Cambridge Additional Mathematics

If the domain is restricted,

we need to check the value

of the function at the

endpoints of the domain.

Applications of differential calculus (Chapter 14) 379

Example 11 Self Tutor

Find the exact position and nature of the stationary point of y=(x¡2)e¡x.

dy

dx

= (1)e¡x+(x¡2)e¡x(¡1) fproduct ruleg

=e¡x(1¡(x¡2))

=

3 ¡x

ex

where ex is positive for allx

So,

dy

dx

=0when x=3.

The sign diagram of

dy

dx

is:

) at x=3we have a local maximum.

But when x=3, y= (1)e¡^3 =^1

e^3

) the local maximum is at (3,

1

e^3

).

6 Find the position and nature of the stationary point(s) of:

a y=xe¡x b y=x^2 ex c y=e

x

x

d y=e¡x(x+2)

7 Consider f(x)=xlnx.

a For what values ofxisf(x)defined? b Show that the global minimum value off(x)is¡

1

e

.

8 Find the greatest and least value of:

a x^3 ¡ 12 x¡ 2 for ¡ 36 x 65 b 4 ¡ 3 x^2 +x^3 for ¡ 26 x 63

9 The cubic polynomial P(x)=ax^3 +bx^2 +cx+d touches the line with equationy=9x+2at the

point (0,2), and has a stationary point at (¡ 1 ,¡7). Find P(x).

Example 12 Self Tutor

Find the greatest and least value of y=x^3 ¡ 6 x^2 +5on the interval ¡ 26 x 65.

Now dy

dx

=3x^2 ¡ 12 x

=3x(x¡4)

)

dy

dx

=0when x=0or 4

The sign diagram of

dy

dx

is:

) there is a local maximum at x=0,

and a local minimum at x=4.

The greatest of these values is 5 when

x=0.

The least of these values is¡ 27 when

x=¡ 2 and when x=4.

Critical value (x) f(x)

¡ 2 (endpoint) ¡ 27

0 (local max) 5

4 (local min) ¡ 27

5 (endpoint) ¡ 20

0

+ - +

4 x

dy

dx

+ -

3 x

dy

dx

To determine the nature

of a stationary point, we

can use a sign diagram or

the second derivative.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_14\379CamAdd_14.cdr Monday, 7 April 2014 2:07:02 PM BRIAN