Answer a. Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 a. Graph ƒ ( x ) = − 3 x + 9 To determine To solve: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , b. Partition [ 0 , 3 ] into three subintervals of equal length and choose u as the left endpoint of each subinterval. Answer b. 18 Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 b. Partition [ 0 , 3 ] into three subintervals of equal length 1 and choose u as the left endpoint of each subinterval. The area A is approximated as A = f ( 0 ) 1 + f ( 1 ) 1 + f ( 2 ) 1 = 9 ( 1 ) + 6 ( 1 ) + 3 ( 1 ) = 9 + 6 + 3 = 18 To determine To solve: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , c. Partition [ 0 , 3 ] into three subintervals of equal length and choose u as the right endpoint of each subinterval. Answer c. 9 Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 c. Partition [ 0 , 3 ] into three subintervals of equal length 1 and choose u as the right endpoint of each subinterval. The area A is approximated as A = f ( 1 ) 1 + f ( 2 ) 1 + f ( 3 ) 1 = 6 ( 1 ) + 3 ( 1 ) + 0 ( 1 ) = 6 + 3 + 0 = 9 To determine To solve: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , d. Partition [ 0 , 3 ] into six subintervals of equal length and choose u as the left endpoint of each subinterval. Answer d. 63 4 Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 d. Partition [ 0 , 3 ] into six subintervals of equal length 0.5 and choose u as the left endpoint of each subinterval. The area A is approximated as A = f ( 0 ) ( 0.5 ) + f ( 0.5 ) ( 0.5 ) + f ( 1 ) ( 0.5 ) + f ( 1.5 ) ( 0.5 ) + f ( 2 ) ( 0.5 ) + f ( 2.5 ) ( 0.5 ) = 9 ( 0.5 ) + ( 7.5 ) ( 0.5 ) + 6 ( 0.5 ) + ( 4.5 ) ( 0.5 ) + 3 ( 0.5 ) + ( 1.5 ) ( 0.5 ) = 4.5 + 3.75 + 3 + 2.25 + 1.5 + 0.75 = 15.75 = 63 4 To determine To solve: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , e. Partition [ 0 , 3 ] into six subintervals of equal length and choose u as the right endpoint of each subinterval. Answer e. 45 4 Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 e. Partition [ 0 , 3 ] into six subintervals of equal length 0.5 and choose u as the right endpoint of each subinterval. The area A is approximated as A = f ( 0.5 ) ( 0.5 ) + f ( 1 ) ( 0.5 ) + f ( 1.5 ) ( 0.5 ) + f ( 2 ) ( 0.5 ) + f ( 2.5 ) ( 0.5 ) + f ( 3 ) ( 0.5 ) = ( 7.5 ) ( 0.5 ) + 6 ( 0.5 ) + ( 4.5 ) ( 0.5 ) + 3 ( 0.5 ) + ( 1.5 ) ( 0.5 ) + 0 ( 0.5 ) = 3.75 + 3 + 2.25 + 1.5 + 0.75 + 0 = 11.25 = 45 4 To determine To solve: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , f. What is the actual area A ? Answer f. 27 2 Explanation Given: The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] . Calculation: ƒ ( 0 ) = − 3 ( 0 ) + 9 = 9 ; ƒ ( 0.5 ) = − 3 ( 0.5 ) + 9 = 7.5 ; ƒ ( 1 ) = − 3 ( 1 ) + 9 = 6 ; ƒ ( 1.5 ) = − 3 ( 1.5 ) + 9 = 4.5 ƒ ( 2 ) = − 3 ( 2 ) + 9 = 3 ; ƒ ( 2.5 ) = − 3 ( 2.5 ) + 9 = 1.5 ; ƒ ( 3 ) = − 3 ( 3 ) + 9 = 0 f. The actual area under the graph of ƒ ( x ) = − 3 x + 9 from 0 to 3 is the area of a right triangle whose base is of length 3 and whose height is 9. The actual area A is Therefore A = 1 2 b a s e × h e i g h t A = 1 2 ( 3 ) ( 9 ) = 27 2

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

See similar textbooks

Question

Chapter 14.5, Problem 11AYU

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

a. Graph $ƒ$ .

In (b)–(e), approximate the area $A$ under $ƒ$ from 0 to 3 as follows:

Expert Solution

Answer to Problem 11AYU

Precalculus Enhanced with Graphing Utilities, Chapter 14.5, Problem 11AYU , additional homework tip 1

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

a. Graph $ƒ (x) = - 3 x + 9$

Precalculus Enhanced with Graphing Utilities, Chapter 14.5, Problem 11AYU , additional homework tip 2

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

b. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Expert Solution

Answer to Problem 11AYU

b. 18

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

b. Partition $[0, 3]$ into three subintervals of equal length 1 and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0) 1 + f (1) 1 + f (2) 1$

$= 9 (1) + 6 (1) + 3 (1)$

$= 9 + 6 + 3$

$= 18$

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

c. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Expert Solution

Answer to Problem 11AYU

c. 9

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

c. Partition $[0, 3]$ into three subintervals of equal length 1 and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A = f (1) 1 + f (2) 1 + f (3) 1$

$= 6 (1) + 3 (1) + 0 (1)$

$= 6 + 3 + 0$

$= 9$

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

d. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Expert Solution

Answer to Problem 11AYU

d. $\frac{63}{4}$

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

d. Partition $[0, 3]$ into six subintervals of equal length $0.5$ and choose $u$ as the left endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0) (0.5) + f (0.5) (0.5) + f (1) (0.5) + f (1.5) (0.5) + f (2) (0.5) + f (2.5) (0.5)$

$= 9 (0.5) + (7.5) (0.5) + 6 (0.5) + (4.5) (0.5) + 3 (0.5) + (1.5) (0.5)$

$= 4.5 + 3.75 + 3 + 2.25 + 1.5 + 0.75$

$= 15.75 = \frac{63}{4}$

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

e. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Expert Solution

Answer to Problem 11AYU

e. $\frac{45}{4}$

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

e. Partition $[0, 3]$ into six subintervals of equal length $0.5$ and choose $u$ as the right endpoint of each subinterval.

The area $A$ is approximated as

$A = f (0.5) (0.5) + f (1) (0.5) + f (1.5) (0.5) + f (2) (0.5) + f (2.5) (0.5) + f (3) (0.5)$

$= (7.5) (0.5) + 6 (0.5) + (4.5) (0.5) + 3 (0.5) + (1.5) (0.5) + 0 (0.5)$

$= 3.75 + 3 + 2.25 + 1.5 + 0.75 + 0$

$= 11.25 = \frac{45}{4}$

To determine

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

f. What is the actual area $A$ ?

Expert Solution

Answer to Problem 11AYU

f. $\frac{27}{2}$

Explanation of Solution

Given:

The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ .

Calculation:

$ƒ (0) = - 3 (0) + 9 = 9; ƒ (0.5) = - 3 (0.5) + 9 = 7.5; ƒ (1) = - 3 (1) + 9 = 6; ƒ (1.5) = - 3 (1.5) + 9 = 4.5$

$ƒ (2) = - 3 (2) + 9 = 3; ƒ (2.5) = - 3 (2.5) + 9 = 1.5; ƒ (3) = - 3 (3) + 9 = 0$

f. The actual area under the graph of $ƒ (x) = - 3 x + 9$ from 0 to 3 is the area of a right triangle whose base is of length 3 and whose height is 9. The actual area $A$ is

Therefore

$A = \frac{1}{2} b a s e \times h e i g h t$

$A = \frac{1}{2} (3) (9) = \frac{27}{2}$

Chapter 14.5, Problem 10AYU

Chapter 14.5, Problem 12AYU

Chapter 14 Solutions

Precalculus Enhanced with Graphing Utilities

Ch. 14.1 - Graph f( x )={ 3x2ifx2 3ifx=2 (pp.100-102)Ch. 14.1 - If f( x )={ xifx0 1ifx0 what is f( 0 ) ?...Ch. 14.1 - The limit of a function f( x ) as x approaches c...Ch. 14.1 - If a function f has no limit as x approaches c ,...Ch. 14.1 - True or False lim xc f( x )=N may be described by...Ch. 14.1 - True or False lim xc f( x ) exists and equals some...Ch. 14.1 - lim x2 ( 4 x 3 )Ch. 14.1 - lim x3 ( 2 x 2 +1 )Ch. 14.1 - lim x0 x+1 x 2 +1 Ch. 14.1 - lim x0 2x x 2 +4

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Knowledge Booster

The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Solution Summary: The author calculates the area A under from 0 to 3 by dividing the function into three subintervals of equal length.

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

a. Graph $ƒ$ .

In (b)–(e), approximate the area $A$ under $ƒ$ from 0 to 3 as follows:

Answer to Problem 11AYU

Explanation of Solution

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

b. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

c. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

d. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

e. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

f. What is the actual area $A$ ?

Answer to Problem 11AYU

Explanation of Solution

Chapter 14 Solutions

Additional Math Textbook Solutions

The function ƒ ( x ) = − 3 x + 9 is defined on the interval [ 0 , 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Solution Summary: The author calculates the area A under from 0 to 3 by dividing the function into three subintervals of equal length.

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , a. Graph ƒ . In (b)–(e), approximate the area A under ƒ from 0 to 3 as follows:

Answer to Problem 11AYU

Explanation of Solution

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , b. Partition [ 0, 3 ] into three subintervals of equal length and choose u as the left endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , c. Partition [ 0, 3 ] into three subintervals of equal length and choose u as the right endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , d. Partition [ 0, 3 ] into six subintervals of equal length and choose u as the left endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , e. Partition [ 0, 3 ] into six subintervals of equal length and choose u as the right endpoint of each subinterval.

Answer to Problem 11AYU

Explanation of Solution

To solve: The function ƒ( x ) = −3x + 9 is defined on the interval [ 0, 3 ] , f. What is the actual area A ?

Answer to Problem 11AYU

Explanation of Solution

Chapter 14 Solutions

Additional Math Textbook Solutions

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

a. Graph $ƒ$ .

In (b)–(e), approximate the area $A$ under $ƒ$ from 0 to 3 as follows:

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

b. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

c. Partition $[0, 3]$ into three subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

d. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the left endpoint of each subinterval.

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

e. Partition $[0, 3]$ into six subintervals of equal length and choose $u$ as the right endpoint of each subinterval.

To solve: The function $ƒ (x) = - 3 x + 9$ is defined on the interval $[0, 3]$ ,

f. What is the actual area $A$ ?