MATH 201 - Week 2 · 2008. 1. 16. · MATH 201 - Week 2 MATH 201 - Week 2 Ferenc Balogh Concordia University 2008 Winter Based on the textbook J. Stuart, L. Redlin, S. Watson, Precalculus

MATH 201 - Week 2

MATH 201 - Week 2

Ferenc Balogh

Concordia University

2008 Winter

Based on the textbook

J. Stuart, L. Redlin, S. Watson, Precalculus - Mathematics for Calculus, 5th Edition, Thomson

MATH 201 - Week 2

Overview

Lines (Chapter 1, Section 1.10)

Slope of a LineEquations of Lines

Point-Slope FormSlope-Intercept FormGeneral Equation

Parallel and Perpendicular LinesSlope as Rate of Change

What is a Function? (Chapter 2, Section 1)

Definition of FunctionEvaluation of a FunctionThe Domain of a Function

Graphs of Functions (Chapter 2, Section 2)

Graphing FunctionsGraphing Piecewise Defined FunctionsThe Vertical Line TestEquations That Define Functions

MATH 201 - Week 2

Overview









MATH 201 - Week 2

Overview









MATH 201 - Week 2

Lines

Slope of a Line

The slope gives the ”steepness” of a straight line as we move fromthe left to the right.It is defined as

slope =riserun

.

The Slope of a Line

If P(x1, y1) and Q(x2, y2) are on theline l , then the slope of l is given by

m =y2 − y1

x2 − x1

The slope can be positive, negative or zero.

Important: The slope of a vertical line is not defined!

MATH 201 - Week 2

Lines

Slope of a Line

The slope gives the ”steepness” of a straight line as we move fromthe left to the right.It is defined as

slope =riserun

.

The Slope of a Line

If P(x1, y1) and Q(x2, y2) are on theline l , then the slope of l is given by

m =y2 − y1

x2 − x1

The slope can be positive, negative or zero.

Important: The slope of a vertical line is not defined!

MATH 201 - Week 2

Lines

Slope of a Line

Problem. Find the slope of the line that passes through the pointsP(2,−1) and Q(4, 2).

Solution.

x1 = 2, y1 = −1, x2 = 4, y2 = 2.

m =2− (−1)

4− 2=

3

2

MATH 201 - Week 2

Lines

Slope of a Line

Problem. Find the slope of the line that passes through the pointsP(2,−1) and Q(4, 2).Solution.

x1 = 2, y1 = −1, x2 = 4, y2 = 2.

m =2− (−1)

4− 2=

3

2

MATH 201 - Week 2

Lines

Equations of Lines: the Point-Slope Form

Question. Suppose that we are given a point P(x0, y0) in theplane and the slope m.What is the equation of the line which passes through P and hasslope m?

Answer. The moving point Q(x , y) is on the line if and only if

m =y − y0

x − x0.

The point-slope form

If m is the slope of a line and it passes through P(x0, y0) then theequation of the line is

m(x − x0) = y − y0.

MATH 201 - Week 2

Lines


Question. Suppose that we are given a point P(x0, y0) in theplane and the slope m.What is the equation of the line which passes through P and hasslope m?Answer. The moving point Q(x , y) is on the line if and only if

m =y − y0

x − x0.

The point-slope form

If m is the slope of a line and it passes through P(x0, y0) then theequation of the line is

m(x − x0) = y − y0.

MATH 201 - Week 2

Lines


Problem. Find the equation of the line which passes throughP(1, 1) and has slope 2.

Solution. Using the point-slope form ofthe equation,

2(x − 1) = y − 1

2x − 2 = y − 1

2x − y − 1 = 0

So the equation of the line is

2x − y − 1 = 0.

MATH 201 - Week 2

Lines


Problem. Find the equation of the line which passes throughP(1, 1) and has slope 2.

Solution. Using the point-slope form ofthe equation,

2(x − 1) = y − 1

2x − 2 = y − 1

2x − y − 1 = 0

So the equation of the line is

2x − y − 1 = 0.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line through the pointsP(−2, 1) and Q(3,−1).

Solution. We calculate the slope of theline first:

m =(−1)− 1

3− (−2)=−2

3 + 2= −2

5.

Now the point-slope form gives

−2

5(x − (−2)) = y − 1

−2(x + 2) = 5(y − 1)

−2x − 4 = 5y − 5

0 = 2x + 5y − 1

So the equation of the line is 2x + 5y − 1 = 0.

MATH 201 - Week 2

Lines




m =(−1)− 1

3− (−2)=−2

3 + 2= −2

5.


−2

5(x − (−2)) = y − 1

−2(x + 2) = 5(y − 1)

−2x − 4 = 5y − 5

0 = 2x + 5y − 1


MATH 201 - Week 2

Lines




m =(−1)− 1

3− (−2)=−2

3 + 2= −2

5.


−2

5(x − (−2)) = y − 1

−2(x + 2) = 5(y − 1)

−2x − 4 = 5y − 5

0 = 2x + 5y − 1


MATH 201 - Week 2

Lines

Equations of Lines: The Slope-Intercept Form

Question. Suppose that we know the y -intercept of a line and itsslope m.What is the equation of the line?

Answer. The line intersects the y -axis at the point P(0, b). So wecan use the point-slope form of the equation:

m(x − 0) = y − b.

This gives

The Slope-Intercept Form

If a line has y -intercept b its slope is m then

y = mx + b

is an equation of the line.

MATH 201 - Week 2

Lines


Question. Suppose that we know the y -intercept of a line and itsslope m.What is the equation of the line?Answer. The line intersects the y -axis at the point P(0, b). So wecan use the point-slope form of the equation:

m(x − 0) = y − b.

This gives

The Slope-Intercept Form

If a line has y -intercept b its slope is m then

y = mx + b

is an equation of the line.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line with slope 2 andy -intercept −3.

Solution. The slope-intercept form ofthe equation is

y = 2x − 3.

So the equation of the line is 2x − y − 3 = 0.

MATH 201 - Week 2

Lines




y = 2x − 3.


MATH 201 - Week 2

Lines




y = 2x − 3.


MATH 201 - Week 2

Lines

Vertical and Horizontal Lines

Equation of a horizontal line: y = a.

Equation of a vertical line: x = b.

MATH 201 - Week 2

Lines


Equation of a horizontal line: y = a.

Equation of a vertical line: x = b.

MATH 201 - Week 2

Lines


Problem. Find equations of the horizontal line and the verticalline through the point R

(2, 3

4

).

Solution. The horizontal line is givenby

y =3

4.

The vertical line is given by

x = 2.

MATH 201 - Week 2

Lines


Problem. Find equations of the horizontal line and the verticalline through the point R

(2, 3

4

).

Solution. The horizontal line is givenby

y =3

4.

The vertical line is given by

x = 2.

MATH 201 - Week 2

Lines

General Equation of a Line

A linear equation is of the form

Ax + By + C = 0

where A, B, C are constants and A and B are not both 0.

Examples.A non-vertical line has an equation of the form

mx − y + b = 0 (A = m, B = −1, C = b).

A vertical line has an equation of the form

x − a = 0 (A = 1, B = 0, C = −a).

MATH 201 - Week 2

Lines


A linear equation is of the form

Ax + By + C = 0

where A, B, C are constants and A and B are not both 0.

Examples.A non-vertical line has an equation of the form

mx − y + b = 0 (A = m, B = −1, C = b).

A vertical line has an equation of the form

x − a = 0 (A = 1, B = 0, C = −a).

MATH 201 - Week 2

Lines


Conversely, the graph of every linear equation

Ax + By + C = 0

is a line:

If B 6= 0 then the equation is

y = −A

Bx − C

B

which is the equation of a non-vertical line with slope −A/Band y -intercept −C/B.

If B = 0 then A 6= 0 and the equation is

x = −C

A

which is the equation of a vertical line.

MATH 201 - Week 2

Lines



The graph of every linear equation

Ax + By + C = 0

(where A and B are not both 0) is a line.Conversely, every line is a graph of a linear equation.

ALGEBRA GEOMETRY

MATH 201 - Week 2

Lines


Problem. Sketch the graph of the linear equation x − 2y + 3 = 0.

Solution. The x-intercept of the line isgiven by the equation

x − 2 · 0 + 3 = 0

x = −3

The y -intercept is given by

0− 2y + 3 = 0

−2y = −3

y =3

2The line is passing through the points (−3, 0) and

(0, 3

2

).

MATH 201 - Week 2

Lines


Problem. Sketch the graph of the linear equation x − 2y + 3 = 0.Solution. The x-intercept of the line isgiven by the equation

x − 2 · 0 + 3 = 0

x = −3

The y -intercept is given by

0− 2y + 3 = 0

−2y = −3

y =3

2The line is passing through the points (−3, 0) and

(0, 3

2

).

MATH 201 - Week 2

Lines


Problem. Sketch the graph of the linear equation5x + 2y − 1 = 0.

Solution. The second way is to use theslope-intercept form:

5x + 2y − 1 = 0

2y = −5x + 1

y = −5

2x +

1

2.

MATH 201 - Week 2

Lines


Problem. Sketch the graph of the linear equation5x + 2y − 1 = 0.

Solution. The second way is to use theslope-intercept form:

5x + 2y − 1 = 0

2y = −5x + 1

y = −5

2x +

1

2.

MATH 201 - Week 2

Lines

Parallel and Perpendicular Lines

Parallel Lines

Two non-vertical lines are parallel if and only if their slopes are thesame.

Note that two vertical lines are always parallel.

Perpendicular Lines

Two lines with slopes m1 and m2 are perpendicular if and only if

m1m2 = −1.

Note that a horizontal line is always perpendicular to a vertical one.

MATH 201 - Week 2

Lines


Parallel Lines

Two non-vertical lines are parallel if and only if their slopes are thesame.

Note that two vertical lines are always parallel.

Perpendicular Lines

Two lines with slopes m1 and m2 are perpendicular if and only if

m1m2 = −1.

Note that a horizontal line is always perpendicular to a vertical one.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line through the point T (−1, 2)which is parallel to the line 3x − 2y + 4 = 0.

Solution. The slope-intercept form of the equation of the givenline:

2y = 3x + 4

y =3

2x + 2.

The slope of this line is 32 . The parallel line has the same slope

therefore the point-slope form of its equation gives

3

2(x − (−1)) = y − 2

3(x + 1) = 2(y − 2)

3x + 3 = 2y − 4

3x − 2y + 7 = 0.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line through the point T (−1, 2)which is parallel to the line 3x − 2y + 4 = 0.Solution. The slope-intercept form of the equation of the givenline:

2y = 3x + 4

y =3

2x + 2.

The slope of this line is 32 . The parallel line has the same slope

therefore the point-slope form of its equation gives

3

2(x − (−1)) = y − 2

3(x + 1) = 2(y − 2)

3x + 3 = 2y − 4

3x − 2y + 7 = 0.

MATH 201 - Week 2

Lines


Problem. Show that the points P(1, 0), Q(3, 1) and R(1, 5) arethe vertices of a right triangle.

Solution. The slope of the line passingthrough P and Q is

m1 =1− 0

3− 1=

1

2.

The slope of the line passing through Qand R is

m2 =5− 1

1− 3= −4

2= −2.

The product of the slopes is

m1m2 =1

2(−2) = −1.

We conclude that the triangle with vertices P, Q and R is a righttriangle with right angle at Q.

MATH 201 - Week 2

Lines


Problem. Show that the points P(1, 0), Q(3, 1) and R(1, 5) arethe vertices of a right triangle.Solution. The slope of the line passingthrough P and Q is

m1 =1− 0

3− 1=

1

2.

The slope of the line passing through Qand R is

m2 =5− 1

1− 3= −4

2= −2.

The product of the slopes is

m1m2 =1

2(−2) = −1.

We conclude that the triangle with vertices P, Q and R is a righttriangle with right angle at Q.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line that is perpendicular to theline given by the equation 2x − y + 1 = 0 and passes through thepoint P(−1,−2).

Solution. The slope-intercept form of the equation of the givenline:

y = 2x + 1

The slope m2 of the yet unknown line is given by 2m1 = −1 so

m2 = −1

2.

The point-slope form provides the equation of the perpendicularline:

−1

2(x − (−1)) = y − (−2)

−(x + 1) = 2(y + 2)

−x − 1 = 2y + 4

0 = x + 2y + 5.

MATH 201 - Week 2

Lines


Problem. Find an equation of the line that is perpendicular to theline given by the equation 2x − y + 1 = 0 and passes through thepoint P(−1,−2).Solution. The slope-intercept form of the equation of the givenline:

y = 2x + 1

The slope m2 of the yet unknown line is given by 2m1 = −1 so

m2 = −1

2.

The point-slope form provides the equation of the perpendicularline:

−1

2(x − (−1)) = y − (−2)

−(x + 1) = 2(y + 2)

−x − 1 = 2y + 4

0 = x + 2y + 5.

MATH 201 - Week 2

Lines

Slope as Rate of Change

Example.The water level in a lake is changingaccording to the graph.It is easy to see that the water levelchanges linearly.

For example, it increases 4 inches in each 6 year-long period. Inother words, the rate of change is the ratio

4 inches6 years

=2

3

inchyear

.

The rate of change between two linearly related quantities is givenby the slope of the corresponding graph.

MATH 201 - Week 2

Lines


Example.The water level in a lake is changingaccording to the graph.It is easy to see that the water levelchanges linearly.

For example, it increases 4 inches in each 6 year-long period. Inother words, the rate of change is the ratio

4 inches6 years

=2

3

inchyear

.

The rate of change between two linearly related quantities is givenby the slope of the corresponding graph.

MATH 201 - Week 2

Lines


Problem. The tank of a car is full containing 40 litres of gas. After 572kilometres of (monotonous) driving along the Trans-Canada Highwayfrom Winnipeg to Regina, the tank contains 3 litres only.

What is the rate of change between the amount of gas consumedand the distance driven?

Find an equation which gives the amount of gas in terms of thedistance from Winnipeg (assuming linearity).

Solution. We can assume that the amount of gas is changing linearly interms of the distance (Trans-Canada highway).The rate of change is given by the slope of that line:

m =3− 40

572− 0

lkm

= − 37

572

lkm≈ 0.065

lkm

Using the slope-intercept form, we have the equation

y = − 37

572x + 40.

MATH 201 - Week 2

What is a Function?

Definition of Function

Examples.

temperature as a function of time

source: www.theweathernetwork.com

postage as a function of weight

source: www.canadapost.ca

MATH 201 - Week 2

What is a Function?


Function

Let A and B denote two sets.A function from the set A to the function B is a rule that assignsto each element x in A exactly one element, called f (x) in B.We write

f : A→ B.

MATH 201 - Week 2

What is a Function?


Some terminology for a function f : A→ B

f (x) is called the value of f at x or the image of x under f .

The set A is called the domain of the function f .

The range of the function f is the set of all possible values off (x) as x runs through the domain of f :

Range(f ) = {f (x) | x ∈ A}.

The symbol that represents an element in the domain of f iscalled an independent variable.

The symbol that represents an element in the range of f iscalled a dependent variable.

For example, if one writes

v = f (u),

then u is the independent variable, v is the dependent variable.

MATH 201 - Week 2

What is a Function?






Range(f ) = {f (x) | x ∈ A}.




v = f (u),


MATH 201 - Week 2

What is a Function?






Range(f ) = {f (x) | x ∈ A}.




v = f (u),


MATH 201 - Week 2

What is a Function?

Evaluation of a Function

Problem. Let f (x) = 2x2 − 4x + 3.Calculate f (2), f (−3), f (0), f

(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?



(35

), f (√

2), f (a), f (b + 2).

Solution.

f (2) = 2 · 22 − 4 · 2 + 3 = 8− 8 + 3 = 3

f (−3) = 2 · (−3)2 − 4 · (−3) + 3 = 18 + 12 + 3 = 33

f (0) = 2 · 02 − 4 · 0 + 3 = 0 + 0 + 3 = 3

f(

35

)= 2 ·

(35

)2 − 4 · 35 + 3 = 18

25 −125 + 3 = 18−60+75

25 = 3325

f (√

2) = 2 · (√

2)2 − 4 · (√

2) + 3 = 4− 4√

2 + 3 = 7− 4√

2

f (a) = 2 · (a)2 − 4 · (a) + 3 = 2a2 − 4a + 3

f (b + 2) = 2 · (b + 2)2 − 4 · (b + 2) + 3 =2b2 + 8b + 8− 4b − 4 + 3 = 2b2 + 4b + 7

MATH 201 - Week 2

What is a Function?


Problem.Write down a formula for the postage tobe paid for a letter of type

other lettermail

as a function p of the weight w . source: www.canadapost.ca

Solution. The domain of the function is the interval [0, 500] (theunit is grams). In other words, w should be between 0 and 500.

p(w) =

1.15 if 0 ≤ w ≤ 100

1.92 if 100 < w ≤ 200

2.65 if 200 < w ≤ 500

MATH 201 - Week 2

What is a Function?


Problem.Write down a formula for the postage tobe paid for a letter of type

other lettermail

as a function p of the weight w . source: www.canadapost.ca

Solution. The domain of the function is the interval [0, 500] (theunit is grams). In other words, w should be between 0 and 500.

p(w) =

1.15 if 0 ≤ w ≤ 100

1.92 if 100 < w ≤ 200

2.65 if 200 < w ≤ 500

MATH 201 - Week 2

What is a Function?

The Domain of a Function

The domain of the function may be stated explicitly:In the previous example, p was defined on the domain0 ≤ w ≤ 500.

Convention. If the domain is given by an algebraic expression andthe domain is not stated explicitly, then the domain of the functionis the the set of all real numbers for which the expression is defined.Examples.

h(x) =1

x − 2r(x) =

√x − 1

The function h is given by an expression which is not defined atx = 2 (the denumerator is 0 there). So the domain of h is{x | x 6= 2}.The square root is not defined for negative numbers: the domainof g is therefore {x | x ≥ 1}.

MATH 201 - Week 2

What is a Function?



Convention. If the domain is given by an algebraic expression andthe domain is not stated explicitly, then the domain of the functionis the the set of all real numbers for which the expression is defined.

Examples.

h(x) =1

x − 2r(x) =

√x − 1


MATH 201 - Week 2

What is a Function?



Convention. If the domain is given by an algebraic expression andthe domain is not stated explicitly, then the domain of the functionis the the set of all real numbers for which the expression is defined.Examples.

h(x) =1

x − 2r(x) =

√x − 1


MATH 201 - Week 2

What is a Function?


Problem. Find the domain for each function.

f (x) =x − 2

x + 3g(s) =

√4− s2 h(r) =

r2

√r − 1

.

Solution.

The denumerator of f cannot be zero therefore the domain off is {x | x 6= −3}.

The expression 4− s2 in g is under a square root so it mustbe nonnegative: the domain of g is {s | − 2 ≤ s ≤ 2}.The expression r − 1 in h is under a square root so it must benonnegative and it cannot be zero. The domain of h is{r | r > 1}.

MATH 201 - Week 2

What is a Function?



f (x) =x − 2

x + 3g(s) =

√4− s2 h(r) =

r2

√r − 1

.

Solution.

The denumerator of f cannot be zero therefore the domain off is {x | x 6= −3}.The expression 4− s2 in g is under a square root so it mustbe nonnegative: the domain of g is {s | − 2 ≤ s ≤ 2}.

The expression r − 1 in h is under a square root so it must benonnegative and it cannot be zero. The domain of h is{r | r > 1}.

MATH 201 - Week 2

What is a Function?



f (x) =x − 2

x + 3g(s) =

√4− s2 h(r) =

r2

√r − 1

.

Solution.

The denumerator of f cannot be zero therefore the domain off is {x | x 6= −3}.The expression 4− s2 in g is under a square root so it mustbe nonnegative: the domain of g is {s | − 2 ≤ s ≤ 2}.The expression r − 1 in h is under a square root so it must benonnegative and it cannot be zero. The domain of h is{r | r > 1}.

MATH 201 - Week 2

Graphs of Functions

The Graph of a Function

Let f : A→ R be a function on the domain A ⊆ R.The graph of f is the set of points

{(x , f (x)) | x in A}

in the coordinate plane.

Example.

f (x) = 4x3 + 2x2 − 3x + 2

MATH 201 - Week 2

Graphs of Functions

Graphing Functions

Examples.

A linear function is of the form

f (x) = mx + b

where m and b are some real numbers.The graph of this linear function is theline y = mx + b.

As a special case, if m = 0 we have aconstant function

f (x) = b

whose graph is the horizontal liney = b.

MATH 201 - Week 2

Graphs of Functions

Graphing Functions

Problem. Sketch the graph of the function

g(x) = (x − 1)2 − 1.

Solution.We make a table of values first for somespecific values of x :

x g(x) (x , g(x))

0 0 (0, 0)

1 −1 (1,−1)

−1 3 (−1, 3)

2 0 (2, 0)

−2 8 (−2, 8)

Then we plot the points and try to jointhem to obtain a curve.

MATH 201 - Week 2

Graphs of Functions

Graphing Functions


g(x) = (x − 1)2 − 1.


x g(x) (x , g(x))

0 0 (0, 0)

1 −1 (1,−1)

−1 3 (−1, 3)

2 0 (2, 0)

−2 8 (−2, 8)


MATH 201 - Week 2

Graphs of Functions

Graphing Functions


g(x) = (x − 1)2 − 1.


x g(x) (x , g(x))

0 0 (0, 0)

1 −1 (1,−1)

−1 3 (−1, 3)

2 0 (2, 0)

−2 8 (−2, 8)


MATH 201 - Week 2

Graphs of Functions

Graphing Functions


w(x) =√

x + 1.

Solution.We make a table of values first for somespecific values of x :(the domain of w is {x | x ≥ −1})

x w(x) (x , w(x))

0 1 (0, 1)

−1 0 (−1, 0)

1√

2 (1,√

2)

2√

3 (2,√

3)

3 2 (3, 2)


MATH 201 - Week 2

Graphs of Functions

Graphing Functions


w(x) =√

x + 1.


x w(x) (x , w(x))

0 1 (0, 1)

−1 0 (−1, 0)

1√

2 (1,√

2)

2√

3 (2,√

3)

3 2 (3, 2)


MATH 201 - Week 2

Graphs of Functions

Graphing Functions


w(x) =√

x + 1.


x w(x) (x , w(x))

0 1 (0, 1)

−1 0 (−1, 0)

1√

2 (1,√

2)

2√

3 (2,√

3)

3 2 (3, 2)


MATH 201 - Week 2

Graphs of Functions

Graphing Piecewise Defined Functions


f (x) =

{−x + 2 if x < 0x2 + 1 if x ≥ 0

Solution. We expect the behaviour of f to be different for x < 0and x ≥ 0.

x f (x) (x , f (x))

0 1 (0, 1)

−1 3 (−1, 3)

−2 4 (−2, 4)

1 2 (1, 2)

2 5 (2, 5)

Then we plot the points and try to jointhem to obtain a curve (be careful at thepoint x = 0).

MATH 201 - Week 2

Graphs of Functions

Graphing Piecewise Defined Functions


f (x) =

{−x + 2 if x < 0x2 + 1 if x ≥ 0

Solution. We expect the behaviour of f to be different for x < 0and x ≥ 0.

x f (x) (x , f (x))

0 1 (0, 1)

−1 3 (−1, 3)

−2 4 (−2, 4)

1 2 (1, 2)

2 5 (2, 5)

Then we plot the points and try to jointhem to obtain a curve (be careful at thepoint x = 0).

MATH 201 - Week 2

Graphs of Functions

The Vertical Line Test

Vertical Line Test

A curve in the coordinate plane is the graph of a function if andonly if no vertical line intersects it more that once.

Examples.

graph of a function

NOT graph of a function

MATH 201 - Week 2

Graphs of Functions

Equations That Define Functions

Problem. Does the equation

2y − x2 − 3 = 0

define y as a function of x?Solution. Yes, because we can solve for y and it gives a uniquesolution in terms of the variable x :

2y − x2 − 3 = 0

2y = x2 + 3

y =1

2x2 +

3

2.

In other words, y is given by the function

f (x) =1

2x2 +

3

2.

MATH 201 - Week 2

Graphs of Functions

Equations That Define Functions

Problem. Does the equation

2y2 − x2 − 3 = 0

define y as a function of x?Solution. No, because we can solve for y and but it gives twopossible solutions in terms of the variable x :

2y2 − x2 − 3 = 0

2y2 = x2 + 3

y2 =1

2x2 +

3

2

y = ±√

1

2x2 +

3

2.

MATH 201 - Week 2

Assignments

Assignments:

Linesp. 120, # 4, 16, 20, 28, 32

What is a Function?p. 155, # 4, 16, 30, 44, 48

Graphs of Functionsp. 167, #8, 10, 28, 30, 32, 34, 36, 38, 44

The assignments are due to Wednesday, January 23.

Documents

MATH 201 - Week 2 · 2008. 1. 16. · MATH 201 - Week 2 MATH 201 - Week 2 Ferenc Balogh Concordia University 2008 Winter Based on the textbook J. Stuart, L. Redlin, S. Watson, Precalculus