all functions graphs

1. Graphs of
Functions
By Sanaullah Memon
Chapter One

2. CONTAINTS
Graphs of Functions 8
PhysicalApplicationsof Functions
Functionsin Economics

3. Linear Function
y = mx + c
The general form of a
linear function
ax + by + c = 0

4. Draw the graph of the linear functions: y = x - 1, y = x, y = x + 1 on the same graph
paper.
Solution: See the table below where tabular values of y = x – 1, y = x and y = x + 1 are
given.
x 0 1
y = x – 1 -1 0
y = x 0 1
y = x + 1 1 2
y = x + 1
y = x
y = x - 1
-1 0 1

5. REMARK: Look at the three graphs. Graph of y = x – 1 is shifted one unit to the
right and graph of y = x + 1 is shifted one unit left. The shape of each graph is
same.
Generally, for a given function y = f(x), the shape of graphs of f(x – a) and f(x + a)
will be same as that of f(x) but graph of f(x – a) is shifted `a` units to the right and
that of f(x + a) `a` units to the left of x-axis.
On the contrary, the graphs of f(x) – a and f(x) + a will have the same shape as that
of f(x) but they will be shifted `a` units down and up on the y-axis respectively,
where a > 0.

6. Quadratic Function
A function defined by the equation y = ax2
+ bx + c, a ≠ 0 where a, b, c are constants, is
called a quadratic function. This equation always represents a parabola. The graph of
parabola will have one of the shapes as shown below:
when a < 0 when a < 0
Similarly the equation x = ay2
+ by + c, where a ≠ 0 represents the parabola having one of
the following shapes:
when a > 0 when a < 0
Thus, if one of the variables in the equation appears with single power and the other in its
square form, then the graph is a parabola.

7. Examples
Draw the graphs of y = x2 – 1, y = x2 and y = x2 + 1 on the same graph paper.
Solution: See the table below where tabular values of y = x2
– 1, y = x2
and y
= x2
+ 1 are given.
x -1 0 1
y = x2
– 1 0 -1 0
y = x2
1 0 1
y = x2
+ 1 2 1 2
y = x2 + 1
y = x2
y = x2 -1
Observe that graphs of y = x2
– 1 and y = x2
+ 1 are similar to the graph of y = x2
but are
shifted one unit down and one unit up respectively on the y-axis.

8. Draw the graphs of y = (x – 1)2, y = x2 and y = (x + 1)2 on the same graph paper.
Solution: See the table below where tabular values of y = x2
– 1, y = x2
and y
= x2
+ 1 are given.
x -1 0 1
y = (x – 1)2
4 1 0
y = x2
1 0 1
y = (x + 1)2
0 1 4
y = (x + 1)2 y = x2 y = (x – 1)2
-1 0 1
Observe that graphs of y = (x – 1)2
and y = (x +1)2
are similar to the graph of y = x2
but are
shifted one unit right and one left respectively on the x-axis.

9. The circleare shifted one unit right and one left respectively on the x-axis.
3. The Circle
The equation of a circle with radius `r` and center (0, 0)
is x2
+ y2
= r2
so that 2 2 2 2 2
y r x y r x      
This equation does not represent a function because we see
that for one value of x, there are two values of y. Moreover, Df = [- r, r].
(i) 2 2
y r x  is the upper semi – circle and the equation
2 2
y r x   represents the lower semi – circle.
Both 2 2
y r x  and 2 2
y r x   are functions and
each one is called the branch of circle x2
+ y2
= r2
.
(ii) The equation x2
+ y2
= r2
can also be written as
x2
= - y2
+ r2
so that 2 2
x r y   . Since x is
positive on the right of y – axis, therefore graph of
2 2
x r y  is the right semi – circle.
Similarly 2 2
x r y   represents the left semi–circle.
Remark: The circle x2
+ y2
= 1 whose center is at the origin
and the radius is 1, is called the unit circle.

10. 4.SquareRootFunction
Squarerootfunctionisdefinedby  f x x,x 0   y x
Thedomainofthesquarerootfunctionconsistsof
allnonnegativerealnumbers,thatis,(x≥0).
becausesquarerootofanegativenumberisnot
arealnumber.Graphofthisfunctionisshownhere.
Square root
function

11. The Cube function is defined as f(x)=x3
, x R .Thedomainof the cube function
consistsofalltherealnumbers.
The reciprocal function is defined as: f(x) = 1/x, x ≠ 0. The graph of reciprocal function is
shown here and is known as rectangular hyperbola.
Cube Function Reciprocal Function
Cube and Reciprocal Function

12. Constant Function
Let f :R R be defined by f(x) = c for all x R , c being a fixed real number. Such
function is known as constant function.
c f(x) = c, c > 0
O
c f(x) = c, c <
(i)Functionf(x)=cisastraightlineparalleltox–axis(ii)f(x)=0representsthex–axis.

13. Identity FunctionREMARK: (i) Function f(x) = c is a straight line parallel to x – axis (ii) f(x) = 0
represents the x–axis. f(x) = x
8. Identity Function
A function f :R R defined by f(x) = x for all x R is called
an identity function. Its graph is a straight line passing through
through the origin and making angle of 45o
with the x and y–axes.
Here Df = R.
9. The Absolute Function

14. The Absolute Function
A function f :R R defined by
 
x, x 0
f x
x, x 0

 
 
f(x) = -x f(x) = x
is called the absolute function.
The graph of f consists of parts of the lines
f(x) = x and f(x) = -x above the x – axis.
It may be noted that this function is usually denoted by f(x) = | x |. Hence, it is also known
as modulus function.

15. 1| are shown as under. You may observe the shifting along y-axis and along x-axis
respectively.
y = |x| + 1
y = |x| y = |x + 1| y = |x| y = |x – 1|
y = |x| - 1

16. The Bracket Function
Greatest Integer Function: A function whose value at any number x is the greatest
integer less than or equal to x, is called ‘Greatest Integer Function’ and is usually denoted
by  f x x    . It is also known as “floor function”.
The graph is shown here.
 f x 0, 0 x 1
1, 1 x 2
2, 2 x 3
3, 3 x 4
.............................
1, 1 x 0
2, 2 x 1
3, 3 x 2
  
  
  
  
    
     
     
Notice that,      f 2.4 2.4 2, f 1.9 1.9 1, f 0.3 0.3 1,                  
     f 1.2 1.2 2, f 2 2 2, f 2 2 2                      and so on. It may be
noted that Df = R.

17. The Bracket Function
(ii) The Least Integer Function
A function whose value at any number x is the smallest integer greater than or equal to x,
denoted by  f x x    . It is also known as “ Ceiling function”. To draw the graph, we
find the points given below:
f(x) = -1, -2 < x ≤ -1
= 0, -1 < x ≤ 0
= 1, 0 < x ≤ 1
= 2, 1 < x ≤ 2
= 3, 2 < x ≤ 3
It may be noted that:
     f 3 3 3, f 5 5 5, f 2.4 2.4 3,                  
     f 1.9 1.9 2, f 0.3 0.3 0, f 1.2 1.2 1                     and so on. Also Df = R.

18. Piecewise Function
Sometimes a function uses different formulas on different parts of its domain. Such types
of functions are called piecewise functions. For example, consider a function defined as
under.
f(x) = -x f(x) = 1
  2
x, x 0
f x x , 0 x 1
1, x 1
 

  
 
f(x) = x2
is defined on the entire real line but has values given by different formulas depending on
the value of x.
The graph of this function is drawn by applying different formulas as given in the function
f(x).

19. Polynomial Function
An expression of the form an xn
+ an-1 xn-1
+ …+ a2 x2
+ a1 x + a0 is known as a polynomial
of degree n. Hence, the function of the form
f(x) = an xn
+ an-1 xn-1
+ …+ a2 x2
+ a1 x + a0
is known as a polynomial function. The domain of a polynomial function is a set of all
real numbers

20. function is called an “Algebraic Function” function if it can be expressed as the sum,
difference, products, quotients, powers or roots of polynomials.
For example, f(x) = 2x3
+ 5x – 7 is polynomial function and g(x) = (x2
+ 1)/(x-2),
h(x) = 2
x 2x 5  and k(x) = x (x2
+ 5) + 2x 5 are algebraic functions.

21. Transcendental Function
Functions that are not algebraic are known as transcendental functions. The combination
of algebraic and transcendental functions is also a transcendental function. Such functions
contain any trigonometric, inverse trigonometric, exponential or logarithmic functions. For
example, the function:
f(x) = x3
+ x 6 + cos x – ex
+ log x is a transcendental function.

22. Bounded Functions
A function f(x) from R to R is said to be bounded if range of f is bounded otherwise it is
unbounded function. For example, f(x) = 2
4 x is bounded function because its range is
[-2, 2]. This function represents the upper half of the circle centered at origin and radius 2.
Graph is shown on page 10.
The functions f(x) = x2
, g((x) = x + 1 are unbounded because the range of f(x) is the set of
non-negative real numbers and that of g(x) is the set of all real numbers. Both sets are
unbounded hence f(x) and g(x) are unbounded functions.

23. Circular functions

24. This Photo by Unknown Author is licensed under CC BY-SA-NC

25. Hyperbolic Functions
In the following figures, we have shown the graphs of the exponential functions x
e and
x
e
respectively. Moreover, the graphs of the curve  x x
y e e / 2
  and that of
 x x
y e e / 2
  are also shown below:
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
-2
0
2
4
6
8
10
12
14
16
18
20
22
Exponential function, y = ex
X - axis
Y-axis
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
-2
0
2
4
6
8
10
12
14
16
18
20
Exponential function y = e-x
X - axis
Y-axis
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
Hyperbolic function y = cosh(x)
X - aixs
Y-axis
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3
-15
-12
-9
-6
-3
0
3
6
9
12
15
Hyperbolic function y = sinh(x)
X - aixs
Y-axis

27. Vertical Line Test
The definition of a function says that for every x there is one y. The graphical interpretation
of this idea is considered in the vertical line test. If a vertical line (VL) intersects a curve
in two or more points, then the graph of the given curve does not represent a function. In
this case, there would be two or more y values corresponding to a particular value of x. See
the following figures:
VL VL
VL VL
The graphs of first two figures do not represent functions because VL cuts the graphs at
two points, whereas, the last two graphs represent the function because VL cuts the graphs
at only one point.

28. PHYSICALAPPLICATIONS OF
FUNCTIONS
There is perhaps no field or area where functions are not used. In real
life, where there is a relation between two variables, the application of
function is must. These applications are found in social and natural
sciences, engineering and medical sciences, etc. Applications of
functions are also known as mathematical modeling of functions.

29. Example 01: A rectangular fence is to be constructed so that its length is 3x + 2 and
itswidthisxmeters.IfPisthefunctionthatgivesperimeter,determineP(x)?
Solution:Perimeteristhesumofthesidesofany polygon. 3x+2
Nowforrectangleasshowninthefigure,
P(x) =2(3x+2)+2x=8x+4meters. x

30. Example 02: A colony of bacteria is placed into a
growth inhabiting environment. The number of bacteria present at any time t (
in hours) is given by n(t) = 1000 + 20t + t2. Find: n(0), n(1) and n(10). Interpret your
results.
Solution: n(0) = 1000 + 0 + 0 = 1000. This means that at the beginning (t = 0) the number
of bacterial in the colony is 1000.
n(1) = 1000 + 20 + 1 = 1021. This means that after one hour, number of bacteria in the
bacterial colony is 1021.
n(10) = 1000 + 20 (10) + (10)2
= 1000 + 200 + 100 = 1300. This means that number of
bacteria in the colony after 10 hours is 1300.

31. Example03:Whenacarismovingatxmilesperhourandthedriverdecidestoslam
on the brakes, the car will travel x + 0.5 x2 ft. If car travels 175 ft after the driver
decidestostop,howfastwascarmoving?
Solution: The answer to this problem is very simple. Equating both conditions, we get:
x+0.5x2
=175
Multiplyingby2,weget: x2
+2x–350=0.
Thisisquadraticequation.Solving,weobtain:x=17.8 ; 18miles/h.

32. Example 04: A person weighing 150 pounds on earth has weight given by:
2
2,400,000,000
w(d)
(4,000 d)


miles above the earth surface.
a. Find w(0) and interpret this?
b. How much this person weigh while flying in air plane at 29,000 feet?
c. An astronaut orbits the earth at an average of 80 miles above the surface. If
he weighs 150 pounds on earth, how much does he weigh while in orbit?
Solution: (a) w(0) = (2, 400, 000, 000)/(4, 000)2
= 150. This means that weight of a person
on the earth is 150 lbs.
(b) d = 29, 000 ft = 29, 000/(5280) = 5.5 miles. (NOTE: 1 mile = 5280 ft)
Putting this in given equation, we get:
w = (2, 400, 000, 000)/(4005.5)2
= 149.6
This means the weight of person at the height of 29, 000 ft above the ground is 149.6 lbs.
(c) Putting d = 80 in the formula, we get:
w = (2, 400, 000, 000)/(4080)2
= 144.2 lbs

33. Example 05
A number of degrees d in each interior angle of a region of polygon of n sides are:
 d 180n 360 / n 
Use this formula to compute the number of degrees for a polynomial of sides 3, 4, 5
and 6.
Solution: If n = 3, d = [180(3) - 360]/3 = 60o
 Each angle of equilateral triangle is 60o
.
If n = 4, d = [180(4) - 360]/4 = 90o
 Each angle of square is 90o
.
If n = 5, d = [180(5) - 360]/5 = 108o
 Each angle of regular pentagon is 108o
.
If n = 6, d = [180(6) - 360]/5 = 144o
 Each angle of regular hexagon is 144o
.
REMARK: This formula is valid only for regular polygons.

34. Example 06: One of the methods to determine the children’s dosage D of medicine is
given by D(c) = [c – 1]a/24 where, c is the child age and `a` is the dosage of adult. If a
child is 8 year old, what is his dosage if adult dosage is 400 mg? Interpret the result.
Solution: Putting c = 8 and a = 400, we get: D = (8 – 1).400/24 = 116.7 mg. Thus according
to formula given, if the adult dosage is 400 mg and the child is 8 year old then his dosage
will be 116.7 mg.
Example 07: The following formula is used to know the depreciated value of an item:
D = C – [(C – S)t]/n where t is time in years, D the depreciated value after t years, C
is the original cost, n is the useful life in years and S is the scrape value (Resale value).
What will be the depreciated value of the machine after 8 years when it was purchased
for 3400 dollars and its present value (scrape value) is 400 dollars and the useful time
is 15 years?
Solution: Here, C = $3400, S = $400, n = 15 years, t = 8. Thus,
D = 3400 - [8(3400 – 400)]/15 = $1800
Thus depreciated value of the machine after 8 years was $1800.

36. Example 12: It costs a manufacturer C(x) = 0.4x2 + 7x + 95 dollars to produce x
computer chips. They can be sold at $40 each; that is, revenue from the sale of x chips
is R(x) = 40x dollars.
(a) Determine the profit function
(b) What is the profit on the manufacture and sale of 25 chips?
(c) What is the profit on the manufacture and sale of 25th chip?
(d) What is the profit on the manufacture and sale of 2 chips?
Solution: (a) Using P(x) = R(x) – C(x), we have
P(x) = 40x – (0.4x2
+ 7x + 95) = -0.4x2
+ 33x – 95$
(b) P(25) = -0.4(25)2
+ 33(25) – 95 = $480.
This means on manufacturing and sale of 25 chips, the profit will be $480.
(c) The profit on the sale of 25th
chip = P(25) – P(24) = 480 – 466.6 = $13.4
(d) P(2) = -0.4(4) +33(2) – 95 = -30.6
The negative sign shows that there is a loss of 30.6 dollars. This means that the company
would loss $30.6 on the production and sale of only 2 chips.

37. Workers at a fast food restaurant earn $5 per hour for the first 40 hours in a week
and then $7.5 per hour for additional hours. Let x be the number of hours worked in
a week, write a two piece function P that describes a worker’s pay. What would be
the pay if a worker works for (a) 35 hours (b) 45 hours.
Solution: If a worker works for 0 up to 40 hours, he will be paid $5 per hour. Hence his
pay is a function of x (hours), which is
P(x) 5x for 0 x 40  
When x is greater than 40, the worker makes $5 per hour for 40 hours ($200 total) plus
$7.5 per hour for each extra hour over 40. Now the extra hours are (x – 40) and then earning
would be {200 + 7.5(x – 40)} dollars. After simplification, we have:
P(x) = 7.5x – 100 for x > 40
Thus, the two – piece function P is given by the formula
 
5x, for 0 x 40
P x
7.5x 100, for x 40
 
 
 
Now P(35) = 5(35) = $175, and P(45) = 7.5(45) – 100 = $237.5

38. : The monthly charge for water in a small town is given by
 
18, for 0 x 20
f x
18 0.1(x 20) for x 20
 
 
  
where x is in hundreds of gallons and f is in dollars. Find the monthly charge for each
of the following usages.
(i) 30 gallons (ii) 3000 gallons (iii) 4000 gallons
Solution: (i) Since, x is in hundreds of gallons, hence 1 unit of gallon will be
1/100 = 0.01. Thus 30 gallons will be equivalent to 0.30 units. Now, according to the given
domain: f(0.3) = $18
(ii) Since x is in hundreds of gallons, 3000 gallons will be equivalent to 301003000  of
units. Now, according to given domain:
f(30) = 18 + 0.1(30 – 20) = $19
(iii) Since x is in hundreds of gallons, 4000 gallons will be equivalent to 401000004 
of units. Now, according to given domain:
f(40) = 18 + 0.1(40 – 20) = $20
EXAMPLE

39. Exercise 01
1. Draw the graphs of the following functions. Also mention the domain and range of these
functions.
(a) y = 2x + 7 (b) y = 2x2
+ 1 (c) y = (x2
+ 1)/(x – 1)
2. Solve each of the following inequalities:
(a) |2x + 5| > |2 – 5x| (b) |x| + |x – 1| > 1 (c) 12x2
– 25x + 12 > 0
(d) |x2
+ x + 1| > 1 (e) x-2
– 4x-1
+ 4 > 0 (f) 2x/(x + 2) ≥ x/(x – 2)
3. If 2
2
1
f (x) x 1 and g(x)
4 x
  

, show that fog≠ gof.
4. In 1998, a patient paid $300 per day for a semiprivate hospital room and $1500 for an
appendectomy operation. Express the total amount for an appendectomy as a function of
number of days of hospital confinement.
5. In some cities you can rent a car for $18 per day and $ 0.20 per miles.
a. Find the cost of renting the car for one day and driving 200 miles.
b. If the car is rented for one day then express the total rental expenses as a function of
the number x miles driven.
6. Suppose the longer side of a rectangle has twice the length of shorter side, and if x is
length of shorter side, express the perimeter of the rectangle as a function of x.
7. The monthly charge (in dollars) for x kilowatt hours (KWH) of electricity used by a
commercial customer is given by the following function:
 
7.52 0.1079x, 0 x 5
19.22 0.1079x, 5 x 750
C x
20.795 0.1058x, 750 x 1500
131.345 0.0321x, x 1500
  
   
 
  
  
Find the monthly charge for the following usages.

40. (a) 5 KWH (b) 6 KWH (c) 3000 KWH.
8. The pressure P of a certain gas is related to volume V according to: P = 100/V
(a) Is 0 in the domain of this function?
(b) What are P(100) and P(50)?
(c) As volume decreases, what happens to pressure?
9. The total cost of producing a product is given by
C(x) = 300 x + 0.1 x2
+ 1, 200 dollars
where x represents the number of units produced.
(a) What is the total cost of producing 10 units?
(b) What is the average cost per unit when 10 units are produced?
10. A cigar box distributor’s revenue is R(x) = 1.35 x dollars. Where x is the number of
boxes sold.
(a) How much revenue is obtained from selling 5 boxes?
(b) How much revenue is obtained from the sale of 5th
box?
(c) How much revenue is obtained from the sale of 8th
box?
11. It costs a TV manufacturer C(x) = 0.1 x2
+ 150x +1, 000 dollars to produce x TV sets.
The revenue from the sale of x TV sets is R(x) = 280x dollars.
(a) Determine the profit function.
(b) What is the profit on the manufacture and sale of 50 TV sets?
12. The cost on producing x radios is C(x) = 0.4 x2
+ 7x + 95 $. The revenue received is
R(x) = 40x $. What is the profit function? Find P(24) and P(25). What is the profit on the
sale of 25th
radio?
13. A psychologist needs volunteers for an experiment. She offers to pay $ 8 per hour for
volunteer who works up to 5 hours. Those who work more than 5 hours are paid $10 per
hour for additional hour.
(a) Write down the function that represents the volunteer’s pay V(x), where x represents
the hours worked. (b) Also find V(5), V(10) and V(15).
14. If p = 0.01x + 19 dollars is the price of one jacket and each jacket is sold for $80,
determine the profit function and then find P(2) and (50). What is BEP?

41. 15. Sketch the graph of following piecewise functions.
(a)  
2 for x 0
f x
x for x 40

 
 
(b)   2
x for x 0
f x
x for x 0

 

(c)  
x for x 1
f x
x for x 1

 

(d)  
| x | for x 0
f x
2x for x 0

 

16. A motorbike was purchased 2 years ago for Rs. 50, 000 is now worth Rs. 25, 000. Find
the linear relationship between its value (y) and the life (x) in years.
(a) How many years from the time of purchase will it be before motorbike is worth
Rs. 35, 000.
(b) What would be its cost after 10 years?
17. A house was purchased 5 years ago for Rs. 5 million is now worth Rs. 8 million. Find
the linear relationship between its value (y) and the life (x) in years.
(a) How many years from the time of purchase will it be before house is worth Rs. 6
millions.
(b) What would be its cost after 8 years?
18. State which of the following functions is even or odd.
(a) f(x) = x4
+ x2
– 1
(b) g(x) = x3
+ sin x + 1
(c) h(x) = x3
– x
(d) k(x) = |x| + cos x – 2
19. What is the domain and range of the following functions?
(i) f(x) = 1/(x2
– 1)
(ii) g(x) = (x + 1)/(2x – 1)
(iii) h(x) = 1/ 2
x 9
(iv) k(x) =
2
1
16 x
(v) m(x) =
2
1
x 16
(vi) n(x) = 2
16 x

42. (vii) p(x) = 2
x 16
(viii) q(x) = 2
x 16 / 2
x 9

43. THE END