Mathematics Winter School worksheet

CALCULUS - MEMO

Sarel Frederick van Greunen

QUESTION 1

1.1

ü ü correct substitution into formula

ü

ü

ü (5)

1.2

ü multiplying

ü

ü

ü (4)

1.3.1

ü answer (1)

1.3.2

ü

ü

ü (3)

2 2

02 2 2

02 2 2

02

0

0

0

[3( ) ( ) ] [3 ]( ) lim

3 3 ( 2 ) 3lim

3 3 2 3lim

3 2lim

(3 2 )lim

lim (3 2 )

3 2

h

h

h

h

h

h

x h x h x xf xh

x h x xh h x xh

x h x xh h x xh

h xh hh

h x hhx h

x

®

®

®

®

®

®

+ - + - -¢ =

+ - + + - +=

+ - - - - +=

- -=

- -=

= - -

= -

23( ) ( )x h x h+ - +

2

0

3 2limh

h xh hh®

- -

0lim (3 2 )h

x h®

- -

3 2x-

24

4

24

4

88

8 8

2 424

1 221 2 441 2 44

xyx

xx

xx

x x-

æ ö= +ç ÷ç ÷è ø

æ ö= +ç ÷è ø

= + +

= + +

9 7 79

1 2( 8) 0 32 324

dy x x xdx x

- -= - + + = +

8 81 2 44x x- + +

92x-

732x

2 3 4(4) 44

f += =

12

3 32 2

32

32

2 3( )

( ) 2 31 1( ) 22

1 1(4)8(4)

xf xx x

f x x

f x x xx

f

-

- -

= +

\ = +

¢\ = ´- = - = -

¢\ = - = -

12( ) 2 3f x x-= +

32x--

18

-

1.3.3

ü substitution ü equation in the form (2)

QUESTION 2

2.1.1

Based on the shape of the cubic graph, there is a local max at

and a local min at .

and

Alternatively:

is an x-intercept of f.

At ,

ü

ü

ü

ü

ü

ü (6)

2.1.2

ü turning points ü x-intercepts ü shape (3)

1(4 ; 4) lies on and 8

14 ( 4)8

1 1 48 21 98 2

f m

y x

y x

y x

= -

\ - = - -

\ = - + +

\ = - +

....y =

3 2

2

2

( ) 4 4

( ) 3 8 4

0 3 8 40 (3 2)( 2)

2 or 23

f x x x x

f x x x

x xx x

x x

= - +

¢\ = - +

\ = - +\ = - -

\ = =

2 3

x = 2x =

2a\ = 23

b =

( ; 0)a3 2

2

0 4 4

0 ( 4 4)0 ( 2)( 2)

0 or 2

x x x

x x xx x x

x x

= - +

\ = - +\ = - -\ = =

2a\ =

23

x =3 22 2 2 2 324 4

3 3 3 3 27f æ ö æ ö æ ö æ ö= - + =ç ÷ ç ÷ ç ÷ ç ÷è ø è ø è ø è ø

3227

c\ =

20 3 8 4x x= - +2 or 23

x x= =

2a =23

b =

2 323 27

f æ ö =ç ÷è ø3227

c =

(2 ; 0)

3223 27( ; )

2.1.3 ü

ü (2) 2.2.1 At the turning point of the graph of the parabola

there is a point of inflection, since the gradient of is zero. This means that at the turning point of . Therefore, at , we know that , which means that the graph of f has a point of inflection at .

is the point of inflection of f.

ü (1)

2.2.2 ü (1) 2.2.3 The graph of f decreases for all .

This happens for all real values of x excluding . ü (1)

QUESTION 3

ü

ü

ü

ü üü (6)

32 or 027

k k< - >32 27

k < -

0k >( )y f x¢=

f ¢( ) 0f x¢¢ = f ¢0x = (0) 0f ¢¢ =

0x =(0 ; 5)\ -

(0 ; 5)-

( 2 ; 0)- (2 ; 0)( ) 0f x¢ <

0x =R {0}-

2

2 2

2

Surface area 4

Cost of material 75 50(4 ) 75 200

C( ) 75 200

x hx

x hx x hx

x x hx

= +

= + = +

\ = +

2

2

11

x h

hx

=

\ =

2

22

2

2 1

C( ) 75 200175 200

20075

75 200

x x hx

x xx

xx

x x-

= +

æ ö= + ç ÷è ø

= +

= +

2C ( ) 150 200x x x-¢ = -

2

2

3

3

3

3

0 150 2002000 150

0 150 200

200 150200150

200 1,1150

x x

xx

x

x

x

x

-= -

\ = -

\ = -

\ =

\ =

\ = =

2C( ) 75 200x x hx= +

21hx

=

2 1C( ) 75 200x x x-= +2C ( ) 150 200x x x-¢ = -

1,1x =

QUESTION 4

4.1.1

ü ü

ü

ü (4)

4.1.2 ü

ü (1) 4.1.3

ü

ü

ü ü ü ü (6)

4.1.4

ü ü ü ü (4)

4.1.5

ü ü

ü (3)

3

3

(2) 1(2) 72 1

(3) 1(3) 133 1

(2 ; 7) (3 ;13)13 7Average gradient 63 2

f

f

-= =

--

= =-

-= =

-

(2 ; 7)(3 ;13)13 73 2--

6

22( 1)( 1)( ) 1

( 1)x x xf x x x

x- + +

= = + +-

2( 1)( 1)( 1)

x x xx

- + +-

2 1x x+ +2 2

02 2 2

02

0

0

0

( ) ( ) 1 ( 1)( ) lim

2 ( ) 1 1lim

2lim

(2 1)lim

lim (2 1)

2 1(2) 2(2) 1 5

h

h

h

h

h

x h x h x xf xh

x xh h x h x xh

xh h hh

h x hh

x h

xf

®

®

®

®

®

+ + + + - + +¢ =

+ + + + + - - -=

+ +=

+ +=

= + +

= +¢\ = + =

2( ) ( ) 1x h x h+ + + +2( 1)x x- + +

22xh h h+ +(2 1)x h+ +2 1x +(2) 5f ¢ =

2

5

(2) (2) (2) 1 7Substitute (2 ; 7)

7 5( 2)7 5 105 3

tm

f

y xy xy x

=

= + + =

- = -\ - = -\ = -

5tm =(2) 7f =7 5( 2)y x- = -5 3y x= -

2( ) 1( ) 2 1

2 1 012

f x x xf x xx

x

= + +¢\ = +

\ + >

\ > -

( ) 2 1f x x¢ = +2 1 0x + >

12

x > -

4.2

ü and

ü

ü ü

ü

ü (6)

[25] QUESTION 5

5.1

ü Substituting into ü Substituting into ü Showing that ü Showing that (4)

5.2

ü Showing that the point of inflection is ü Stating that this point is also the y-int (2)

5.3 The graph is concave down for all x such that

Alternatively: The shape of the graph is as follows: The graph is concave down for all

ü (1)

52

12

12

12

12

2 32

2 3

4

4

4

1 2 321 2 321 1(2 ) 2 3( 3)2 2

9

1 9

xy x xx

y x x x

dy x x xdxdy x x xdxdy xdx xx

-

-

- -

- -

= + -

\ = + -

æ ö\ = + - -ç ÷è ø

\ = + +

\ = + +

212x 33x-

522x122x

x

12

1

x

49x

3

3

2

2

( )

( 1) ( 1) ( 1)0 1

1

( ) 3

( 1) 3( 1)0 3

33 1

2

f x x ax b

f a ba b

a b

f x x a

f aa

ab

b

= + +

- = - + - +\ = - - +\ - = -

¢ = +

¢\ - = - +\ = +\ = -- - = -\ = -

( )f x

( )f x¢3a = -2b = -

3

2

( ) 3 2

( ) 3 3( ) 6

For a point of inflection ( ) 66 0

0(0) 2Point of inflection is (0 ; 2)

The -intercept is also (0 ; 2)

f x x x

f x xf x x

f x xxxf

y

= - -

¢ = -¢¢ =

¢¢ =\ =\ =

= -\ -

-

(0 ; 2)-

( ) 0f x¢¢ <( ) 66 00

f x xxx

¢¢ =\ <\ <

0x <

0x <

(0 ; 2)-

5.4

üü turning points üü x-intercepts ü point of inflection ü y-intercept ü shape (7)

5.5 ü (1) [17]

2

2

2

3

3

2

-intercept point of inflection: (0 ; 2)Turning points:

( ) 3 3

0 3 3

0 11

( 1) 0( 1; 0)

(1) (1) 3(1) 2 4(1; 4)

-intercepts:

0 3 2

0 ( 1)( 2)0 ( 1)( 1)( 2)

1 or 2

y

f x x

x

xxf

f

x

x x

x x xx x x

x x

= -

¢ = -

\ = -

\ = -\ = ±

- =-

= - - = --

= - -

\ = + - -\ = + + -\ = - =

4 0k- < < 4 0k- < <

(0 ; 2)-

(2 ; 0)( 1; 0)-

(1; 4)-

QUESTION 6 6.1

ü ü

ü (3)

6.2

ü ü

ü (3)

6.3

ü ü ü ü ü (5)

[11] QUESTION 7

7.1

The rate of change is negative indicating that the depth is decreasing.

ü

ü ü decreasing (3)

7.2

At 09h00 ( ), the inflow of petrol be the same as the outflow.

If hours, then

80 minutes later will be at 10h10.

ü

ü

ü 10h10 (3)

2

2

( ) 3 8 3

0 3 8 30 (3 1)( 3)

1 or 33

f x x x

x xx x

x x

¢ = + -

\ = + -\ = - +

\ = = -

20 3 8 3x x= + -0 (3 1)( 3)x x= - +

1 or 33

x x= = -

23 8 3 013 or 3

x x

x x

+ - >

\ < - >

23 8 3 0x x+ - >3 x < -13

x >

3 2

2

2

3 2

( ) 18

( ) 3 2

( ) 3 8 33 3 and 2 8 and 3

1 and 4

( ) 4 3 18

f x ax bx cx

f x ax bx c

f x x xa b ca b

f x x x x

= + + -

¢ = + +

¢ = + -\ = = = -\ = =

\ = + - -

2( ) 3 2f x ax bx c¢ = + +1a =4 b =3c = -

3 2( ) 4 3 18f x x x x= + - -

2 3

2

2

2

1 1D( ) 64 8

1 1D ( ) 2 34 81 3D ( )2 81 3 15D (3) (3) (3) 1,88 /2 8 8

t t t

t t t

t t t

m s

= + -

¢\ = ´ - ´

¢\ = -

¢\ = - = - = -

21 3D ( )2 8

t t t¢ = -

1,88 /m s-

2

2

D ( ) 01 302 8

0 4 30 (4 3 )

40 or 3

t

t t

t tt t

t t

¢ =

\ = -

\ = -\ = -

\ = =

0t =

43

t = 4 60 80 minutes3

t = ´ =

21 302 8t t= -

40 or 3

t t= =

QUESTION 8

8.1

ü

ü

ü

ü

ü

ü

ü (7)

8.2

ü

ü

ü

ü (4)

4x

4x

4x

4x

4x

5248x

-

5248x

-

2

2 2

2 2

2

ABCD

AB BC CD AD4

EF4

BE EF CF 6

2BE 64

2BE 64

52BE 64

5 24 5BE 38 8

Area of ABCD Area BCFE

24 54 8 4

24 516 322 24 5

3224 3

32

xx

x

xx x

xx

x

x x

x x x

x x x

x x x

x x

=

\ = = = =

\ =

+ + = -

\ + = -

\ = - -

\ = -

-\ = - =

+

-æ ö æ öæ ö= +ç ÷ ç ÷ç ÷è ø è øè ø

-= +

+ -=

-=

AB BC CD AD4x

= = = =

EF4x

=

BE EF CF 6 x+ + = -5 24 5BE 38 8x x-

= - =

2

4xæ ö

ç ÷è ø24 58 4x x-æ öæ ö

ç ÷ç ÷è øè ø

224 332x x-

2224 3 3 3A( )

32 4 323 3A ( )4 16

3 304 164

x xx x x

x x

x

x

-= = -

¢\ = -

\ = -

\ =

23 34 32x x-

3 3A ( )4 16

x x¢ = -

3 304 16

x= -

4x =

[11]

QUESTION 9

9.1.1

ü ü correct substitution into formula

ü

ü

ü (5)

9.1.2(a)

ü

ü

ü (3)

9.1.2(b) At a point of inflection on g:

ü

ü (2)

9.2

üü

ü

ü

ü (4)

[14]

2 2

02 2 2

02 2 2

02

0

0

0

[3( ) ( ) ] [3 ]( ) lim

3 3 ( 2 ) 3lim

3 3 2 3lim

3 2lim

(3 2 )lim

lim (3 2 )

3 2

h

h

h

h

h

h

x h x h x xf xh

x h x xh h x xh

x h x xh h x xh

h xh hh

h x hhx h

x

®

®

®

®

®

®

+ - + - -¢ =

+ - + + - +=

+ - - - - +=

- -=

- -=

= - -

= -

23( ) ( )x h x h+ - +

2

0

3 2limh

h xh hh®

- -

0lim (3 2 )h

x h®

- -

3 2x-

2

( ) ( )At a turning point: ( ) 0

0 30 (3 )

0 or 3

f x g xg x

x xx x

x x

¢=¢ =

\ = -\ = -\ = =

2

0

3 2limh

h xh hh®

- -

0lim (3 2 )h

x h®

- -

3 2x-

( ) 0( ) 0

3 2 02 332

g xf x

xx

x

¢¢ =¢\ =

\ - =\- = -

\ =

3 2 0x- =32

x =

( )

1 12 2

3 12 2

3 122

12

21 1

21 2

3 2

1 1

2

y xx

y xx

y x xdy x xdxdydx xx

-

- -

æ ö= - -ç ÷è ø

\ = - - +

\ = - -

\ = -

\ = -

1 12 23 2y x x-= - -

3 12 21

2x x- --

0lim (3 2 )h

x h®

- -

3 122

1 1

2xx-

QUESTION 10

10.1

ü substitution into formula ü expanding ü simplifying ü a, b and c (4)

10.2

ü ü ü x-values ü (4)

10.3 is the y-intercept of g.

ü ü (2)

10.4.1 üü (2)

10.4.2 Point of inflection:

Concave up for all:

ü

ü

ü (3)

10.5 ü (1) [14]

2

3 2 2

3 2

( 3)( 2)( 5)

( 3)( 7 10)

7 10 3 21 30

4 11 30

y x x x

x x x

x x x x x

x x x

= + - -

= + - +

= - + + - +

= - - +41130

abc

= -= -=

3 2

2

2

3 2

( ) 4 11 30

( ) 3 8 11

0 3 8 110 (3 11)( 1)

11 23 or 13 3

( 1) ( 1) 4( 1) 11( 1) 30 36A( 1; 36)

f x x x x

f x x x

x xx x

x x

f

= - - +

¢\ = - -

\ = - -\ = - +

\ = = = -

- = - - - - - + =-

20 3 8 11x x= - -0 (3 11)( 1)x x= - +

A( 1; 36)-

D(0 ; 30)2(0) 3(0) 8(0) 11 11

( ) 11 30gm f

g x x

¢= = - - = -

\ = - +

11gm = -( ) 11 30g x x= - +

21 33

x- < <21 33

x- < <

2( ) 3 8 11( ) 6 8

0 6 88 46 3

f x x xf x x

x

x

¢ = - -¢¢\ = -

\ = -

\ = =

43

x >

( ) 6 8f x x¢¢ = -8 46 3

x = =

43

x >

30k > 30k >

QUESTION 11 11.1

ü ü

ü

ü substitution ü expansion (5)

11.2

ü

ü

ü

ü ü (5)

[10]

QUESTION 12

2 21Vol H3

r r h= p - p

1H 3H 3

r

r

=

\ =H 23 23 2

hr hh r

= +\ = +\ = -

2 2

2 2

3 3 2

3 2

1Vol H31(3 ) (3 2)3233

223

r r h

r r r r

r r r

r r

= p - p

= p - p -

= p - p + p

= p + p

H 3r=3 2h r= -

2 21Vol H3

r r h= p - p

3 2

2

2

2

2

2

2

2V( ) 23

4V ( ) 634498 63

4498 63

1 494 18 4

747 9 2

0 9 2 7470 (9 83)( 9)

9 cm

r r r

r r r

r r

r r

r r

r r

r rr r

r

= p + p

¢\ = p + p

\ p = p + p

\ = + p

\ = +

\ = +

\ = + -\ = + -\ =

2 4V ( ) 63

r r r¢ = p + p

2 4498 63

r rp = p + p

2 21Vol H3

r r h= p - p

20 9 2 747r r= + -9 cmr =

ü 𝑥-int ü 𝑦-int ü Shape - increasing ü Shape - concavity ü Turning point (5)