BESTSOLUTION1
INSTRUCTORS
SOLUTION MANUAL
FOR CALCULUS
EARLY
TH
TRANSCENDENTS 9
EDITION BY JAMES
STEWART|DANIEL K.
CLEGG|SALEEM
WATSON
1
,BESTSOLUTION1
1 Functions And Models
1.1 Four Ways To Represent A Function
√ √
1. The Functions Ƒ(X) = X + 2 − X And G(U) = U + 2 − U Give Exactly The Same Output Values For Every Input Value, So Ƒ
And G Are Equal.
X2 X X(X − 1) = X For X 1 = 0, So Ƒ And G [Where G(X) = X] Are Not Equal Because Ƒ(1) Is Undefined And
2. Ƒ(X) = − =
—
X −1 X −1
6
G(1) = 1.
3. (A) The Point (−2 2) Lies On The Graph Of G, So G(−2) = 2. Similarly, G (0) = −2, G (2) = 1, And G (3) 2 5.
(b) Only The Point (−4 3) On The Graph Has A Y-Value Of 3, So The Only Value Of X For Which G(X) = 3 Is −4.
(c) The Function Outputs G(X) Are Never Greater Than 3, So G(X) ≤ 3 For The Entire Domain Of The Function. Thus, G(X) ≤ 3
For
−4 ≤ X ≤ 4 (Or, Equivalently, On The Interval [−4 4]).
(d) The Domain Consists Of All X-Values On The Graph Of G: {X | −4 ≤ X ≤ 4} = [−4 4]. The Range Of G Consists Of All The
Y-Values On The Graph Of G: {Y | −2 ≤ Y ≤ 3} = [−2 3].
(e) For Any X1 X 2 In The Interval [0 2], We Have G(X1 ) G(X2 ). [The Graph Rises From (0 −2) To (2 1).] Thus,
G(X) sIIncreasing On [0 2].
4. (A) From The Graph, We Have Ƒ(−4) = −2 And G(3) = 4.
(b) Since Ƒ(−3) = −1 And G(−3) = 2, Or By Observing That The Graph Of G Is Above The Graph Of Ƒ At X = −3, G(−3) Is
Larger Than Ƒ(−3).
(c) The Graphs Of Ƒ And G Intersect At X = −2 And X = 2, So Ƒ(X) = G(X) At These Two Values Of X.
(d) The Graph Of Ƒ Lies Below Or On The Graph Of G For −4 ≤ X ≤ −2 And For 2 ≤ X ≤ 3. Thus, The Intervals On Which
Ƒ (X) ≤ G(X) Are [−4 −2] And [2 3].
(e) Ƒ (X) = −1 Is Equivalent To Y = −1, And The Points On The Graph Of Ƒ With Y-Values Of −1 Are (−3 −1) And (4
−1), So The Solution Of The Equation Ƒ(X) = −1 Is X = −3 Or X = 4.
(f) For Any X1 X 2 In The Interval [−4 0], We Have G(X1 ) G(X2 ). Thus, G(X) Is Decreasing On [−4 0].
(g) The Domain Of Ƒ Is {X | −4 ≤ X ≤ 4} = [−4 4]. The Range Of Ƒ Is {Y | −2 ≤ Y ≤ 3} = [−2 3].
(h) The Domain Of G Is {X | −4 ≤ X ≤ 3} = [−4 3]. Estimating The Lowest Point Of The Graph Of G As Having Coordinates
(0 0 5), The Range Of G Is Approximately {Y | 0 5 ≤ Y ≤ 4} = [0 5 4].
5. From Figure 1 In The Text, The Lowest Point Occurs At About (T A) = (12 −85). The Highest Point Occurs At About
(17 115). Thus, The Range Of The Vertical Ground Acceleration Is −85 ≤ A ≤ 115. Written In Interval Notation, The
Range Is [−85 115].
°
C 2021 Cengage Learning. All Rights Reserved. May Not Be Scanned, Copied, Or Duplicated, Or Posted To A Publicly Accessible Website, In Whole Or In Part. 9
2
,BESTSOLUTION1
10 ¤ Chapter 1 Functions And Models
6. Example 1:
Distance D Traveled By The Car Is A Function Of The Time T. The
Domain Of The Function Is {T | 0 ≤ T ≤ 2}, Where T Is M easured
In Hours. The Range
Of The Function Is {D | 0 ≤ D ≤ 120}, Where D Is M easured In M iles.
Example 2: At A Certain University, The Number Of Students N On
Campus At Any Time On A Particular Day Is A Function Of The
Time T AfterM idnight. The Domain Of The Function Is {T | 0 ≤ T ≤
24}, Where T Is
M easured In Hours. The Range Of The Function Is {N | 0 ≤ N ≤ K},
Where N Is An Integer And K Is The Largest Number Of Students On
Campus At Once.
Example 3: A Certain Employee Is Paid $8 00 Per Hour And Works pay
A M aximum Of 30 Hours Per Week. The Number Of Hours 240
238
Worked Is Rounded Down To The Nearest Quarter Of An Hour. 236
This Employee’s Gross Weekly Pay P Is A Function Of The
4
Number Of Hours Worked H. The Domain Of The Function Is [0 2
30] And The Range Of The Function Is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. We Solve 3x − 5y = 7 For Y: 3x − 5y = 7 ⇔ −5y = −3x +7 ⇔ Y = 3 X − 7 . Since The Equation Determines Exactly
5 5
One Value Of Y For Each Value Of X, The Equation Defines Y As A Function Of X.
8. We Solve 3x2 − 2y = 5 For Y: 3x2 − 2y = 5 ⇔ −2y = −3x2 +5 ⇔ Y = 3 X 2 − 5 . Since The Equation Determines
2 2
Exactly One Value Of Y For Each Value Of X, The Equation Defines Y As A Function Of X.
√
9. We Solve X 2 + (Y − 3)2 = 5 For Y: X2 + (Y − 3)2 = 5 ⇔ (Y − 3)2 = 5 − X 2 ⇔ Y − 3 = ± 5 − X2 ⇔
√
Y = 3 ± 5 − X2. Some Input Values X Correspond To M ore Than One Output Y. (For Instance, X = 1 Corresponds To Y =
1 And To Y = 5.) Thus, The Equation Does Not Define Y As A Function Of X.
10. We Solve 2xy + 5y 2 = 4 For Y: 2xy + 5y 2 = 4 ⇔ 5y 2 + (2x) Y − 4 = 0 ⇔
−2x ± (2x) − 4(5)(−4)
2 √ √
−2x ± 4x2 + 8010 −X ± X2 + 20
y= 2(5) = = (Using The Quadratic Formula). Some Input
5
Values X Correspond To M ore Than One Output Y. (For Instance, X = 4 Corresponds To Y = −2 And To Y = Thus,
TheEquation Does Not Define Y As A Function Of X.
√
11. We Solve (Y + 3)3 + 1 = 2x For Y: (Y + 3)3 + 1 = 2x ⇔ (Y + 3)3 = 2x − 1 ⇔ Y + 3 = 3 2x − 1 ⇔
√
Y = −3 + 3 2x − 1. Since The Equation Determines Exactly One Value Of Y For Each Value Of X, The Equation Defines Y As
A Function Of X.
3
, BESTSOLUTION1
°
C 2021 Cengage Learning. All Rights Reserved. May Not Be Scanned, Copied, Or Duplicated, Or Posted To A Publicly Accessible Website, In Whole Or In Part.
4
INSTRUCTORS
SOLUTION MANUAL
FOR CALCULUS
EARLY
TH
TRANSCENDENTS 9
EDITION BY JAMES
STEWART|DANIEL K.
CLEGG|SALEEM
WATSON
1
,BESTSOLUTION1
1 Functions And Models
1.1 Four Ways To Represent A Function
√ √
1. The Functions Ƒ(X) = X + 2 − X And G(U) = U + 2 − U Give Exactly The Same Output Values For Every Input Value, So Ƒ
And G Are Equal.
X2 X X(X − 1) = X For X 1 = 0, So Ƒ And G [Where G(X) = X] Are Not Equal Because Ƒ(1) Is Undefined And
2. Ƒ(X) = − =
—
X −1 X −1
6
G(1) = 1.
3. (A) The Point (−2 2) Lies On The Graph Of G, So G(−2) = 2. Similarly, G (0) = −2, G (2) = 1, And G (3) 2 5.
(b) Only The Point (−4 3) On The Graph Has A Y-Value Of 3, So The Only Value Of X For Which G(X) = 3 Is −4.
(c) The Function Outputs G(X) Are Never Greater Than 3, So G(X) ≤ 3 For The Entire Domain Of The Function. Thus, G(X) ≤ 3
For
−4 ≤ X ≤ 4 (Or, Equivalently, On The Interval [−4 4]).
(d) The Domain Consists Of All X-Values On The Graph Of G: {X | −4 ≤ X ≤ 4} = [−4 4]. The Range Of G Consists Of All The
Y-Values On The Graph Of G: {Y | −2 ≤ Y ≤ 3} = [−2 3].
(e) For Any X1 X 2 In The Interval [0 2], We Have G(X1 ) G(X2 ). [The Graph Rises From (0 −2) To (2 1).] Thus,
G(X) sIIncreasing On [0 2].
4. (A) From The Graph, We Have Ƒ(−4) = −2 And G(3) = 4.
(b) Since Ƒ(−3) = −1 And G(−3) = 2, Or By Observing That The Graph Of G Is Above The Graph Of Ƒ At X = −3, G(−3) Is
Larger Than Ƒ(−3).
(c) The Graphs Of Ƒ And G Intersect At X = −2 And X = 2, So Ƒ(X) = G(X) At These Two Values Of X.
(d) The Graph Of Ƒ Lies Below Or On The Graph Of G For −4 ≤ X ≤ −2 And For 2 ≤ X ≤ 3. Thus, The Intervals On Which
Ƒ (X) ≤ G(X) Are [−4 −2] And [2 3].
(e) Ƒ (X) = −1 Is Equivalent To Y = −1, And The Points On The Graph Of Ƒ With Y-Values Of −1 Are (−3 −1) And (4
−1), So The Solution Of The Equation Ƒ(X) = −1 Is X = −3 Or X = 4.
(f) For Any X1 X 2 In The Interval [−4 0], We Have G(X1 ) G(X2 ). Thus, G(X) Is Decreasing On [−4 0].
(g) The Domain Of Ƒ Is {X | −4 ≤ X ≤ 4} = [−4 4]. The Range Of Ƒ Is {Y | −2 ≤ Y ≤ 3} = [−2 3].
(h) The Domain Of G Is {X | −4 ≤ X ≤ 3} = [−4 3]. Estimating The Lowest Point Of The Graph Of G As Having Coordinates
(0 0 5), The Range Of G Is Approximately {Y | 0 5 ≤ Y ≤ 4} = [0 5 4].
5. From Figure 1 In The Text, The Lowest Point Occurs At About (T A) = (12 −85). The Highest Point Occurs At About
(17 115). Thus, The Range Of The Vertical Ground Acceleration Is −85 ≤ A ≤ 115. Written In Interval Notation, The
Range Is [−85 115].
°
C 2021 Cengage Learning. All Rights Reserved. May Not Be Scanned, Copied, Or Duplicated, Or Posted To A Publicly Accessible Website, In Whole Or In Part. 9
2
,BESTSOLUTION1
10 ¤ Chapter 1 Functions And Models
6. Example 1:
Distance D Traveled By The Car Is A Function Of The Time T. The
Domain Of The Function Is {T | 0 ≤ T ≤ 2}, Where T Is M easured
In Hours. The Range
Of The Function Is {D | 0 ≤ D ≤ 120}, Where D Is M easured In M iles.
Example 2: At A Certain University, The Number Of Students N On
Campus At Any Time On A Particular Day Is A Function Of The
Time T AfterM idnight. The Domain Of The Function Is {T | 0 ≤ T ≤
24}, Where T Is
M easured In Hours. The Range Of The Function Is {N | 0 ≤ N ≤ K},
Where N Is An Integer And K Is The Largest Number Of Students On
Campus At Once.
Example 3: A Certain Employee Is Paid $8 00 Per Hour And Works pay
A M aximum Of 30 Hours Per Week. The Number Of Hours 240
238
Worked Is Rounded Down To The Nearest Quarter Of An Hour. 236
This Employee’s Gross Weekly Pay P Is A Function Of The
4
Number Of Hours Worked H. The Domain Of The Function Is [0 2
30] And The Range Of The Function Is 0 0.25 0.50 0.75 29.50 29.75 30 hours
{0 2 00 4 00 238 00 240 00}.
7. We Solve 3x − 5y = 7 For Y: 3x − 5y = 7 ⇔ −5y = −3x +7 ⇔ Y = 3 X − 7 . Since The Equation Determines Exactly
5 5
One Value Of Y For Each Value Of X, The Equation Defines Y As A Function Of X.
8. We Solve 3x2 − 2y = 5 For Y: 3x2 − 2y = 5 ⇔ −2y = −3x2 +5 ⇔ Y = 3 X 2 − 5 . Since The Equation Determines
2 2
Exactly One Value Of Y For Each Value Of X, The Equation Defines Y As A Function Of X.
√
9. We Solve X 2 + (Y − 3)2 = 5 For Y: X2 + (Y − 3)2 = 5 ⇔ (Y − 3)2 = 5 − X 2 ⇔ Y − 3 = ± 5 − X2 ⇔
√
Y = 3 ± 5 − X2. Some Input Values X Correspond To M ore Than One Output Y. (For Instance, X = 1 Corresponds To Y =
1 And To Y = 5.) Thus, The Equation Does Not Define Y As A Function Of X.
10. We Solve 2xy + 5y 2 = 4 For Y: 2xy + 5y 2 = 4 ⇔ 5y 2 + (2x) Y − 4 = 0 ⇔
−2x ± (2x) − 4(5)(−4)
2 √ √
−2x ± 4x2 + 8010 −X ± X2 + 20
y= 2(5) = = (Using The Quadratic Formula). Some Input
5
Values X Correspond To M ore Than One Output Y. (For Instance, X = 4 Corresponds To Y = −2 And To Y = Thus,
TheEquation Does Not Define Y As A Function Of X.
√
11. We Solve (Y + 3)3 + 1 = 2x For Y: (Y + 3)3 + 1 = 2x ⇔ (Y + 3)3 = 2x − 1 ⇔ Y + 3 = 3 2x − 1 ⇔
√
Y = −3 + 3 2x − 1. Since The Equation Determines Exactly One Value Of Y For Each Value Of X, The Equation Defines Y As
A Function Of X.
3
, BESTSOLUTION1
°
C 2021 Cengage Learning. All Rights Reserved. May Not Be Scanned, Copied, Or Duplicated, Or Posted To A Publicly Accessible Website, In Whole Or In Part.
4
