Differentiation is Simplified!

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.
i.e. f(x) = x 2 and g(x) = 2x. In this case,
f 5  25 and g 5  10 , f a   a 2 , g k   2k , f a  h  a  h  a  h and so on.
2


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
(c) ht  1  t  1  t  1
2

 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 b) gf
Solution:
a) g x   x 3 f g ( x)   f ( x 3 )  10  x 3  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   x 2 ; f [ g x ]  f x 2   5 x 2  1; fg  5 x 2  1
(b) f x  5x  1; g[ f x]  g 5x  1  5x  1  25x 2  10x  1
2


The Inverse of a function
Consider the function f  x   x 3  1
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.
Example 5: Find the inverse of f  x   x 3  1
1
8
Solution:




1

, 1 3
Let y 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
 f  x   3 8 x  1
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
i.e. f 1 x   3 8x  1 ; f 1 9  3 89  1  3 64  4
Example 6: (beginning of lesson on 16/1/12)
. Find f  x 
5x  7 1
Let f  x  
3x  2
Solution:
5x  7 5x  7
f x   ; y ; y 3x  2  5 x  7; 3xy  2 y  5 x  7
3x  2 3x  2
7  2y 7  2x
3xy  5 x  7  2 y; x3 y  5  7  2 y; x  ; f 1 x  
3y  5 3x  5
Example 7: Given that f(x)=5x+1 ,find the values of (a) f 36, (b) f 1 1
0
Solution:
Let y=5x+1
y 1 x 1
5 x  y  1; x  ; f 1  x  
5 5
1
(a ) f 36   7; (b) f 0  
1 35 1

5 5
Example 8: Given that f(x)=10x and g(x)=x+3, find
a) fg x  b)  fg 1 x
Solution:
a) g x   x  3; f ( g x ]  f x  3  10x  30 ;  fg ( x)  10x  30
b)  fg x   10x  30 ;
y  30 x  30
Let y  10 x  30 y  30  10 x; x  ;  fg 1 x  
x
3
10 10 10



Exercise 1
Find the inverse of the following functions


2

,1. f  x   x  32
5
9
2. f x   180x  2
5 x  7 
3. f  x   9
3
4. f  x   2
1
x
Answers to Exercise 1
1. f  x    x  32
5
9
y   x  32
5
9
9 y  5 x  160
9 y  160
x
5
9 x  160
f x  
1

5
9x
  32
5

2. f  x   180 x  2 
y  180x  360
y  360
x
. 180
x  360 x
f 1 ( x)   2
180 180
5 x  7  5 x  35 5 x  35
3. f x    9; y   9; y9 
3 3 3
3 y  27  35 3y  8
3 y  9  5 x  35; x  ; x ; f 1
x   3x  8
5 5 5

4. f x   2  y  2  x 2   x   x   
1 1 1 1 1 1
; f
x x y y 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  A and x  x ' , then f x   f x ' or
 
If f x   f x ' implies x  x ' x, x '  A
That is distinct elements in the domain have distinct images.
Example 9: Let f :    be defined by f x   x 2 .Is f 1 1 ?
Solution:
f is not 1 1 since  1  1 but f  1  f 1 .
Example 10: Let f :    be defined by f x   2 x  1 .Is f 1 1 ?
Solution:
f is 1 1 since if a, b   such that f a   f b, then



3
T

, 2a  1  2b  1
 2a  2b
ab
Example 11:
S
f
1 a  f is 1 1
2 b
3 c


Onto functions (surjective functions)
Let f : A  B . Then we say that f is an onto (or surjective function) if for every y  B ,there exists x  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 1
b 2
c 3
d 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 1  and  1 has no pre-image in  .




Example 14:


Let
1 f is onto
x
2
y 4
3
z
4
5