Cambridge Additional Mathematics

340 Introduction to differential calculus (Chapter 13)

EXERCISE 13B

1 Use the method inDiscovery 1to answer theOpening Problemon page 334.

2aUse the method inDiscovery 2to find the gradient of the tangent to y=x^2 at the point (2,4).

b Evaluate lim

x! 2

x^2 ¡ 4

x¡ 2

, and provide a geometric interpretation of this result.

For a non-linear function with equation y=f(x), the gradients of

the tangents at various points are different.

Our task is to determine a gradient functionwhich gives the

gradient of the tangent to y=f(x) at x=a, for any pointain

the domain off.

The gradient function of y=f(x) is called itsderivative functionand is labelled f^0 (x).

We read the derivative function as “eff dashedx”.

The value of f^0 (a) is the gradient of the tangent to y=f(x) at the point where x=a.

Example 2 Self Tutor

For the given graph, find f^0 (4) and f(4).

The graph shows the tangent to the curve y=f(x) at the point where x=4.

The tangent passes through (2,0) and (6,4), so its gradient is f^0 (4) =^4 ¡^0

6 ¡ 2

=1.

The equation of the tangent is y¡0=1(x¡2)

) y=x¡ 2

When x=4, y=2, so the point of contact between the tangent and the curve is (4,2).

) f(4) = 2

C The derivative function

x

y

y = f(x)

O

2 4 x

y

(6 4),

y = f(x)

O

cyan magenta yellow black

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 4037 Cambridge
Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_13\340CamAdd_13.cdr Friday, 4 April 2014 5:24:55 PM BRIAN