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Unit 2
Functions
Learning objectives
Function is one of the primitive notions in this course, indeed, in mathematics in general. Throughout this course we shall be manipulating and analyzing functions.
When you have completed this unit, you will be able to:
- understand the notion of a function;
- apply and understand the notation;
- use functions as a tool, find their domain and range, and check for various properties;
- characterize even and odd functions and describe the properties of their graphs; and
- determine the composition of functions.
Preparation
Read Section 1.1. We shall be using primarily visual and algebraic representation of functions (by
graphs and formulas respectively). Do some of problems at the end of that section; include some of problems 31โ67 (domain of functions) and 69โ79 (odd and even functions).
Read section 1.3. Focus on the last section: Combinations of Functions. The notion of composition of functions is particularly indispensable in calculus, and will be used extensively later (notably in
the section on the Chain Rule of differentiation). Do some of problems 1โ57 and 62โ64, regard problems 31โ53 as more important than the others.
Examples and comments
Example 1. Find the domain of the function
a) f ( ) x = 2 x โ 1
b) g ( ) x = 2 x โ 1 , 2 โค x โค 3
c) h ( x ) =
x
x
2
Solution by Mark Markless:
a) Since is defined only for non-negative numbers, it follows that the domain of f ( x )is all x
such that x โฅ 0. (mark: 0.5/2)
Comments:
The above applies to the domain of the function x , not of the function f ( ) x = 2 x โ 1.
As a general rule, if the domain of a given function f ( x )is not explicitly specified (as in the part
(a) here), then it consists of all numbers x , for which f ( x )is well defined.
Introduction to Calculus MATH 1500 Unit 2 1
Correct solution:
a) The radicand 2 x โ 1 must be at least 0 in order 2 x โ 1 to exist. So we need to solve 2 x โ 1 โฅ 0.
2 x โ 1 โฅ 0 โ 2 x โฅ 1 โ x โฅ.
Hence, the domain of the function is the set of all real numbers x that satisfy 2
x โฅ. โ
b) Here we are told that the domain of g ( x ) is the set of all numbers x such that 2 โค x โค 3 (the
closed interval [ 2,3] ). โ
c) The denominator should never be 0 , for division by 0 is not defined in the set of real numbers. This means that (^320) 2 x + x + โ . Using (for example) the quadratic formula we find that the
solutions of 3 2 0
2 x + x + = are -1 and -2. So, we conclude that the function h ( x )is well defined for all x except for x = โ 1 and x = โ 2 (that is, the domain is the union of the
intervals ( โโ, โ 2 ), (โ 2 ,โ 1 )and ( โ 1 ,+โ)). โ
Example 2. Find the domain and sketch the graph of the function ๏ฃด๏ฃณ
x x
x x f x.
Correct solution:
Note first the domain: it is given to consist of all non-zero numbers (simply because f ( x )is defined only for such numbers).
The graphs, apparently, consists of two pieces: it is the line (^) y = โ x when (^) x < 0 , and it is the line
y = x + 1 when x > 0. Note that there is a gap in the graph for x = 0.
The graph of the function f ( x ). โ
Correct solution:
For every number x , we have ( ) ( ) 5 ( ) 5 ( ) 5 5 f โ x = โ x โ โ x =โ x + x =โ f x. So f ( x )is odd.
Mark already established that g ( x ) is neither even nor odd.
Finally, for every x we have ( ) 1
2 2 hx x x
h x =
โ = , so that h ( x ) is indeed even. โ
Below you may find the graphs of all three functions: note the symmetry of the graphs of y = f ( x )
and (^) y = h ( x ).
f ( x )
g ( x )
h ( x )
Only the parabola g ( x )is not symmetric with respect to the y-axis or the origin.
Example 5. Given f ( x )= 3 x and 1
x
g x , find f ๏ฏ g , g ๏ฏ f and g ๏ฏ g.
Solution by Mark Markless:
x x
f ๏ฏ gx f gx f
x
g ๏ฏ f x g f x g x
x
x
x
g ๏ฏ gx g. (mark 1.5/2)
Comments:
The error in this case is a bit subtler and you may wish to try to find it before you proceed further.
The point to note of is that functions are determined not only by the expressions that specify them
but also by their domain. (See example 1 above.)
Correction to the solution by Mark Markless
As it happens, the first two compositions are correctly evaluated, provided we accept the
convention that whenever the domain is not specified explicitly, then it consists of all numbers for which the expression determining the function makes sense. The reason behind this is that the
function f ( ) x is rather tame - the domain of f ( ) x being the set of all numbers. In the third case
though we have to take care of the domain of both of the functions g ( ) x and g ๏ฏ g ( ) x , for, in
order to evaluate the latter for some value of x , this value should also be accepted by the former!
Hence, since g x ( ) is not defined for x = 1 , we conclude that ( ) , 1 , 2
= x x x
x g ๏ฏ gx.
(So, the domain of g ๏ฏ g ( ) x is the set of all numbers other than 1 and 2.)โ
Gems of humor
Schoolmaster: โSuppose x is the number of sheep in the problem.โ Pupil: โBut sir, suppose x is not the number of sheep.โ
Introduction to Calculus MATH 1500 Unit 2 5
While, g ( ) 1 = 1 = 1 g (โ 1 )is not defined at all. Hence g is neither even nor odd.
[Even and odd functions have symmetric domains too.] โ
Problem 3
2 f + g x = x + x โ.
2 f โ
g x = x โ
x โ.
( )( ) ( ) ( )
2 f ๏ฏ g x = f x โ 1 = x โ 1. Be careful when (if) simplifying further, for (^1 )^1
2 x โ โ x โ.
We have (^1 )^11
2 x โ = x โ wherex โฅ.
( g^ ๏ฏ^ g^ )( x^ ) =^ g^ ( x^ โ^1 )=^ x โ^1 โ^1. [What is the domain of the function^ (^ g^ ๏ฏ^ g )( ) x ?]^ โ
Problem 4
For every x we have ( f ๏ฏ f )( โ x ) = f ( f ( โ x )) = f ( f ( ) x ) =( f ๏ฏ f )( ) x , so that f ๏ฏ f is even.
The second equality is true since f ( x )is even, while in the other two we only use the definition of
composition.
For every x we have ( g ๏ฏ g )( โ x ) = g ( g ( โ x )) = g ( โ g ( ) x ) =โ g ( g ( ) x ) =โ( g ๏ฏ g )( ) x.
It follows that g ๏ฏ g is odd. In each of the steps (equalities above) we use the assumption that
g ( x )is odd, i.e., that g ( โ any number )=โ g ( thatnumber ).
For every x we have ( f ๏ฏ g )( โ x ) = f ( g ( โ x )) = f ( โ g ( ) x ) = f ( g ( ) x ) =( f ๏ฏ g )( ) x , so f ๏ฏ g is
even.
For every x we have ( g ๏ฏ f )( โ x ) = g ( f ( โ x )) = g ( f ( ) x ), so that g ๏ฏ f is even. โ
Introduction to Calculus MATH 1500 Unit 2 7
