SMA - SMA 104. Including course work, tests and solutions PDF

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SMA 104. Including course work, tests and solutions...

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SMA 104: CALCULUS I BACKGROUND INFORMATION Functions Example 1: Suppose we have 2 functions, f and g, both having  as a domain and suppose one of them squares each member of a domain and the other doubles each member of a domain. We wrote f(x) to represent the image of x under the function f and g(x) to represent the image of g(x) under g. g(x) = 2x. In this case, i.e. f(x) = x 2 and 2

f (5) = 25 and g(5) = 10 , f (a) = a 2 , g(k ) = 2k , f (a + h) = (a + h ) = ( a + h ) and so on.

Example 2: Given that h( x ) = x 2 - x , find the value of a) h(10) b) h(- 3) c) h(t + 1) Solution: ( a) h ( x ) = x 2 - x h(10 ) = 100 -10 = 90 (b) h(- 3) = 9 + 3 = 12 2

(c ) h (t + 1) = (t + 1) - (t + 1) = t 2 + 2t + 1 - t - 1 = t2 + t Composite functions Example 3: Given that f ( x ) = 10 + x and g (x ) = x 3 . Find (a) fg

Solution: a) g( x) = x 3

f ( g ( x) ) = f ( x3 ) = 10 + x3

b) gf

\ fg = 10 + x 3

b) gf f(x)=10+x ; g{ f ( x)} = g[10 + x] =(10 + x)

3

Example 4: Given that f(x)=5x+1 and that g (x ) = x 2 , express the composite functions (a) fg b) gf in their simplest possible forms. Solution: (a) g( x) = x2; f[ g( x) ] = f ( x2) = 5 x2 +1; fg = 5 x2 +1

(b) f ( x) = 5 x + 1; g[ f ( x )] = g( 5x + 1) = ( 5x + 1)2 = 25 x 2 + 10x + 1 The Inverse of a function 1 Consider the function f ( x) = x 3 + 1 8 If we are given a member of the range say f(x)=9 ,it is possible to find the corresponding member of the domain. 1 Example 5: Find the inverse of f (x ) = x 3 + 1 8 Solution:

1

Let

y=

1 3 x +1 8

1 3 x + 1= y 8 1 3 x = y -1 8 3 x = 8 ( y - 1) x = 3 8 ( y - 1) -1 \ f ( x ) = 3 8( x - 1) e.g. If f(x)=9; , we have y = 9, i.e. 1 3 x +1 = 9 8 1 3 x =8 83 x = 64 x=4 We may also use the inverse function f -1 (9 ) = 3 8(9 - 1 ) = 3 64 = 4 i.e. f - 1(x ) = 3 8(x - 1) ; Example 6: -1 5 x +7 . Find f (x ) Let f ( x) = 3x + 2 Solution: 5x + 7 5x + 7 f ( x) = ; y= ; y (3 x + 2 ) = 5 x + 7; 3 xy + 2 y = 5 x + 7 3x + 2 3x + 2 3xy - 5 x = 7 - 2 y; x (3 y - 5 ) = 7 - 2 y; x =

7 - 2y ; 3y- 5

f

-1

( x ) = 7 - 2x

3x - 5 Example 7: Given that f(x)=5x+1 ,find the values of (a) f (36), ( b) f Solution: Let y=5x+1 y -1 x -1 ; f - 1 (x ) = 5 x = y - 1; x = 5 5 35 -1 -1 -1 = 7; (b ) f (0) = ( a) f (36 ) = 5 5 Example 8: Given that f(x)=10x and g(x)=x+3, find b) ( fg )-1 x a) fg ( x ) Solution: a) g( x) = x + 3; f ( g( x) ] = f ( x + 3) = 10 x + 30 ; ( fg)( x) =10 x + 30 b) ( fg)( x) = 10 x + 30 ; -1

Let

y = 10 x + 30

y - 30 = 10 x; x =

y - 30 ; 10

2

( fg )- 1 x = x - 30 10

=

-1

(0 )

x -3 10

Exercise 1 Find the inverse of the following functions 5 1. f ( x) = ( x - 32) 9 2. f ( x ) = 180( x - 2) 5 ( x + 7) -9 3. f ( x) = 3 1 4. f ( x) = 2 x Answers to Exercise 1 5 1. f ( x) = ( x - 32) 9 5 y = (x - 32 ) 9 9 y = 5 x - 160 9 y + 160 x= 5 9 x + 160 -1 f ( x) = 5 9x = + 32 5 2. f (x ) = 180(x - 2 ) y = 180x - 360 y + 360 x= 180 . f

3.

x + 360 x = +2 180 180 5x + 35 5 (x + 7 ) - 9; y = - 9; f ( x) = 3 3

-1

(x) =

y +9 =

3y + 27 - 35 3y -8 ; x= ; 5 5 1 1 1 1 4. f (x ) = 2 Þ y = 2 Þ x 2 = Þ x = ± ; y y x x

5x + 35 3

3( y + 9 )= 5 x + 35; x =

f f

-1

-1

(x ) = 3x - 8

(x ) = ±

5

1 x

Injections (One-to-one functions) Let f : A ® B be a function, then f is said to be 1- 1 (injective) if x ' , x Î A and x ¹ x ' , then f ( x) ¹ f x ' or

( ) f ( x) = f ( x ) implies x = x

' ' "x, x ' Î A If That is distinct elements in the domain have distinct images. Example 9: Let f : Â ® Â be defined by f (x ) = x 2 .Is f 1- 1 ? Solution:

3

f is not 1 -1 since - 1¹ 1 but f (- 1 ) = f (1 ). Example 10: Let f : Â ® Â be defined by f ( x) = 2 x + 1 .Is f 1- 1? Solution: f is 1- 1since if a, b Î Â such that f (a) = f (b ) , then 2a + 1 = 2 b + 1 Þ 2 a = 2b Þ a=b Example 11: T S f

a b c

1 2 3

\ f is 1- 1

Onto functions (surjective functions) Let f : A ® B . Then we say that f is an onto (or surjective function) if for every y ÎB ,there exists x Î A such that y = f (x ) . i.e f ( A ) = Im f = B i.e the image of f is the entire codomain B .In this case,we say that f is a function from A onto B . Example 12: Let

a b c d

1 2 3 4

This is not an onto function since 4 has no corresponding image. Example 13: Let f :  ®  be defined by f (x ) = x 2 .Is f onto? Solution: f is not onto since - 1Î Â and - 1has no pre-image in  . Example 14: Let

S

1 2 3 4 5

x y z

f is onto T 4

Example 15: Let f : Â ® Â be defined by f ( x) = 2 x + 1 .Is f onto? Solution: f is onto. Bijective functions A bijective function is a function which is both 1 -1 and onto A function f : A ® B is invertible if and only if it is a bijection i.e a 1 - 1and onto function. Example 16: Let f : Â ® Â be defined by f (x ) = x 2 + 1. Is f onto? Is f 1- 1 ? Is f a bijection? Solution: a) f is not onto since for example 0Î Â has no corresponding preimage. b) f is not 1 - 1since f (- 1) = f (1 ) = 2 and - 1 ¹ 1 . c) f is not a bijection since it is not an onto and 1 - 1 function. Example 17: Let f : Â ® Â be defined by f ( x) = 2 x + 1 .Find the formula for f -1 . Solution: Let y = 2 x + 1 ; making x the subject, 1 2x = y - 1; x = ( y - 1) 2 1 Hence f -1 (x ) = (x - 1) 2 3x - 5 .Find a formula for f -1 . Example 18:Let f ( x) = 7 Solution: 3x- 5 , then Let y = 7 7 y- 5 7x - 5 > y = 3 x - 5 Þ 3 x = 7 y - 5; x = ; \ f -1 (x ) = 3 3 Composition of functions Let f : A ® B and g : B ® C. The composite function g o f of is the function from A to C defined by (g of )( x) = g ( f (x )) for all x Î A Example 19: Let A = {1,2}, B = {3,4,5} and C = {6,7} and let f : A ® B and g : B ® C be defined by f( 1) = 3, f( 2) = 5, g( 3) = 6, g( 4) = 7 , g(5) = 6 . Find image i.e. im ( g o f ) Solution: A B C g f

1 2

6 7

3 4 5

Im ( g o f )={6}

5

2 Example 20: Let f : Â ® Â and g :Â ® Â be defined by f ( x) = 2 x + 1 and g( x ) = x - 2 .Find the formula defining the composition 2 (iii) fog (ii) fof = f (iv) ( g of ) (2 ) (i) g of Solution: ( i) (g of )( x ) = g ( f (x ) )= g (2 x +1) = (2 x + 1)2 - 2 = 4 x2 + 4 x+ 1- 2 = 4 x2 + 4 x -1

( ii) ( fof )( x) = f ( f ( x)) = f (2 x + 1) = 2 (2 x + 1) + 1 = 4 x + 2 +1 = 4x+3

(

( iii) ( fog )(x ) = f ( g ( x)) = f x2 - 2 = 2 x2 - 2 +1 = 2 x 2 - 4 +1 = 2 x 2 -3

(

)

)

(iv ) ( g of )(x ) = 4x 2 + 4 x - 1

( )

(g of )( 2) = 4 22 + 4(2) - 1 = 4(4) + 8 - 1 = 16+ 8- 23

Exercise 2 3 1.Let g ( x ) = 5 x - 3 and f ( x) = x + 1 for x Î Â.Find the composite function h = g of . 3 2.Let f : Â ® Â be defined by f (x ) = x - x .Show that f is not injective.

6

Limits The concept of limits of a function is one of the fundamental ideas that distinguishes Calculus from other areas of mathematics e.g. Algebra or Geometry. If f(x) becomes arbitrarily close to a single number L as x approaches a from either side, then the lim f (x ) = L . limit of f(x) as x approaches a is L written as x ®a Consider a function y=f(x) lim f ( x) = L means the limit of f(x) as x approaches a is equal to a number L i.e. as x gets x® a closer and closer to a ( x ¹ a ), f(x) gets closer and closer to L. lim Example 21: Let f ( x ) = x 2 . Find f ( x) x®2 Solution: lim lim x 2 = 22 = 4 f (x ) = x® 2 x ®2 lim Example 22: Let f(x) = 5x – 3. Find 5x - 3 x® 2 Solution: lim 5 x - 3 = (5 ´ 2 -3 ) = 7 x®2 Example 23: Let

f (x ) =

1 x .

Find lim

1 x®¥ x Solution: lim 1 =¥ x® 0 x

(undefined)

Properties of limits lim 1. k=k x ®a lim lim lim [ f (x ) + g (x )] = 2. f ( x )+ g (x ) x®a x ®a x ®a lim lim lim f (x )´ g (x ) = f (x )´ g (x ) 3. x ®a x ®a x® a

1

4. lim x ®a

lim f (x ) f ( x) = x ®a lim g( x) g( x) x®a lim

5. lim

e. g

f (x ) = n x ® a

n

f ( x) 1

1 2

lim

provided that lim g( x) ¹ 0

x ö2 x = æç lim ® x a ÷ø x ®a è

Example 24: lim x 2 - 4 x + 3 lim x2 - lim 4 x + lim 3 x ®5 x® 5 x® 5 = x ®5

= 52 - 4 ´ 5 + 3 = 25- 20+ 3 =8 Example 25: 3x + 5 3 ´ 2 + 5 11 3 x + 5 lim = = x ®2 = x® 2 5 x + 7 lim 5x + 7 5 ´ 2 + 7 17

lim

x ®2

Example 26:

x2 - 4 x 2 - 4 lim x®2 lim ¹ x ®2 x - 2 lim x - 2 x 2

since lim x - 2 = 0 x®2

® 2

Hence

lim x - 4 lim (x + 2)(x - 2 ) lim ( x + 2) = 4 = = x ® 2 x - 2 x ® 2 ( x - 2) x ®2

Example 27: 2

2 3

lim x ®8

lim x 3 + lim 3 x

x + 3 x x ®8 x ®8 = 16 16 lim 4 - lim 4x ®8 x ®8 x x 2

83 + 3 8 16 48 4+6 2 = 2 = 2 +3 2 Example 28: =

2

lim lim 3x + 5 = x ® ¥ 6x -8 x ® ¥

3x 5 5 3 + + x x = lim x 8 8 x ® ¥ 6 6 x x

lim lim 5 3 + x ® ¥ x ® ¥ x 3 + 0 1 = = lim lim 6- 0 2 6 8 x ® ¥ x ® ¥

lim 4x2 - x Divide by the highest power of x. x ® ¥ 2 x3 - 5 æ4 1 ö lim ç x - x 2 ÷ 0 - 0 0 ÷= ç = =0 5 ÷ 2-0 2 x ® ¥ç ç2 3 ÷ x ø è

Example 29:

Example 30: 2 ö æ x 2 ç1 + 2 ÷ x +2 x ®¥ è x ø = x ® ¥ 3x - 6 3x - 6 lim

lim

2

1

2 ö2 æ xç1 + 2 ÷ lim è x ø = x ® ¥ 3x - 6

x = lim

lim

2 ö æ ç1 + 2 ÷ x ø è 6ö æ xç 3 - ÷ xø è

2 x ®¥ x2 = lim æ 6ö ç3- ÷ x ® ¥è xø 1 = 3 1+

Example 31: 3

lim x 3 - 1 =3 x ®1 x- 1

(

)

lim x 2 + x + 1 (x - 1) lim 2 x + x+1 = (x - 1 ) x®1 x®1 = 1+ 1 + 1 =3

Example 32: lim æ x 3 - 8 ö 0 ÷= ç x ® 2 çè x - 2 ÷ø 0 lim

(x - 2)(x 2 + 2x + 4 )

x®2

x-2

lim 2 x + 2 x + 4 = 4 + 4 + 4 = 12 x®2

Exercise 3 lim 5x + 1 1. x ® ¥ 10 + 2x lim x 2 - 4x - 5 2. x® 5 x - 5 lim x 2 - 25 3. x ® 5 x- 5 4.

lim 2 + x - 2 x® 0 x x and g (x ) = 7 - x is  .Write down as simply as possible. 5 d. ( fg )-1( x )

5.The domain of the functions f ( x) = a. f

-1

(x )

b. g -1 (x ) c. fg (x )

Solutions to Exercise 3 1 lim 5 + x 5x + 1 1 1. = =2 2 x ® ¥ 10 + 2x x ® ¥ 10 +2 x lim

4

lim x 2 - 4 x - 5 lim ( x -5 )( x +1) = 2. x ®5 x-5 x ® 5 ( x - 5) lim x +1= 6 = x®5 lim 2 x - 4 Or = 2(5) - 4 = 6 x®5 1 lim ´ 2 - 25 lim ( x + 5 )( x - 5 ) = 10 3. x ® 5 x - 5 = x ® 5 ( x - 5)

Or

4.

lim 2 x = 2 (5 ) =10 x®5 1

2-x - 2 2 -x + 2 ´ x® 0 x 2- x + 2 2 -x -2 = x 2 + 2- x -x -1 = = 2 + 2- x x 2 + 2- x lim

(

)

(

)

-1 = x ® 0 2 + 2- x lim

2 -1 -1 2 2 2 2 = ´ = = 4 2+ 2 2 2 2 2 4´2

L' Hospital Rule lim f (x ) 0 = or ¥ x ® a g (x ) 0 Then

lim f ( x )

x ® a g (x)

=

lim f ' ( x )

x ® a g ' ( x)

e.g 1.

lim x 3 - 1 lim 3 x 2 = x ® 1 x -1 x ® 1 1 = 3´ 1 =3

lim x 3 - 8 0 lim 3x 2 = =12 2. x® 2 x - 2 0 x ® 2 1

3.

lim cos x - 2 x - 1 lim - sin x - 2 = = x® 0 x ®0 3 3x 3 5

Continuity Continuity at a point. A function is considered continous if the following conditions are met. 1. f ( a) is defined. 2. lim f ( x) exists. x® a

3. lim f ( x ) = f ( a ) x® a

Otherwise it is discontinuous. Example 33: Show that f ( x ) =

x2 - 4 is not continous at x=2 x-2

Solution: 4- 4 0 = , which is undefined 2- 2 0 lim lim x2 - 4 Condition 2: f ( x) = x®2 x® 2 x-2 lim ( x + 2)( x - 2) = x ®2 x -2 lim x +2 = 4 = x®2 Therefore, lim f ( x) exists.

Condition 1: f ( 2 ) =

x ®2

Condition 3: lim f ( x) = 4, but f (2) is undefined x® 2

lim \ f (x ) ¹ f ( 2) x®2 Therefore f(x) is not continous at x=2 Note: It is possible to redefine f(x) to make it continous at x=2, as follows:

ì x2 - 4 ïï x - 2 , x ¹ 2 f ( x) = í ï ïî4, x = 2 lim x®2

f ( x) = 4, i.e.lim exists, we redefine f(x) so that x® 2

lim f ( x ) = f ( 2) = 4 x ®2 Example of a continous function. X

y = x2 6

Example of a discontinous function. y=

1 x

0

Remarks 1.Polynomials are always continous functions. e.g f (x ) = x 2 - 2x + 1 at csince Condition 1: f (c ) is defined i.e. f ( c ) = c 2 - 2c + 1 Condition 2: Condition3:

lim 2 lim f ( x) x - 2 x + 1 = c 2 - 2x + 1 exists. = x®c x®c lim f ( x ) 2 = c - 2c + 1 = f ( c) x® c

2 .Discontinuity means a function breaks at a particular point. Example 34: Discuss the continuity of f ( x ) if ì x3 + 27 ; x ¹ -3 ï ï x +3 f ( x) = í ï27; x = - 3 ï î Solution:Condition 1: f(-3)=27, therefore f(x) is defined at x=3 2 x3 + 27 lim ( x + 3 ) x - 3 x + 9 lim Condition 2: = x +3 x ® -3 (x + 3) x ® -3 lim x 2 - 3x + 9 = x ® -3 = 9+ 9+ 9 = 27 lim f ( x ) Condition 3: = f ( -3 ) = 27 x ® -3 \ f ( x) is continous. Example 35:Determine whether or not the function below is continous at x=1

(

)

7

ì x2 - 1 ï x 1 if x ¹ 1 ï í f ( x ) = ï 2 if x = 1 ï î

Solution: Condition 1: f ( 1) = 2 hence f (1) is defined. Condition 2:

lim f ( x ) x ®1

=

lim x 2 - 1 lim ( x + 1)(x - 1) lim f ( x) exists. = = 2 Therefore x ® 1 x - 1 x ® 1 (x - 1) x ®1 .

lim x2 -1 = f (1) , hence f (x )is continous at x=1 x ®1 x-1 Example 36: .Discuss the continuity of f ( x ) if Condition3: lim f ( x )

ì x2 - 4 ,x¹2 ï ï x- 2 f ( x) = í ï3 x = 2 ï î

Solution: Condition 1: f(2) = 3, so f(x) is defined at x=2 Condition 2: lim f ( x) lim x 2 - 4 = x ® 2 x®2 x -2 lim (x + 2 )(x - 2 ) hence lim f (x )exists. = x ® 2 (x - 2 ) =2 \ f ( x) lim f ( x ) lim f ( x) Condition 3: f (2 ) = 3 but ¹ f (2) Thus f(x) is discontinuous =2 \ x® 2 x ®2 at x = 2 Exercise Define the continuity of a real valued function f (x ) at a point x=a. Hence determine if the following function is continous at x=1. 3 ì x -1 , x ¹1 ï ï x -1 f ( x) = í ï3, x = 1 ï î

8

Example37:Show that f ( x ) =

1 for is x-2

x ¹ 2 is not continous at x =2.

Solution:

2 1 f ( x) = x -2

Because f is not defined at the point x = 2 it is not continous there. Moreover f has what might be called an infinite discontinuity at x = 2 Combinations of continous Functions. Any sum or product of continous functions is continous. That is, if the functions f and g are continous at x = a , then so are f + g and f × g e.g if f and g are continous at x = a , then liméë f ( x) + g ( x )ùû = lim f ( x) + lim g ( x) = f ( a ) + g ( a ) x ®a

x ®a

x ®a

Example 38: f ( x ) = x is continous everywhere,i.e. y = f ( x)

It follows that the cubic polynomial function f ( x ) = x3 -3 x2 +1 is continous everywhere. More generally every polynomial function p ( x) = bn xn + bn -1 xn - 1 + K + b1 x + b0 is continous at each point of the real line. If p( x) and q ( x) are polynomials, then the quotient law for limits and the continuity of polynomials imply that p ( x ) p ( a) p( x) lim p( x) = x® a = lim provided q (a ) ¹ 0 . Thus every rational function f ( x ) = is x ®a q ( x ) q (x ) lim q ( x ) q ( a) x® a

continous wherever it is defined. 9

The point x = a where the function f is discontinuous is called a removable discontinuity of f provided that there exists a function F such that F ( x ) = f ( x ) for all x ¹ a in the domain of f , and this new function F is continous at x = a . x- 2 Example 39: Suppose that f (x ) = 2 x - 3x + 2 2 x - 3x + 2 = ( x - 1)( x - 2)

\ f ( x) =

x-2 ( x - 1)( x - 2)

This shows that f is not defined at x = 1 and x = 2 Þ f ( x ) is continous except at these points. 1 x-2 1 . The new function F ( x ) = is continous at x = 2, where But f ( x) = = x x x 1 2 1 x -1 ( )( ) F ( 2) = 1 . Therefore f has a removable discontinuity at x = 2 ; the discontinuity at x = 1is not removable. y=F(x)

1

Composition of Continous Functions Let h ( x) = f ( g ( x) ) where f and g are continous functions.The composition of two continous functions is continous or more precisely, if g is continous at a and f is continous at g (a ) , then

f o g is continous at a where f o g = f ( g (x )) . Proof: The continuity of g at a means that lim g ( x) = g ( a ) ,and the continuity of f at g( a) x ®a

implies

that

lim

g ( x )® g ( a )

(

f ( g (x ) ) = f ( g (a ) )

)

i.e. lim f ( g (x ) ) = f lim g ( x ) = f ( g ( a )) x ®a

x ®a

2

x - 7 ö3 æ Example 40: Show that the function f ( x ) = ç 2 ÷ is continous on the whole real line. è x + 2x + 2 ø Solution: Consider the denominator 10

x 2 + 2 x + 2 = (x + 1 ) + 1 > 0 for all value of x. Hence the rational function 2

(

2 x -7 is defined and continous everywhere. Thus f ( x ) = éër ( x ) ùû 2 x + 2x + 2 continous everywhere. One-sided limits Let S Í ¡ and f : S ® ¡ be a function. If for every x Î S , f ( x ) ® L as x ® a and

r ( x) =

) is 1 3

+

x > a always, then we say that x ® a from the right and write x ® a or we say lim f ( x) = L . +

x® a

Similarly, if f ( x ) ® L as x ® a and x < a always, we say that x ® a from the left and write -

x ® a or we say lim- = L . x ®a

The limits lim+ f ( x) and lim- f ( x) are called one-sided limits of f and a x ®a

x ®a

Remarks 1. lim f ( x) = L x® a

iff

lim+ f ( x ) = lim- f ( x ) = L x® a

x® a

i.e the limit of a function f ( x) exists if the right hand side limit = left-hand side limit. 2. If lim+ ¹ lim- f ( x) , then lim f ( x ) does not exist. x ®a

x ®a

x ®a

Example 41: Given f (x ) =

x ,Find lim+ f (x ) and lim- f (x ) x -1 x®1 x ®1

Solution: 0 0

1 2 -1

1 ¥

1

1 2

3

Also consider the graph of f ( x ) =

2 2 1 x -1

y

0

11

1

x

lim- f ( x ) = -¥ if x < 1 x ®1

lim+ f ( x ) - ¥1 if x > 1 x ®1

\ lim f (x ) = ¥ Þ lim f ( x ) does not exist. x ®1

x ®1

Example 42: Consider the following graph y 1 y = f (x) = 2 x 1 y= 2 x

x

lim+ f ( x) = ¥

lim+ f ( x) = ¥

x ®0

x® 0

Example 43: Draw the graph of ì1, if x = 1 ïï f (x ) = í- x , if - 1 < x < 1 ï ïî -1, if x > 1 Solution: 4 y 3 2 -3 -2 -1 -1 -2 -3

(

Example 44: Evaluate lim+ 1+ x® 2

Solution:

(

lim+ 1 + x - 2 x® 2

1

)

2

3

(

x-2

x - 2 and lim- 1 + x®2

x

)

)

= lim1 + lim+ x - 2 + x ®2

x ®2

=1+ 0 = 1

(

)

On the other hand, lim- 1 + x - 2 does not exist (is no...