Easy Problems
1.
{ 15, x<10
F(x){ 4x-10 10<x<15 solve for f(12)
{ 5(x-1) 15<x
x=12 and fits the signs of the second problem and so we plug 12 in for x in that problem.
4(12)-10=
48-10=
38
2. { 8(x-6) -1<x<2
F(x){ 3x-2 2<x<5 solve for f(6)
{ 2 x>5
x=6 fits into the last domain restriction making it 2
=2
3. { 2 x<3
F(x){ x 3<x<5 Make a table
{ 2x 5<x
Each Equation gets their own table
x | y x | y x | y
0 | 2 3 | 3 5 | 10
1 | 2 4 | 4 6 | 12
3 |2 5 | 5 7 | 14
{ 15, x<10
F(x){ 4x-10 10<x<15 solve for f(12)
{ 5(x-1) 15<x
x=12 and fits the signs of the second problem and so we plug 12 in for x in that problem.
4(12)-10=
48-10=
38
2. { 8(x-6) -1<x<2
F(x){ 3x-2 2<x<5 solve for f(6)
{ 2 x>5
x=6 fits into the last domain restriction making it 2
=2
3. { 2 x<3
F(x){ x 3<x<5 Make a table
{ 2x 5<x
Each Equation gets their own table
x | y x | y x | y
0 | 2 3 | 3 5 | 10
1 | 2 4 | 4 6 | 12
3 |2 5 | 5 7 | 14
Medium Problems
1. { 6 x<0
F(x){ 3x-1 0<x<5 Graph it!
{ 4(x-1) 5<x
x | y x | y x | y
0 | 6 0 | -1 5 | 10
-1 | 6 3 | 8 6 | 12
-2 | 6 5 | 14 7 | 14
F(x){ 3x-1 0<x<5 Graph it!
{ 4(x-1) 5<x
x | y x | y x | y
0 | 6 0 | -1 5 | 10
-1 | 6 3 | 8 6 | 12
-2 | 6 5 | 14 7 | 14
2. { x-5, x<1
F(x){ 2x-1 1<x<6 Graph it!
{ x-1 6<x
x | y x | y x | y
1 |-4 1 | 1 6 | 5
0 |-5 3 | 5 7 | 6
-1 |-6 6 | 11 8 | 7
F(x){ 2x-1 1<x<6 Graph it!
{ x-1 6<x
x | y x | y x | y
1 |-4 1 | 1 6 | 5
0 |-5 3 | 5 7 | 6
-1 |-6 6 | 11 8 | 7
3.
{ 9, x<-1
F(x){ 4x-2 -1<x<2 Graph it!
{ 2(x-1) 2<x
x | y x | y x | y
-1 |-4 -1 | -6 2 | 2
-2 |-5 0 | -2 3 | 4
-3 |-6 2 | 6 4 | 6
{ 9, x<-1
F(x){ 4x-2 -1<x<2 Graph it!
{ 2(x-1) 2<x
x | y x | y x | y
-1 |-4 -1 | -6 2 | 2
-2 |-5 0 | -2 3 | 4
-3 |-6 2 | 6 4 | 6
Hard Problems (Finding the Equation of a Piecewise Function)
(1.) The first portion of the function is a quadratic, the second is linear
The equation of the first function is: f(x)=x^2 +1 We find this using our knowledge of transformations
The equation of the second function is: f(x)= -x+3 We find this using the slope formula and solving for "b"
The first function has no explicit beginning, but ends with an open dot at x=2. This generates the domain restriction: x<2
The second function begins with a closed dot at x=2, but has no explicit end. This generates the domain restriction: x>=2
The final piecewise function is:
f( x ) = x 2 + 1 if x < 2
= - x + 3 if x >= 2
The equation of the first function is: f(x)=x^2 +1 We find this using our knowledge of transformations
The equation of the second function is: f(x)= -x+3 We find this using the slope formula and solving for "b"
The first function has no explicit beginning, but ends with an open dot at x=2. This generates the domain restriction: x<2
The second function begins with a closed dot at x=2, but has no explicit end. This generates the domain restriction: x>=2
The final piecewise function is:
f( x ) = x 2 + 1 if x < 2
= - x + 3 if x >= 2
(2.) This piecewise function consists of three linear functions
The first (leftmost) is constant, at -1. This is the equation, f(x)= -1
The second (middle) is constant, at 1. This is the equation, f(x)=1
The third equation is modeled by the equation f(x)=x
The first equation goes to negative infinity and ends with a closed dot at x= -1
The second equation begins with an open dot at x= -1 and ends with a closed dot at x=1
The third equation goes to infinity and begins at x=1 with an open dot
The final piecewise function is:
f( x ) = -1 if x <= -1
= 1 if -1 < x <= 1
= x if x > 1
The first (leftmost) is constant, at -1. This is the equation, f(x)= -1
The second (middle) is constant, at 1. This is the equation, f(x)=1
The third equation is modeled by the equation f(x)=x
The first equation goes to negative infinity and ends with a closed dot at x= -1
The second equation begins with an open dot at x= -1 and ends with a closed dot at x=1
The third equation goes to infinity and begins at x=1 with an open dot
The final piecewise function is:
f( x ) = -1 if x <= -1
= 1 if -1 < x <= 1
= x if x > 1
(3.) This is the you try, so we only provide the answer:
f( x ) = 1 / x if x < 0
= e -x if x >= 0
f( x ) = 1 / x if x < 0
= e -x if x >= 0
Real World Application Problems
Problem One
Zero photocopies
$0
|
20 photocopies
$3.00
|
50 photocopies
$6.00
|
60 photocopies
$6.00
|
The equation is:
0.15x, if 0<x<=20
(C)x= {0.12x, if 21<=x<=50
0.10x, if x>=51
0.15x, if 0<x<=20
(C)x= {0.12x, if 21<=x<=50
0.10x, if x>=51
Question 1. - $2.55
Question 2. - $6.10
Question 3. - $4.32
Question 4. - 41 copies
Question 2. - $6.10
Question 3. - $4.32
Question 4. - 41 copies
Problem Two
6 years old
$4.50
|
12 years old
$5.00
|
19 years old
$6.00
|
61 years old
$4.50
|
The equation is:
4.50, if 2<=x<=9 or x>= 66
(C)x= { 5.00, if 10<=x<=18
6.00, if 19<=x<=55
Question (1.) $6.00 + $5.00 + $4.50= $15.50
4.50, if 2<=x<=9 or x>= 66
(C)x= { 5.00, if 10<=x<=18
6.00, if 19<=x<=55
Question (1.) $6.00 + $5.00 + $4.50= $15.50
Problem Three (You Try)
1.) Seven miles
2.) $16.75
2.) $16.75