Functions 1
A function a definite realistic rule of association or relation between two or more variables. Example 1: National park overnight camping fees
Child under 12 = Free
Adult = $10
Senior above 60 = $5
No children under 2 year old allowed restriction This table is a rule for the association between the two variables:
Variable one = Age of overnight campers independent variable Variable two = Fee per person per night dependent variable
If you are 10 then you sleep for free. If your age is 13 you pay $10 per night. If you are 80 year old you pay $5 per night.
What are the possible values of these two variables?
Age of overnight campers = 2 to about 125 = {x| 2£x £125} domain Fee per person per night in dollar = {0, 10, 5} = {y| y = 0, 10, 5} range Graphical description of the camp fee in terms of age of the camper:
Camping fee by Math equations
0 2 £t < 12
z = 10 12 £t < 60 piece wise function
5 t ≥60
Where
z = camp fee in $
t = age of campers in years
Example 2: Parking fees First one hour = free
After the 1st hour = $5 per hour After the 4th hour = $3 per hour
After 30 days (720h) = taken away restriction
Hours are rounded up the way they count
This parking fee rule (or function) relates two variables:
Parking hour independent variable
If you park your car there for 20 hours the fee will be = for 1st hour + second 3 hours + the last 16 hours = 0 + (5)(3) + (3)(16) = $63
Possible values of these two variables are: Parking hour = zero to 720 but rounded up
Parking hour = 0, 1, 2, 3, ---, 720 = {x| 0 £ x £720 and x = integer} domain Parking fee in dollar
= 0, 5, 10, 15, and then 18, 21, 24, 27, 30, - - - , maximum at 720 hour = 0, 5, 10, 15, 18, 21, 24, 27, 30, - - - , [0 + (5)(3) + (3)(716)]
= 0, 5, 10, 15, 18, 21, 24, 27, 30, - - - , 2163 range
Value of dependent variable depends on value of independent variable The function is the mapping from independent variable to depende nt variable
f: x ‡y An important property of function
One independent variable can not be associated with more than one dependent variable. If the function f is f: x ‡ y then one x-value produces one unique y- value
However, one y-value may be associated with more than one x- value.
Specifying a function Example 3: by words
A community college collects the fees for a program as follow:
If you are a Surrey resident, the fee is $1000. Residents of BC other than Surrey pay $1300. Students from other provinces pay $1500. Foreigners are required to pay $2000.
Example 4: pattern
+ - x ∏
If the pattern repeats what is next after ∏? It is going to be +
Example 5: ordered pairs and pattern (10, 1), (20, 2), (30, 3), (40, 4), -What is the 8thterm?
The 8thterm should be (80, 8)
What is the general term? relation between the two numbers in the pair for the nthterm Give these two numbers names, (x, y). They have relationship: x = 10y
General term = nthterm where n = any integer x = 10n
General term = nthterm = (10n, n)
Example 6: equation and graph
Graph the relation between x and y given by y = 0.1(x-5)2– 2, and prove that it represents a function. Identify independent and dependent variables. Then find:
a) domain and range
b) where the graph is decreasing or increasing
c) possible maximum and minimum values of f(x) and where they occur d) f(10) and explain its meaning
e) a line of symmetry if any
f) x- and y- intercepts and explain their meaning. g) Is the function continuous, piecewise, or what? Plot the graph by table values:
X -1 2 5 8 11
y = f(x) = 0.1(x-5)2 – 2 1.6 -1.1 -2 -1.1 1.6
We have five points: (-1, 1.6), (2, -1.1), (5, -2), (8, -1.1), (11, 1.6). Plots these points and draw a smooth graph to represent the function f(x).
To prove that the equation y = 0.1(x-5)2– 2 represents a function, we have to find the direction of the function and prove that each independent variable produces only one unique dependent variable.
We assume that the function is f: x ‡y direction of the function x = independent variable
y = dependent variable which is = 0.1(x-5)2– 2
Graphical method:
If you draw a vertical line anywhere (in the domain) it will pass the graph of y = 0.1(x-5)2 – 2 only once. It implies that for each x-value there is only one y- value. The graph passes the vertical line test. The graph represents a function.
a)
There is no restriction on what must be the x-value. For any real input x-value the function f(x) = 0.1(x-5)2– 2 produces a real number outcome.
Domain: all possible x values: The set of all real numbers What are the possible y- values from this: y = 0.1(x-5)2– 2 ?
(x-5)2≥0 square of (x-5) is at least zero whatever the x-value is 0.1(x-5)2 ≥0 multiply by (0.1) both sides
0.1(x-5)2 -2 ≥0 – 2 Subtract (2) from both sides 0.1(x-5)2 -2 ≥– 2
y ≥– 2
Range: all possible y-values: y ≥– 2 we can also see this from the above diagram b)
The graph is decreasing for x < 5
The value of f(x) or y decreases when you go right in that region. The graph is increasing for x > 5
The value of f(x) or y increases when you go right in that region. c)
Minimum value of f(x) or y = -2. This occurs when x = 5
There is no maximum value because when you go left (move x toward -•) or when you go to right (increase x toward +•) the y-value increase indefinitely.
You can explain this by plugging in some big x-values into the function formula. y = f(x) = 0.1(x-5)2 – 2
If x = 20, y = 0.1(20-5)2– 2 = 20.5 If x = 200, y = 0.1(200-5)2– 2 = 3800.5
If x = 2000, y = 0.1(2000-5)2– 2 = 398000.5 y gets bigger without bound Also
If x = -20, y = 0.1(-20-5)2– 2 = 60.5 If x = -200, y = 0.1(-200-5)2– 2 = 4200.5
If x = -2000, y = 0.1(-2000-5)2– 2 = 402000.5 y gets bigger without bound d)
f(10) = = 0.1(10-5)2 – 2 = 0.5
When the x-value is 10, the function value f(x) or y is 0.5. e)
f)
y- intercept: where the graph of f(x) crosses y-axis. At these points x = 0 y = f(x) = 0.1(x-5)2 – 2
y = f(x) = 0.1(0-5)2 – 2 = 0.5
y- intercept (0, 0.5) point A sits on both the graph and y-axis x- intercept: where the graph crosses x-axis. At these points y = 0
y = f(x) = 0.1(x-5)2 – 2
0 = 0.1(x-5)2– 2 this equation represents the x- intercept points B & C
2 = 0.1(x-5)2 add 2 to both sides
20 = (x-5)2 divide both sides by 0.1
20
± = x – 5 square root both sides
5 ± 20 = x add 5 to both sides
x- intercepts are: (5 + 20, 0) and (5 - 20, 0) two points B & C sitting on both the graph and x-axis
g) The graph of the function is continuous. There is no discontinuity on the graph. So the function is continuous. It is defined using a single set of rule [y = 0.1(x-5)2– 2] for the whole domain. So it is not piecewise.
Some terms
Variable = a quantity that changes
Constant = a quantity that does not change
Function = a definite meaningful relation between two variables Piecewise function = behaves differently on different intervals (pieces)
Vertical line test = is used to confirm if a graph represents a function between two variables Independent variable = a variable that changes itself not depending on others
Dependent variable = a variable that changes depending on another variable Domain = the set of all possible independent variable in a function
Range = the set of all possible dependent variable in a function
Restriction = limitation or condition on the possible variables (independent or dependent or both)
Example 7: testing for function
Does the graph of the relation: x2 + y2= 16 represent a function? Explain. Algebraic method: We will test for f: x ‡ y
x2 + y2= 16 y2 = 16 - x2
y = + 16- x2or y = - 16- x2
If x = +4, or x = -4 the equations produce single y-value which is zero. For all other x-values between -4 and +4, we get two different y-values. For example if x = 3
Therefore x2+ y2= 16 does not represent a function. Its graph will be a graph but not a function.
Example (8): don’t get lost or confused with symbols.
y as a function of x can be written sometimes as y(x) to emphasize that y is a variable which depends on x- values.
Suppose that we are given the functions:
y = k(x) = 2x2+ 5x +1 y is a function of x given by k(x) x = h(t) = 2t + 1 x is a function of t given by h(t) The question asks us to find
a) y(0), and k(1) b) x(0), and h(1)
c) the value of y when t = 2 d) Express y in terms of t a)
y(0) is the value of y or k(x) when x = 0 y(0) = k(0) = 2x2 + 5x +1
y(0) = k(0) = 2(0)2+ 5(0) +1 = 1
k(1) is the value of k(x) or y when x = 1 k(1) = 2x2+ 5x +1
k(1) = 2(1)2 + 5(1) +1 = 8 b)
x(0) is the value of x or h(t) when t = 0 x(0) = h(0) = 2t + 1
x(0) = h(0) = 2(0) + 1 = 1
h(1) is the value of h(t) or x when t = 1 h(1) = 2t + 1 = 2(1) + 1 = 3
Graphical method:
c) the value of y when t = 2 When t = 2,
x = h(2) = 2t + 1 = 2(2) + 1 = 5 when t = 2, x = 5
y = k(5) = 2x2+ 5x +1 when x = 5, what is y?
y = k(5) = 2(5)2 + 5(5) +1 = 50 + 25 + 1 = 76 d) Express y in terms of t
y = k(x) = 2x2+ 5x +1 x = h(t) = 2t + 1 y = 2x2+ 5x +1
y = 2(2t +1) 2+ 5(2t + 1) +1 replace ‘x’ with 2t + 1 y = 2(4t2+ 4t + 1) + 10t + 5 + 1
y = 8t2+ 8t + 2 + 10t + 5 + 1
y = 8t2+ 18t + 8 y in terms of t
y in terms of t is a composite function of k and h. y = k(x) and x = h(t)
y = k[h(t)] OR y = (koh)(t)= y = 8t2+ 18t + 8
Questions to practice Question (1)
Test for function.
a) R = t2+ t - 10 for R in terms of t b) x = 10t + xt for x in terms of t
Question (2)
Two functions are given: y = f(x) = x + 1
y = g(x) = x + x2
Use the below definitions to write down the new expressions, and inspect whether they are functions. If they are functions what are their domains and ranges?
(f + g)(x) = f(x) + g(x) (f – g)(x) = f(x) – g(x)
(fg)(x) = [f(x)][g(x)] product function
f[g(x)] = (fog)(x) composite function f-of-g-x g[f(x)] = (gof)(x) composite function: g-of- f-x
) x ( g
) x ( f ) x )( g f
( quotient function
