Worksheet 48 (9
Worksheet 48 (9.1)
Chapter 9 Functions
9.1 Relations and Functions
Summary 1:
A relation is a set of ordered pairs.
The domain of a relation is the set of first components of the ordered pairs.
The range of a relation is the set of second components of the ordered pairs.
The ordered pairs of a relation are commonly generated from graphs, charts, and use of equations.
Warm-up 1. Specify the domain for each relation:
a) {(2, 5), (3, 8), (4, 11), (5, 12)}
The domain is { , , , }.
b) {(1, -2), (1, -1), (1, 0), (1, 1), (1, 2)}
The domain is { }.
c) { (x, y) | y = 2x + 3 }
The variable x can be replaced with any __________ number.
The domain is ____________________.
d) { (x, y) | y = x2 + 1 }
The variable x can be replaced with any __________ number.
The domain is ____________________.
e) { (x, y) | x = y2 + 1 }
The equation x = y2 + 1 is equivalent to[pic].
For the radical to be defined in the real numbers, then
x - 1 _____ 0.
The domain is { x | x ≥ }.
Worksheet 48 (9.1)
f) [pic]
The rational expression must be defined in the real numbers; therefore, x - 1 _____ 0.
The domain is { x | x ≠ _____ }.
Warm-up 2. Specify the range for each relation:
a) {(2, 5), (3, 8), (4, 11), (5, 12)}
The range is { , , , }.
b) {(1, -2), (1, -1), (1, 0), (1, 1), (1, 2)}
The range is { , , , , }.
c) { (x, y) | y = 2x + 3 }
Since x is any real number, then y can be any __________ number.
The range is ____________________.
d) { (x, y) | y = x2 + 1 }
For all real x, x2 ≥ _____. Therefore, y ≥ _____.
The range is { y | y ≥ _____ }.
e) { (x, y) | x = y2 + 1 }
Since x ≥ 1, y can be any __________ number.
The range is ____________________.
f)[pic]
Since x ≠ 1 and[pic]because the numerator is nonzero, y can be any real number except __________.
The range is { y | y ≠ _____ }.
Problems - Specify the domain and range for each relation:
1. {(1, 1), (2, 1), (3, 1), (4, 1)} 2. { (x, y) | y = x + 3 }
3. [pic] 4. [pic]
Worksheet 48 (9.1)
Summary 2:
A function is a relation in which no two ordered pairs have the same first component.
Function notation: f(x)
1. The single letter f is the common symbol used to denote a function. Any other single letter can be used.
2. The variable x represents an element of the domain.
3. f(x) is translated as "f of x" or "the value of f at x". Within the context of functions, this notation will never mean f times x.
4. f(x) replaces the variable y.
Evaluating a function means to substitute the variable x with a given number to determine a corresponding functional value. It means the same as finding y given a value for x.
The domain of a function is the set of all possible values for x.
The range of a function is the set of all corresponding values for f(x).
Warm-up 3. Specify those relations that are functions:
a) {(2, 5), (3, 8), (4, 11), (5, 12)}
b) {(1, -2), (1, -1), (1, 0), (1, 1), (1, 2)}
c) { (x, y) | y = 2x + 3 }
d) { (x, y) | y = x2 + 1 }
e) { (x, y) | x = y2 + 1 }
f)[pic]
In the above list of relations, warm-up _____, _____, _____, and _____ are functions.
Worksheet 48 (9.1)
Warm-up 4. Evaluate:
a) If f(x) = 2x + 3, find f(-5).
f(x) = 2x + 3
f( ) = 2( ) + 3
=
f(-5) = _____
ΦNote: -5 is an element of the domain and -7 is an element of the range.
b) If f(x) = x2 + 1, find f(-4).
f(x) = x2 + 1
f( ) = ( )2 + 1
=
f(-4) = _____
c) If f(x) =[pic], find[pic].
f(x) =[pic]
f( ) =[pic]
=[pic]
=
[pic]
Warm-up 5. Specify the domain and range for each function:
a) f(x) = x2 - 6x + 9
The domain is ____________________.
Worksheet 48 (9.1)
Since this is a parabola opening upwards with vertex (3, 0),
f(x) ≥ _____.
The range is { f(x) | f(x) ≥ _____ }.
b) f(x) =[pic]
The rational expression is undefined when __________ = 0.
The domain is { x | x ≠ _____ , x ≠ _____ }.
The numerator is nonzero; therefore, f(x) ≠ _____.
The range is { f(x) | f(x) ≠ _____ }.
c) [pic]
The radical is undefined when _______________ < 0.
The domain is { x | x ≤ _____ or x ≥ _____ }.
The values for f(x) will be positive or _____.
The range is { f(x) | f(x) ≥ _____ }.
Problems
5. Is the relation { (x, y) | y = (x - 1)2 + 3 } a function?
6. Is the relation { (x, y) | x = (y - 1)2 + 3 } a function?
7. If[pic], find f(-9).
8. If[pic], find f(-5).
9. Find the domain and range for[pic]
Worksheet 49 (9.2)
9.2 Functions: Their Graphs and Applications
Summary 1:
A linear function is in the form f(x) = ax + b where a and b are real numbers.
The graph of a linear function is a straight line.
Graphing a linear function:
1. Determine two points of the graph.
2. Draw the line determined by the two points.
3. Use a third point as a checkpoint.
A constant function is a linear function where a = 0.
It is a horizontal line.
Note: Vertical lines are not functions.
Warm-up 1. Graph:
a) f(x) = -2x + 4
| | |
|x |f(x) |
| | |
|0 | |
| | |
|2 | |
f(x)
x[pic]
ΦNote: The vertical axis replaces y with f(x) notation.
Worksheet 49 (9.2)
b) f(x) = -2
f(x)
x[pic]
ΦNote: This is a constant function graphed by a horizontal line through the vertical axis at -2.
Problems - Graph:
1. f(x) = 2x - 3 2. f(x) = 2
[pic] [pic]
Worksheet 49 (9.2)
Summary 2:
A quadratic function is in the form f(x) = ax2 + bx + c where a, b, and c are real numbers and a ≠ 0.
The graph of a quadratic function is a parabola. Its vertex is [pic]and its axis of symmetry is[pic].
Graphing a quadratic function using the form f(x) = ax2 + bx + c:
Determine how the parabola opens:
1. If a > 0, then the parabola opens upward.
If a < 0, then the parabola opens downward.
Determine the vertex of the parabola using a, b, and c:
2. Evaluate[pic]. This is the x-coordinate of the vertex.
3. Find[pic]. This is the y-coordinate of the vertex.
Find another point to help accurately graph the parabola:
4. Choose an appropriate x and find f(x) to find another point on the parabola.
Sketch the graph and label:
5. Plot the vertex. Plot the point from step 4 and its image across the axis of symmetry. Use a smooth curve to draw in the parabola.
Worksheet 49 (9.2)
Warm-up 2. Graph:
a) f(x) = x2 + 4x + 3
Determine how the parabola opens:
The parabola opens _______________.
Determine the vertex of the parabola:
Evaluate[pic]when a = _____ and b = _____.
[pic]=[pic]
= _____
Find[pic]or f(-2).
f(x) = x2 + 4x + 3
f(-2) = ( )2 + 4( ) + 3
= _____
The vertex is ( , ).
Find another point:
Let x = -1.
f(x) = x2 + 4x + 3
f(-1) = ( )2 + 4( ) + 3
= _____
Another point is ( , ).
Its image across the axis of symmetry is ( , ).
Sketch the graph:
f(x)
[pic]x
Worksheet 49 (9.2)
b) f(x) = -2x2 + 4x
Determine how the parabola opens:
The parabola opens _______________.
Determine the vertex of the parabola:
Evaluate[pic]when a = _____ and b = _____.
[pic]=[pic]
= _____
Find[pic]or f(1).
f(x) = -2x2 + 4x
f(1) = -2( )2 + 4( )
= _____
The vertex is ( , ).
Find another point:
Let x = -1.
f(x) = -2x2 + 4x
f(-1) = -2( )2 + 4( )
= _____
Another point is ( , ).
Its image across the axis of symmetry is ( , ).
Sketch the graph:
f(x)
[pic]x
Worksheet 49 (9.2)
Problems - Graph:
3. f(x) = x2 + x - 2 4. f(x) = -2x2 - 6x
[pic] [pic]
Summary 3:
The vertex of a quadratic function is either the highest or lowest point on the graph.
The maximum value of a function is the vertex of a parabola opening downwards. It tells where the maximum occurs, x, and the maximum value of the function, f(x).
The minimum value of a function is the vertex of a parabola opening upwards. It tells where the minimum occurs, x, and the minimum value of the function, f(x).
Worksheet 49 (9.2)
Warm-up 3. a) A farmer has 148 rods of fencing and wants to enclose a rectangular plot of land that needs only three sides fenced because the fourth side is bounded by a stone wall. Find the length and width of the plot that will maximize the amount of area to be fenced.
Let x = width
________ = length
________ = area
The functional representation of area is:
f(x) = _______________
= -2x2 + 148x
ΦNote: Since this can be graphed as a parabola opening downwards, its vertex represents a maximum.
a = _____ and b = _____
[pic]=[pic]
= _____
The maximum area occurs when the width is _____ rods.
The length, 148 - 2x, is _____ rods.
Summary 4:
Other Functions
We will restate some of the graphing suggestions discussed in Chapter 7, to help us graph unfamiliar functions:
1) Determine the domain of the function.
2) Determine any types of symmetry that the equation possesses. (Note that the definition of a function rules out the possibility that the graph of a function has x-axis symmetry.
3) Find the y-intercept and the x-intercept.
4) Set up a table of ordered pairs that satisfy the equation.
5) Plot the points associated with the ordered pairs and connect them with a smooth curve. If appropriate, reflect this part of the curve according to any symmetries possessed by the graph.
Worksheet 49 (9.2)
Warm-up 4. a) Graph [pic]
Determine the domain of the function:
The radicand must be nonnegative, therefore x - 4 ≥ 0.
x ≥ 4
Domain = {x| }
There is no symmetry for this graph.
Find the y-intercept: [pic]
This equation has no real solution, therefore the graph has no y-intercept.
Find the x-intercept: [pic]
The x-intercept is ( , )
Set up a table of values:
| | |
|x |f(x) |
| | |
|4 |0 |
| | |
|5 | |
| | |
|6 |√2 |
| | |
|8 | |
Plot the points and connect with a smooth curve.
f(x)
[pic]x
Worksheet 49 (9.2)
Problem - Use the techniques discussed to graph:
5. f(x) = |x| + 2
y
[pic]x
Worksheet 50 (9.3)
9.3 Graphing Made Easy Via Transformations
Summary 1:
Graphs of 5 Basic Functions
[pic]
f(x) = x2
Domain = { x | x is a real number } Range = { f(x) | f(x) ≥ 0 }
[pic]
f(x) = x3
Domain = { x | x is a real number }
Range = { f(x) | f(x) is a real number }
[pic]
[pic]
Domain = { x | x ≠ 0 }
Range = { f(x) | f(x) ≠ 0 }
[pic]
[pic]
Domain = { x | x ≥ 0 }
Range = { f(x) | f(x) ≥ 0 }
[pic]
f(x) = |x|
Domain = { x | x is a real number }
Range = { f(x) | f(x) ≥ 0 }
Worksheet 50 (9.3)
Summary 2:
Translations are rigid transformations of a given curve. The basic shape of the curve is not changed by shifting up, down, left, or right.
Vertical Translation:
The graph of y = f(x) + k is the graph of y = f(x) shifted k units upward if k > 0 or shifted |k| units downward if k < 0.
Horizontal Translation:
The graph of y = f(x - h) is the graph of y = f(x) shifted h units to the right if h > 0 or shifted |h| units left if h < 0.
Warm-up 1. Graph using either a vertical or horizontal translation:
a) f(x) = x2 + 2
This is a _______________ translation of f(x) = x2.
f(x)
[pic]
x
Each point in f(x) = x2 is shifted _____ units __________.
f(x)
[pic]
x
Worksheet 50 (9.3)
b)[pic]
This is a _______________ translation of f(x) =[pic].
f(x)
[pic]
x
Each point in f(x) =[pic]is shifted _____ units to the __________.
f(x)
[pic]
x
Problems - Graph using either vertical or horizontal translation:
1. f(x) = x3 - 3 2. f(x) =[pic]
[pic] [pic]
Worksheet 50 (9.3)
Summary 3:
Reflections are rigid transformations of a given curve. The basic shape of the curve is not changed by finding its mirror image with respect to a given line.
X-axis Reflection:
The graph of y = -f(x) is the graph of y = f(x) reflected through the x-axis.
Y-axis Reflection:
The graph of y = f(-x) is the graph of y = f(x) reflected through the y-axis.
Warm-up 2. Graph using either x-axis or y-axis reflection:
a) f(x) =[pic]
The graph of f(x) =[pic]is a reflection of f(x) = _______.
f(x)
[pic]
x
Each point of f(x) =[pic]is reflected through the _______________.
f(x)
[pic]
x
Worksheet 50 (9.3)
b) f(x) =[pic]
The graph of f(x) =[pic]is a reflection of f(x) = __________. f(x)
[pic]
x
Each point of f(x) =[pic]is reflected through the _______________.
f(x)
[pic]
x
Problems - Graph using either x-axis or y-axis reflection:
3. f(x) = (-x)2 4. f(x) = -|x|
[pic] [pic]
Worksheet 50 (9.3)
Summary 4:
A transformation that narrows or widens the basic shape of a curve is done by vertical stretching or shrinking.
The graph of y = cf(x) is the graph of y = f(x) where the y-coordinate is multiplied by a factor of c. If c > 1, then the graph is stretched by a factor of c. If 0 < c < 1, then the graph is shrunk by a factor of c.
Warm-up 3. Graph by vertical stretching or shrinking:
a) [pic]
The graph of this function is a vertical _______________ of
f(x) = x3.
f(x)
[pic]
x
Every y-coordinate is to be shrunk by a factor of _____.
f(x)
[pic]
x
Worksheet 50 (9.3)
b) f(x) =[pic]
The graph of this function is a vertical _______________ of
f(x) =[pic].
f(x)
[pic]
x
Every y-coordinate is to be stretched by a factor of _____.
f(x)
[pic]
x
Problems - Graph by vertical stretching or shrinking:
5. f(x) =[pic] 6. f(x) = 3x2
[pic] [pic]
Worksheet 50 (9.3)
Warm-up 4. Graph using all appropriate transformations:
a) f(x) =[pic]
The basic curve for this is f(x) = __________.
f(x)
[pic]
x
List all the transformations that apply: ____________________
____________________
____________________
Each point of the basic curve is to be shifted __________ by 2 units and right by _____ unit. The y-coordinate of the point will then be stretched by a factor of _____.
f(x)
[pic]
x
Problem - Graph using all appropriate transformations:
7. f(x) =[pic]
[pic]
Worksheet 51 (9.4)
9.4 Composition of Functions
Summary 1:
Composition of Functions
The composition of functions f and g is defined by (f ° g)(x) = f( g(x) ), for all x in the domain of g such that g(x) is in the domain of f.
(f ° g)(x) is translated as "the composition of f and g."
f ( g(x) ) is translated as "f of g of x."
Warm-up 1. Find (f ° g)(x), given f(x) and g(x):
a) Let f (x) = x2 and g(x) = x - 2.
(f ° g)(x) = f ( g(x) )
= f (x - 2)
Since f (x) = x2
f (x - 2) = ( )2
= _______________
Therefore, (f ° g)(x) = _______________
Find (f ° g)(-2).
(f ° g)(x) = x2 - 4x + 4
(f ° g)(-2) = ( )2 - 4( ) + 4
= _____
Worksheet 51 (9.4)
b) Let f (x) =[pic]and g(x) = x + 5.
(f ° g)(x) = f ( g (x) )
= f (x + 5)
Since f (x) =[pic]
f(x + 5) =[pic]
=__________
Therefore, (f ° g)(x) = _______________
Find (f ° g)(-½).
(f ° g)(x) =[pic]
(f ° g)(-½) =[pic]
= __________
Warm-up 2. Find (g ° f)(x), given f(x) and g(x):
a) Let f (x) = x2 and g(x) = x - 2.
(g ° f)(x) = g( f (x) )
= g( )
Since g(x) = x - 2
g(x2) = ( ) - 2
Therefore, (g ° f)(x) = _______________
Worksheet 51 (9.4)
Find (g ° f)(0).
(g ° f)(x) = x2 - 2
(g ° f)(0) = ( )2 - 2
= _____
b) Let f(x) =[pic]and g(x) = x + 5.
(g ° f)(x) = g( f (x) )
= g[pic]
Since g(x) = x + 5
g[pic] = [pic] + 5
= _______
Therefore, (g ° f)(x) = _______________
Find (g ° f)(3).
(g ° f)(x) =[pic]
(g ° f)(3) =[pic]
= _____
ΦNote: Compare the order of the composition of functions in 1a & 2a and 1b & 2b. Composition of functions is not a commutative operation.
Worksheet 51 (9.4)
Problems - Determine (f ° g)(x) and (g ° f)(x) for each of the following:
1. f (x) =[pic] and g(x) = 2x + 5
Find (f ° g)(9) and (g ° f)(-2)
Worksheet 51 (9.4)
2. f (x) = -2x and g(x) =[pic]
Find (f ° g)(5) and (g ° f)(-3).
Worksheet 52 (9.5)
9.5 Inverse Functions
Summary 1:
The vertical line test for functions is used on graphs to determine when a relation is a function. The graph is said to pass the vertical line test when any vertical line drawn through the graph does not intersect the graph in more than one point. Every value of f(x) has only one value of x associated with it.
The horizontal line test for one-to-one functions is used on the graphs of functions to determine when a function is one-to-one. The graph of the function is said to pass the horizontal line test when any horizontal line drawn through the graph does not intersect the graph in more than one point. There is exactly one value of x associated with each value of f(x).
A relation is a one-to-one function if and only if the graph passes both the vertical and horizontal line tests.
Note: In a set of ordered pairs, a one-to-one function contains no ordered pairs with the same first components or second components.
Warm-up 1. Identify each graph that passes the vertical line test and is a function:
a)[pic] b)[pic]
c)[pic] d)[pic]
Worksheet 52 (9.5)
Warm-up 2. a) Determine from the functions identified above those that also pass the horizontal line test and are one-to-one.
Problems - Determine which of the following are functions and those functions that are one-to-one.
1. [pic] 2. [pic]
3. [pic] 4. [pic]
Worksheet 52 (9.5)
Summary 2:
An inverse function results when the components of each ordered pair of a given one-to-one function are interchanged.
The inverse function of f(x) is denoted by f -1 (x):
1. f -1 (x) is read "f inverse of x" or "the inverse of f at x."
2. If (a, b) is an ordered pair of f(x), then (b, a) is an ordered pair of
f -1 (x).
Relationships between f(x) and f -1 (x):
1. The graphs of f (x) and f -1 (x) are mirror images with respect to the line y = x.
2. Two functions are inverses when (f ° f -1 )(x) = x and (f -1 ° f)(x) = x.
Finding the inverse of a given function:
1. Let y = f (x)
2. Interchange variables x and y.
3. Solve for y in terms of x.
4. Replace y with f -1 (x) notation.
Warm-up 3. Find the inverse function:
a) {(1, 3), (2, 5), (3, 7)}
The inverse function is {(3, ), (5, ), (7, )}.
b) f (x) = 2x - 6
y = 2x - 6
x = ______ - 6
y = __________
f -1(x) = __________
Worksheet 52 (9.5)
c) f (x) =[pic]
y =[pic]
x =[pic]
[pic] =[pic]
[pic]= _______________
y = _______________
f -1(x) = _______________
Problems - Find the inverse function:
5. {(-2, 2), (-1, 3), (0, 4), (1, 5)}
6. f (x) = x3 + 1
7. f (x) =[pic]
Warm-up 4. Verify that (f ° f -1)(x) = x and (f -1 ° f)(x) = x for the previously worked warm-up examples:
a) f (x) = 2x - 6 and f -1(x) =[pic] (See warm-up 3b above.)
(f ° f -1)(x) = f ( f -1(x) )
Worksheet 52 (9.5)
Since f (x) = 2x - 6
f ([pic]) = 2( ) - 6
= _____
Therefore, (f ° f -1)(x) = _____
(f -1 ° f)(x) = f -1( f (x) )
Since f -1(x) =[pic]
f -1(2x - 6) =[pic]
= _____
Therefore, (f -1 ° f)(x) = _____
b) f (x) =[pic] and f -1(x) =[pic] (See warm-up 3c above.)
(f ° f -1)(x) = f ( f -1(x) )
Since f (x) =[pic]
f ([pic]) =[pic]
= _____
Therefore, (f ° f -1)(x) = _____
(f -1 ° f)(x) = f -1( f (x) )
Since f -1(x) =[pic]
f -1([pic]) =[pic]
= _____
Therefore, ( f -1 ° f)(x) = _____
Worksheet 52 (9.5)
Problems - Verify that (f ° f -1)(x) = x and (f -1 ° f)(x) = x for the previously worked problems:
8. f (x) = x3 + 1 and f -1(x) =[pic] (See problem 6 above.)
9. f (x) =[pic] and f -1(x) = 2x + 1 (See problem 7 above.)
Warm-up 5. Graph f(x) and f -1(x) on the same set of axes for the previously worked warm-up examples:
a) (See warm-up 3b and 4a.)
f (x) = 2x - 6
| | |
|x |f (x) |
| | |
|1 | |
| | |
|4 | |
[pic]
f -1(x) =[pic]
| | |
|x |f -1(x) |
| | |
|-4 | |
| | |
|2 | |
ΦNote: After graphing both straight lines, graph the line y = x. Notice how f -1(x) is the mirror image of f (x) through this line.
Worksheet 52 (9.5)
b) (See warm-up 3c and 4b.)
f (x) =[pic]
| | |
|x |f (x) |
| | |
|3/2 | |
| | |
|3 | |
[pic]
f -1(x) =[pic]
| | |
|x |f -1(x) |
| | |
|0 | |
| | |
|-1/2 | |
ΦNote: After graphing both straight lines, graph the line y = x. Notice how f -1(x) is the mirror image of f (x) through this line.
Problems - Graph f(x) and f -1(x) on the same set of axes for the previously worked problems:
10. (See problem 6 and 8.)
f (x) = x3 + 1 and f -1(x) =[pic]
[pic]
Worksheet 52 (9.5)
11. (See problem 7 and 9.)
f (x) =[pic] and f -1(x) = 2x + 1
[pic]
Worksheet 53 (9.6)
9.6 Direct and Inverse Variation
Summary 1:
Any function that can be written in the form y = kx where k is a nonzero constant is called a direct variation. The constant of variation is k.
Finding the constant of variation, k, when given values of the variables:
1. Translate into an algebraic statement of the form y = kx.
2. Substitute x and y with the known values.
3. Solve for k.
Finding additional values for the variables:
1. Replace k into the form y = kx.
2. Substitute the value given for one of the variables.
3. Solve for the other variable.
Warm-up 1. a) The circumference of a circle varies directly as its diameter. Find the constant of variation when the circumference is 22 cm and the diameter is 7 cm.
Direct variation: y = kx
y can be replaced with C for circumference.
x can be replaced with d for diameter.
C = kd
22 = k( )
k = _____
The constant of variation is _____.
b) Find the circumference of a circle that has a diameter 21 cm long. Use the constant of variation found in a above.
C = kd
C =( )( )
C =
The circumference is _____ cm, when the diameter is 21 cm.
Worksheet 53 (9.6)
c) y varies directly as the cube of x. Find the constant of variation at y = 9 when x = 3.
Direct variation: y = kx3
9 = k( )3
k = _____
The constant of variation is _____.
d) Find x when y = 8. Use the constant of variation in c above.
y =[pic]
______ =[pic]
x = ________
When y = 8, x = __________.
Problems
1. The distance a car travels varies directly as the time. Find the constant of variation if a car travels 204 miles in three hours. Use the constant of variation to find how long it takes to travel 34 miles.
2. y varies directly as the square root of x. Find the constant of variation at y = 10 when x = 25. Use the constant of variation to find y when x = 64.
Worksheet 53 (9.6)
Summary 2:
Any function that can be written in the form[pic]where k is a nonzero constant is called an inverse variation. The constant of variation is k.
Warm-up 2. a) The volume of a gas at a constant temperature varies inversely as the pressure. Find the constant of variation when volume is 15 cubic centimeters at 25 pounds of pressure.
Inverse variation: [pic]
y can be replaced with V for volume.
x can be replaced with P for pressure.
[pic]
[pic]
[pic]
b) Find the pressure when the volume is 125 cubic centimeters. Use the constant of variation from a above.
[pic]
[pic]
[pic]
The pressure is _____ pounds.
Worksheet 53 (9.6)
Problem
3. y varies inversely as the square of x. Find the constant of variation at y = 5 when x = 2. Use the constant of variation to find y when x = 10.
Summary 3:
Joint and Combined Variation
A joint variation is a direct variation relating more than two variables with the same constant: EX. y = kxz; V = khr2
A combined variation is a combination of direct and inverse variation relating more than two variables with the same constant:
EX. [pic]
Warm-up 3. Write a variation equation for each of the following:
a) z varies jointly as y and the cube of z.
z = k( )( )
b) a varies directly as b and inversely as the square of c.
[pic]
Problems - Write a variation equation for each of the following:
4. T varies jointly as V and P.
5. x varies directly as the square root of y and inversely as z.
Worksheet 53 (9.6)
Warm-up 4. a) Solve: If V varies jointly as B and h, and V = 20 when B = 5 and h = 2, find V when B = 25 and h = 4.
V = kBh Write the joint variation equation.
20 = k(5)(2) Substitute in the first set of values for V, B, and h.
20 = k
= k Solve for k.
V = kBh Use the joint variation equation again.
V = 2( )( ) Use the k find in the above step and the second
set of values for B and h.
V = Solve for V.
Problem - Solve:
6. If A varies jointly as b and h, and A = 300 when b = 40 and h = 15, find A when b = 10 and h = 8.
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