

(x 2 −x 1 )

For the curvef(x)=x^2 shown in Figure 34.3,

(a) the gradient of chordAB

=

f( 3 )−f( 1 )

3 − 1

=

9 − 1

2

= 4

0 1 1.5 2 3

2

4

6

8

10

f(x)

x

A D

C

B f(x) 5 x 2

Figure 34.3

(b) the gradient of chordAC

=

f( 2 )−f( 1 )

2 − 1

=

4 − 1

1

= 3

(c) the gradient of chordAD

=

f( 1. 5 )−f( 1 )

1. 5 − 1

=

2. 25 − 1

0. 5

= 2. 5

(d) ifEis the pointon the curve( 1. 1 ,f( 1. 1 ))then the

gradient of chordAE

=

f( 1. 1 )−f( 1 )

1. 1 − 1

=

1. 21 − 1

0. 1

= 2. 1

(e) ifFis the point on the curve( 1. 01 ,f( 1. 01 ))then

the gradient of chordAF

=

f( 1. 01 )−f( 1 )

1. 01 − 1

=

1. 0201 − 1

0. 01

= 2. 01

Thus, as pointBmoves closer and closer to pointA,

the gradient of the chord approaches nearer and nearer

to the value 2. This is called thelimiting valueof the

gradient of the chordABand whenBcoincides withA

the chord becomes the tangent to the curve.

Now try the following Practice Exercise

PracticeExercise 132 The gradient of a

curve (answers on page 354)

1. Plot the curve f(x)= 4 x^2 −1forvalues
   of x fromx=−1tox=+4. Label the
   co-ordinates ( 3 ,f( 3 )) and ( 1 ,f( 1 )) as J
   andK, respectively. Join pointsJ andKto
   form the chordJK. Determine the gradient of