graphs and graphing utilities. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 origin (0, 0) definitions the...

Graphs and Graphing Utilities

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Origin (0, 0)

Definitions• The horizontal number line is the x-axis.• The vertical number line is the y-axis.• The point of intersection of these axes is

their zero points, called the origin.

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2nd quadrant 1st quadrant

3rd quadrant 4th quadrant

Definitions• The axes divide the plane into four quarters,

called quadrants.• Each point in the rectangular coordinate

system corresponds to an ordered pair of real numbers, (x, y).

Example

Plot the point (3,2).Start at the origin and move 3 units to the

right.From that point, move 2 units up.Now plot your point.

Text ExampleSketch the graph of y = x2 – 4.

Let x = 3, then y = x2 – 4 = 9 – 4 = 5. The ordered pair (3, 5) is a solution to the

equation y = x2 – 4. We also say that (3, 5) satisfies the equation.

(3, 5)y = 32 – 4 = 9 – 4 = 53

(2, 0)y = 22 – 4 = 4 – 4 = 02

(1, -3)y = 12 – 4 = 1 – 4 = -31

(0, -4)y = (0)2 – 4 = 0 – 4 = -40

(-1, -3)y = (-1)2 – 4 = 1 – 4 = -3-1

(-2, 0)y = (-2)2 – 4 = 4 – 4 = 0-2

(-3, 5)y = (-3)2 – 4 = 9 – 4 = 5-3

Ordered Pair (x, y)y = x2 – 4x

Text Example Cont.

• First, find several ordered pairs that are solutions to the equation.

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Text Example Cont.• Now, we plot these ordered pairs as points

in the rectangular coordinate system.

Example

Graph 4y + 5x = 20.Substitute zero for x:

4y = 20 or y = 5.Hence, the y-intercept is (0,5).Substitute zero for the y:

5x = 20 or x = 4.Hence, the x-intercept is (4,0).

Graphs and Graphing Utilities

Lines and Slope

Definition of Slope

Definition of SlopeThe slope of the line through the distinct points (x1, y1) and (x2, y2) is

where x2 – x1 = 0.

Change in yChange in x

=RiseRun =

y2 – y1

x2 – x1

(x1, y1)

x1

y1

x2

y2(x2, y2)Rise

y2 – y1

Runx2 – x1

x

y

The Possibilities for a Line’s Slope

Positive Slope

x

y

m > 0

Line rises from left to right.

Zero Slope

x

y

m = 0

Line is horizontal.m is

undefined

Undefined Slope

x

y

Line is vertical.

Negative Slope

x

y

m < 0

Line falls from left to right.

Point-Slope Form of the Equation of a Line

The point-slope equation of a nonvertical line of slope m that passes through the point (x1, y1) is

y – y1 = m(x – x1).

Text ExampleWrite the point-slope form of the equation of the line passing through (-1,3) with a slope of 4. Then solve the equation for y.

Solution We use the point-slope equation of a line with m = 4, x1= -1, and y1 = 3.

This is the point-slope form of the equation.y – y1 = m(x – x1)

Substitute the given values.y – 3 = 4[x – (-1)]We now have the point-slope form of the equation for the given line.y – 3 = 4(x + 1)

We can solve the equation for y by applying the distributive property.y – 3 = 4x + 4y = 4x + 7 Add 3 to both sides.

Slope-Intercept Form of the Equation of a Line

The slope-intercept equation of a nonvertical line with slope m and y-intercept b is

y = mx + b.

Graphing y=mx+b Using the Slope and y-Intercept

• Graphing y = mx + b by Using the Slope and y-Intercept

• Plot the y-intercept on the y-axis. This is the point (0, b).

• Obtain a second point using the slope, m. Write m as a fraction, and use rise over run starting at the y-intercept to plot this point.

• Use a straightedge to draw a line through the two points. Draw arrowheads at the ends of the line to show that the line continues indefinitely in both directions.

Text ExampleGraph the line whose equation is y = 2/3 x + 2.

Solution The equation of the line is in the form y = mx + b. We can find the slope, m, by identifying the coefficient of x. We can find the y-intercept, b, by identifying the constant term.

y = 2/3 x + 2 23

The slope is 2/3.

The y-intercept is 2.

Text Example cont.Graph the line whose equation is y = 2/3x + 2.

We plot the second point on the line by starting at (0, 2), the first point. Then move 2 units up (the rise) and 3 units to the right (the run). This gives us a second point at (3, 4). -5 -4 -3 -2 -1 1 2 3 4 5

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Solution:We need two points in order to graph the line. We can use the y-intercept, 2, to obtain the first point (0, 2). Plot this point on the y-axis.

Equation of a Horizontal Line

•A horizontal line is given by an equation of the form•y = b•where b is the y-intercept.

Equation of a Vertical Line

•A vertical line is given by an equation of the form•x = a•where a is the x-intercept.

General Form of the Equation of a Line

Every line has an equation that can be written in the general form

Ax + By + C = 0Where A, B, and C are three real numbers, and

A and B are not both zero.\

Text ExampleFind the slope and the y-intercept of the line whose equation is 2x – 3y + 6 = 0.Solution The equation is given in general form. We begin by rewriting it in the form y = mx + b. We need to solve for y.

This is the given equation.2x – 3y + 6 = 0

To isolate the y-term, add 3 y on both sides.2x + 6 = 3y

Reverse the two sides. (This step is optional.)3y = 2x + 6

The coefficient of x, 2/3, is the slope and the constant term, 2, is the y-intercept.

Divide both sides by 3.y = 2/3x + 2

Equations of Lines

Point-slope form: y – y1 = m(x – x1)Slope-intercept form: y = m x + bHorizontal line: y = bVertical line: x = aGeneral form: Ax + By + C = 0

Slope and Parallel Lines

• If two nonvertical lines are parallel, then they have the same slope.

• If two distinct nonvertical lines have the same slope, then they are parallel.

• Two distinct vertical lines, both with undefined slopes, are parallel.

Text ExampleWrite an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

Solution We are looking for the equation of the line shown on the left. Notice that the line passes through the point (-3, 2). Using the point-slope form of the line’s equation, we have x1 = -3 and y1 = 2.

y = 2x + 1

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(-3, 2)

Rise = 2

Run = 1

y – y1 = m(x – x1)

y1 = 2 x1 = -3

Text Example cont.Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.

Solution Parallel lines have the same slope. Because the slope of the given line is 2, m = 2 for the new equation.

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(-3, 2)

Rise = 2

Run = 1

y = 2x + 1y – y1 = m(x – x1)

y1 = 2 m = 2 x1 = -3

Text Example cont.Write an equation of the line passing through (-3, 2) and parallel to the line whose equation is y = 2x + 1. Express the equation in point-slope form and y-intercept form.Solution The point-slope form of the line’s equation is

y – 2 = 2[x – (-3)]y – 2 = 2(x + 3)

Solving for y, we obtain the slope-intercept form of the equation.

y – 2 = 2x + 6 Apply the distributive property.

y = 2x + 8 Add 2 to both sides. This is the slope-intercept form of the equation.

Slope and Perpendicular Lines

Slope and Perpendicular Lines• If two nonvertical lines are perpendicular, then the product of their

slopes is –1.• If the product of the slopes of two lines is –1, then the lines are

perpendicular.• A horizontal line having zero slope is perpendicular to a vertical line

having undefined slope.

Two lines that intersect at a right angle (90°) are said to be perpendicular. There is a relationship between the slopes of perpendicular lines.

90°

ExampleFind the slope of any line that is perpendicular to the line whose equation is 2x + 4y – 4 = 0.Solution We begin by writing the equation of the given line in slope-intercept form. Solve for y.

2x + 4y – 4 = 0 This is the given equation.

4y = -2x + 4 To isolate the y-term, subtract x and add 4 on both sides.

Slope is –1/2.

y = -2/4x + 1 Divide both sides by 4.

The given line has slope –1/2. Any line perpendicular to this line has a slope that is the negative reciprocal, 2.

Lines and Slope

Distance and Midpoint Formulas;

Circles

The Distance Formula

• The distance, d, between the points (x1, y1) and (x2,y2) in the rectangular coordinate system is

d (x2 x1)2 (y2 y1)

2

ExampleFind the distance between (-1, 2) and (4, -3).

Solution Letting (x1, y1) = (-1, -3) and (x2, y2) = (2, 3), we obtain

d (x2 x1)2 (y2 y1)

2

d (4 ( 1))2 (( 3) 2)2

d (5)2 ( 5)2

d 25 25

d 50

d 5 2

The Midpoint Formula

• Consider a line segment whose endpoints are (x1, y1) and (x2, y2). The coordinates of the segment's midpoint are

• To find the midpoint, take the average of the two x-coordinates and of the two y-coordinates.

x1 x2

2,y1 y2

2

Text Example

Find the midpoint of the line segment with endpoints (1, -6) and (-8, -4).

Solution To find the coordinates of the midpoint, we average the coordinates of the endpoints.

(-7/2, -5) is midway between the points (1, -6) and (-8, -4).

1 ( 8)2

, 6 ( 4)

2

72

, 102

72

, 5

Definition of a CircleA circle is the set of all points in a plane that are equidistant from a fixed point called the center. The fixed distance from the circle’s center to any point on the circle is called the radius.

Radius: r

(x, y)

(h, k)

Center

Any point on the circle

The Standard Form of the Equation of a Circle

Radius: r

(x, y)

(h, k)

Center

Any point on the circle

The standard form of the equation of a circle with center (h, k) and radius r is(x – h)2 + (y – k)2 = r2.

ExampleFind the center and radius of the circle whose equation is

(x – 2)2 + (y + 3)2 = 9 and graph the equation.Solution In order to graph the circle, we need to know its center, (h, k), and its radius r. We can find the values of h, k, and r by comparing the given equation to the standard form of the equation of a circle.

(x – 2)2 + (y + 3)2 = 9

h is 2.

(x – 2)2 + (y – (-3))2 = 32

k is –3. r is 3.

Example cont.Find the center and radius of the circle whose equation is

(x – 2)2 + (y + 3)2 = 9 and graph the equation.

SolutionWe see that h = 2, k = -3, and r = 3. Thus, the circle has center (2, -3) and a radius of 3 units. Plot the center, (2, -3), and find 3 additional points by going up, right, down, and left of the center by 3 units.

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321

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3

(2, -3)

General Form of the Equation of a Circle

The general form of the equation of a circle is x2 + y2 + Dx + Ey + F = 0.

Distance and Midpoint Formulas;

Circles

Basics of Functions

Definition of a Relation

• A relation is any set of ordered pairs. The set of all first components of the ordered pairs is called the domain of the relation, and the set of all second components is called the range of the relation.

Example

Solution The domain is the set of all first components. Thus, the domain is

{1994,1995,1996,1997,1998}.The range is the set of all second components. Thus, the range is

{56.21, 51.00, 47.70, 42.78, 39.43}.

Find the domain and range of the relation{(1994, 56.21), (1995, 51.00), (1996, 47.70),

(1997, 42.78), (1998, 39.43)}

Definition of a Function

• A function is a correspondence between two sets X and Y that assigns to each element x of set X exactly one element y of set Y. For each element x in X, the corresponding element y in Y is called the value of the function at x. The set X is called the domain of the function, and the set of all function values, Y, is called the range of the function.

Text Example

Solution We begin by making a figure for each relation that shows set X, the domain, and set Y, the range, shown below.

Determine whether each relation is a function.a. {(1, 6), (2, 6), (3, 8), (4, 9)} b. {(6,1),(6,2),(8,3),(9,4)}

1234

689

Domain Range

(a) Figure (a) shows that every element in the domain corresponds to exactly one element in the range. No two ordered pairs in the given relation have the same first component and different second components. Thus, the relation is a function.

689

1234

Domain Range

(b) Figure (b) shows that 6 corresponds to both 1 and 2. This relation is not a function; two ordered pairs have the same first component and different second components.

Text Example

Solution We substitute 2, x + 3, and -x for x in the definition of f. When replacing x with a variable or an algebraic expression, you might find it helpful to think of the function's equation as

f (x) = x2 + 3x + 5.

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x)

a. We find f (2) by substituting 2 for x in the equation.

f (2) = 22 + 3 • 2 + 5 = 4 + 6 + 5 = 15

Thus, f (2) = 15.

Text Example cont.

Solutionb. We find f (x + 3) by substituting x + 3 for x in the equation.

f (x + 3) = (x + 3)2 + 3(x + 3) + 5


Equivalently,

f (x + 3) = (x + 3)2 + 3(x + 3) + 5= x2 + 6x + 9 + 3x + 9 + 5= x2 + 9x + 23.

Square x + 3 and distribute 3 throughout the parentheses.

Text Example cont.

Solutionc. We find f (-x) by substituting -x for x in the equation.

f (-x) = (-x)2 + 3(-x) + 5


Equivalently,

f (-x) = (-x)2 + 3(-x) + 5 = x2 –3x + 5.

Definition of a Difference Quotient

• The expression

for h not equal to 0 is called the difference quotient.

f (x h) f (x)h

Example

}1,1:{

)1)(1()2(

12)(

2

2

3

xxdomain

xxxx

xxxxf

function

Find the domain of the function.

32)( xxg

Example

• Find the domain of the function.

}23:{

23

032

xxanswer

x

sox

Example cont.

Solution

Basics of Functions

Graphs of Functions

Text Example

Solution The graph of f (x) = x2 + 1 is, by definition, the graph of y = x2 + 1.We begin by setting up a partial table of coordinates.

Graph f (x) = x2 + 1. To do so, use integer values of x from the set {-3, -2, -1, 0, 1, 2, 3} to obtain seven ordered pairs. Plot each ordered pair and draw a smooth curve through the points. Use the graph to specify the function's domain and range.

(-1, 2)f (-1) = (-1)2 + 1 = 2-1

(0, 1)f (0) = 02 + 1 = 10

(-2, 5)f (-2) = (-2)2 + 1 = 5-2

(-3, 10)f (-3) = (-3)2 + 1 = 10-3

(x, y) or (x, f (x))f (x) = x2 + 1x

Text Example cont.

Now, we plot the seven points and draw a smooth curve through them, as shown. The graph of f has a cuplike shape. The points on the graph of f have x-coordinates that extend indefinitely far to the left and to the right. Thus, the domain consists of all real numbers, represented by (-oo, oo). By contrast, the points on the graph have y-coordinates that start at 1 and extend indefinitely upward. Thus, the range consists of all real numbers greater than or equal to 1, represented by [1, oo).

Solution

(3, 10)f (3) = (3)2 + 1 = 103

(2, 5)f (2) = (2)2 + 1 = 52

(1, 2)f (1) = 12 + 1 = 21

(x, y) or (x, f (x))f (x) = x2 + 1x

Domain: all Reals

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Ran

ge: [

1, o

o)

Obtaining Information from GraphsYou can obtain information about a function from its graph. At the right or left of a graph, you will find closed dots, open dots, or arrows.

• A closed dot indicates that the graph does not extend beyond this point and the point belongs to the graph.

• An open dot indicates that the graph does not extend beyond this point and the point does not belong to the graph.

• An arrow indicates that the graph extends indefinitely in the direction in which the arrow points.

Text Example

Solutiona. Because (-1, 2) is a point on the graph of f, the

y-coordinate, 2, is the value of the function at the x-coordinate, -1. Thus, f (-l) = 2. Similarly, because (1, 4) is also a point on the graph of f, this indicates that f (1) = 4.

Use the graph of the function f to answer the following questions.

• What are the function values f (-1) and f (1)?• What is the domain of f (x)?• What is the range of f (x)?

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Solutionb. The open dot on the left shows that x = -3 is not in

the domain of f. By contrast, the closed dot on the right shows that x = 6 is. We determine the domain of f by noticing that the points on the graph of f have x-coordinates between -3, excluding -3, and 6, including 6. Thus, the domain of f is { x | -3 < x < 6} or the interval (-3, 6].

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Text Example cont.Use the graph of the function f to answer the

following questions.• What are the function values f (-1) and f (1)?• What is the domain of f (x)?• What is the range of f (x)?

Solutionc. The points on the graph all have y-coordinates

between -4, not including -4, and 4, including 4. The graph does not extend below y = -4 or above y = 4. Thus, the range of f is { y | -4 < y < 4} or the interval (-4, 4].

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Use the graph of the function f to answer the following questions.

• What are the function values f (-1) and f (1)?• What is the domain of f (x)?• What is the range of f (x)?

Text Example cont.

The Vertical Line Test for Functions

• If any vertical line intersects a graph in more than one point, the graph does not define y as a function of x.

Text Example

Solution y is a function of x for the graphs in (b) and (c).

Use the vertical line test to identify graphs in which y is a function of x.

x

y a.

x

y b.

x

yc.

x

yd.

x

y a.

y is not a function since 2 values of y correspond to an x-value.

x

y b.

y is a function of x.

x

yc.

y is a function of x.

x

yd.

y is not a function since 2 values of y correspond to an x-value.

Increasing, Decreasing, and Constant Functions

Constantf (x1) < f (x2)

(x1, f (x1))

(x2, f (x2))

Increasingf (x1) < f (x2)

(x1, f (x1))

(x2, f (x2))

Decreasingf (x1) < f (x2)

(x1, f (x1))

(x2, f (x2))

A function is increasing on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) < f (x2).A function is decreasing on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) > f (x2). A function is constant on an interval if for any x1, and x2 in the interval, where x1 < x2, then f (x1) = f (x2).

Solutiona. The function is decreasing on the interval (-oo, 0), increasing on the

interval (0, 2), and decreasing on the interval (2, oo).

Describe the increasing, decreasing, or constant behavior of each function whose graph is shown.

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a.

b.

Text Example

Solutionb. Although the function's equations are not given, the graph indicates that

the function is defined in two pieces. The part of the graph to the left of the y-axis shows that the function is constant on the interval (-oo, 0). The part to the right of the y-axis shows that the function is increasing on the interval (0, oo).

Describe the increasing, decreasing, or constant behavior of each function whose graph is shown.

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a.

b.

Text Example cont.

Definitions of Relative Maximum and Relative Minimum

1. A function value f(a) is a relative maximum of f if there exists an open interval about a such that f(x) > f(x) for all x in the open interval.

2. A function value f(b) is a relative minimum of f if there exists an open interval about b such that f(b) < f(x) for all x in the open interval.

The Average Rate of Change of a Function

• Let (x1, f(x1)) and (x2, f(x2)) be distinct points on the graph of a function f.The average rate of change of f from x1 to x2 is f (x2 ) f (x1)

x2 x1

Definition of Even and Odd Functions

• The function f is an even function if• f (-x) = f (x) for all x in the domain of f.• The right side of the equation of an even function

does not change if x is replaced with -x.

• The function f is an odd function if• f (-x) = -f (x) for all x in the domain of f.• Every term in the right side of the equation of an

odd function changes sign if x is replaced by -x.

Example

Identify the following function as even, odd, or neither: f(x) = 3x2 - 2.

Solution:We use the given function’s equation to find f(-x).f(-x) = 3(-x) 2-2 = 3x2 - 2.The right side of the equation of the given function

did not change when we replaced x with -x. Because f(-x) = f(x), f is an even function.

Even Functions and y-Axis Symmetry

• The graph of an even function in which f (-x) = f (x) is symmetric with respect to the y-axis.

Odd Functions and Origin Symmetry

• The graph of an odd function in which f (-x) = - f (x) is symmetric with respect to the origin.

Graphs of Functions

Transformations of Functions

Vertical Shifts

y = f (x) + c

y = f (x) y = f (x) - c

y = f (x)

cc

Let f be a function and c a positive real number.•The graph of y = f (x) + c is the graph of y = f (x) shifted c units vertically upward.•The graph of y = f (x) – c is the graph of y = f (x) shifted c units vertically downward.

Example

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-10

-8

-6

-4

-2

2

4

6

8

10

• Use the graph of y = x2 to obtain the graph of y = x2 + 4.

Example cont.

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-10

-8

-6

-4

-2

2

4

6

8

10

• Use the graph of y = x2 to obtain the graph of y = x2 + 4.Step 1 Graph f (x) = x2. The graph of the standard quadratic

function is shown.Step 2 Graph g(x) = x2+4. Because we add 4 to each value of x2 in

the range, we shift the graph of f vertically 4 units up.

Horizontal Shifts

y = f (x + c)y = f (x) y = f (x - c)

y = f (x)cc

Let f be a function and c a positive real number.•The graph of y = f (x + c) is the graph of y = f (x) shifted to the left c units.•The graph of y = f (x + c) is the graph of y = f (x) shifted to the right c units.

Text Example

Use the graph of f (x) to obtain the graph of h(x) = (x + 1)2 – 3.

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SolutionStep 1 Graph f (x) = x2. The graph of the

standard quadratic function is shown.Step 2 Graph g(x) = (x + 1)2. Because we

add 1 to each value of x in the domain, we shift the graph of f horizontally one unit to the left.

Step 3 Graph h(x) = (x + 1)2 – 3. Because we subtract 3, we shift the graph vertically down 3 units.

Reflection about the x-Axis

• The graph of y = - f (x) is the graph of y = f (x) reflected about the x-axis.

Reflection about the y-Axis

• The graph of y = f (-x) is the graph of y = f (x) reflected about the y-axis.

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f (x) = x2

h(x) = 2x2

g(x) = ›x2

Stretching and Shrinking GraphsLet f be a function and c a positive real number.•If c > 1, the graph of y = c f (x) is the graph of y = f (x) vertically stretched by multiplying each of its y-coordinates by c.•If 0 < c < 1, the graph of y = c f (x) is the graph of y = f (x) vertically shrunk by multiplying each of its y-coordinates by c.

Sequence of TransformationsA function involving more than one transformation

can be graphed by performing transformations in the following order.

1. Horizontal shifting2. Vertical stretching or shrinking3. Reflecting4. Vertical shifting

Example

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-10

-8

-6

-4

-2

2

4

6

8

10

• Use the graph of f(x) = x3 to graph g(x) = (x+3)3 - 4

Solution:

Step 1: Because x is replaced with x+3, the graph is shifted 3 units to the left.

Example

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-10

-8

-6

-4

-2

2

4

6

8

10


Solution:

Step 2: Because the equation is not multiplied by a constant, no stretching or shrinking is involved.

Example

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-10

-8

-6

-4

-2

2

4

6

8

10


Solution:

Step 3: Because x remains as x, no reflecting is involved.

Example


Solution:

Step 4: Because 4 is subtracted, shift the graph down 4 units.

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-10

-8

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-4

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Transformations of Functions

Combinations of Functions; Composite Functions

The Sum of Functions

Let f and g be two functions. The sum f + g is the function defined by

(f + g)(x) = f(x) + g(x)The domain of f + g is the set of all real

numbers that are common to the domain of f and the domain of g.

ExampleLet f(x) = 2x+1 and g(x) = x2-2. Find f + g,

f - g, fg, and f/g.Solution:f+g = 2x+1 + x2-2 = x2+2x-1f-g = (2x+1) - (x2-2)= -x2+2x+3fg = (2x+1)(x2-2) = 2x3+x2-4x-2f/g = (2x+1)/(x2-2)

Definitions: Sum, Difference, Product, and Quotient of Functions

Let f and g be two functions. The sum of f + g, the difference f – g, the product fg, and the quotient f /g are functions whose domains are the set of all real numbers common to the domains of f and g, defined as follows:

Sum: (f + g)(x) = f (x)+g(x)Difference: (f – g)(x) = f (x) – g(x)Product: (f • g)(x) = f (x) • g(x)Quotient: (f / g)(x) = f (x)/g(x), provided g(x) does

not equal 0

The Composition of Functions

• The composition of the function f with g is denoted by f o g and is defined by the equation

• (f o g)(x) = f (g(x)).• The domain of the composite function f o g is the

set of all x such that• x is in the domain of g and• g(x) is in the domain of f.

Text ExampleGiven f (x) = 3x – 4 and g(x) = x2 + 6, find:

a. (f o g)(x) b. (g o f)(x)

Solution a. We begin with (f o g)(x), the composition of f with g. Because (f o g)(x)

means f (g(x)), we must replace each occurrence of x in the equation for f by g(x).

f (x) = 3x – 4 This is the given equation for f.

(f o g)(x) = f (g(x)) = 3g(x) – 4

= 3(x2 + 6) – 4 = 3x2 + 18 – 4 = 3x2 + 14

Replace g(x) with x2 + 6.Use the distributive property.Simplify.

Thus, (f o g)(x) = 3x2 + 14.

Solution b. Next, we find (g o f )(x), the composition of g with f. Because (g o f )(x)

means g(f (x)), we must replace each occurrence of x in the equation for g by f (x).

g(x) = x2 + 6 This is the given equation for g.

(g o f )(x) = g(f (x)) = (f (x))2 + 6= (3x – 4)2 + 6= 9x2 – 24x + 16 + 6= 9x2 – 24x + 22

Replace f (x) with 3x – 4.Square the binomial, 3x – 4.Simplify.

Thus, (g o f )(x) = 9x2 – 24x + 22. Notice that (f o g)(x) is not the same as (g o f )(x).

Text Example cont.Given f (x) = 3x – 4 and g(x) = x2 + 6, find:

a. (f o g)(x) b. (g o f)(x)

Combinations of Functions; Composite Functions

Inverse Functions

Definition of the Inverse Function

• Let f and g be two functions such that• f (g(x)) = x for every x in the domain of g• and • g(f (x)) = x for every x in the domain of f.• The function g is the inverse of the function f,

and denoted by f -1 (read “f-inverse”). Thus, f ( f -

1(x)) = x and f -1( f (x)) = x. The domain of f is equal to the range of f -1, and vice versa.

Text ExampleShow that each function is the inverse of the other:

f (x) = 5x and g(x) = x/5.Solution To show that f and g are inverses of each other, we must show that f (g(x)) = x and g( f (x)) = x. We begin with f (g(x)).

f (x) = 5x

f (g(x)) = 5g(x) = 5(x/5) = x.

Next, we find g(f (x)).g(x) = 5/x

g(f (x)) = f (x)/5 = 5x/5 = x.

Notice how f -1 undoes the change produced by f.

Finding the Inverse of a Function• The equation for the inverse of a function f can be found

as follows:1. Replace f (x) by y in the equation for f (x).2. Interchange x and y.3. Solve for y. If this equation does not define y as a function

of x, the function f does not have an inverse function and this procedure ends. If this equation does define y as a function of x, the function f has an inverse function.

4. If f has an inverse function, replace y in step 3 with f -1(x). We can verify our result by showing that f ( f -1(x)) = x and f -1( f (x)) = x.

Find the inverse of f (x) = 7x – 5.Solution Step 1 Replace f (x) by y.

y = 7x – 5

Step 2 Interchange x and y.x = 7y – 5 This is the inverse function.

Step 3 Solve for y.x + 5 = 7y Add 5 to both sides.

x + 5 = y7

Divide both sides by 7.

Step 4 Replace y by f -1(x).

x + 57f -1(x) = Rename the function f -1(x).

Text Example

The Horizontal Line Test For Inverse Functions

• A function f has an inverse that is a function, f –1, if there is no horizontal line that intersects the graph of the function f at more than one point.

Example

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• Does f(x) = x2+3x-1 have an inverse function?

Example

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Solution:This graph does not pass the horizontal line

test, so f(x) = x2+3x-1 does not have an inverse function.

Inverse Functions

Modeling with Functions

Text ExampleYou are choosing between two long-distance telephone plans.

Plan A has a monthly fee of $20 with a charge of $0.05 per minute for all long-distance calls. Plan B has a monthly fee of $5 with a charge of $0.10 per minute for all long-distance calls.

a. Express the monthly cost for plan A, f, as a function of the number of minutes of long-distance calls in a month, x.

b. Express the monthly cost for plan B, g, as a function of the number of minutes of long-distance calls in a month, x.

c. For how many minutes of long-distance calls will the costs for the two plans be the same?

Text Example cont.You are choosing between two long-distance telephone plans.


a. Express the monthly cost for plan A, f, as a function of the number of minutes of long-distance calls in a month, x.

Solution:a. The monthly cost for plan A is the monthly fee, $20, plus

the per-minute charge, $0.05, times the number of minutes of long-distance calls, x.

f(x) = 20 + 0.05 * x

Text Example cont.You are choosing between two long-distance telephone plans.


b. Express the monthly cost for plan B, g, as a function of the number of minutes of long-distance calls in a month, x.

Solution:b. The monthly cost for plan B is the monthly fee, $5, plus

the per-minute charge, $0.10, times the number of minutes of long-distance calls, x.

f(x) = 5 + 0.10 * x

Text Example cont.c. For how many minutes of long-distance calls will the costs

for the two plans be the same?

Solution:

c. We are interested in how many minutes of long-distance calls, x, result in the same monthly costs, f and g, for the two plans. Thus, we must set the equations for f and g equal to each other. We then solve the resulting linear equation for x.

0.05x + 20 = 0.10x + 520 = 0.05x + 5

15 = 0.05x300 minutes = x

Area of a Rectangle and a SquareThe area A of a rectangle with length l and width w is given by:

A = lw.

The area A of a square with one side measuring s linear units is given by:

A = s2

w

l

s

Example

You decide to cover the path shown with bricks.

a. Find the area of the path.b. If the path requires four bricks

for every square foot, how many bricks are needed for the project?

3ft

13ft

6ft

3ft

Example cont.

Because the area of a rectangle is A=lw, we divide the figure into two rectangles. Then we use the length and width of each rectangle to determine the total area.

A = (3ft)(13ft) = 39ft2

A = (3ft)(6ft+3ft) = (3ft)(9ft) = 27 ft2

Total Area = 39ft2 + 27ft2 = 66ft2

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Solution:

Area of a Triangle

The area A of a triangle with height h and base b is given by A = (1/2)bh.

h

b

Area of a Trapezoid

The area A of a trapezoid with parallel bases a and b and height h is given by the formula A = (1/2)h(a+b)

b

h

a

CirclesA circle is a set of points in the plane equally distant

from a given point, its center.The radius, r, is a line segment from the center to

any point on the circle.The diameter, d, is a line segment through the

center whose endpoints lie on the circle.The circumference, C, of a circle with diameter d

and radius r is C = d or C = 2r.

The area, A, of a circle with radius r is A = r2

Example

Which of the following is the better buy: a large pizza with a 16-inch diameter for $15.00 or a medium pizza with an 8-inch diameter for $7.50?

Example cont.

The better buy is the pizza with the cheaper price per square inch.

Large pizza: A = r2 = (8in)2 = 64in2 =201in2

Small pizza: A = r2 = (4in)2 = 16in2 =50in2

Large pizza = $15.00 / 201in2 = $0.07/in2

Small pizza = $7.50 / 50in2 = $0.15/in2

The larger pizza is the better buy.

Solution:

Volume of Rectangular Solid: V = lwh

Volume of a Right Circular Cylinder: V = r2h

Volume of a Cone: V = (1/3)()r2h

Volume of a Sphere:V = (4/3)()r3

Modeling with Functions

graphs and graphing utilities. -5-4-3-212345 5 4 3 2 1 -2 -3 -4 -5 origin (0, 0) definitions the...

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