Section 1.4 Exponential Functions 3. (a) 8 (2) 4 = 8 · 2 4 4 = 1612 (b) (63 ) 4 25 = 6 4 (3 ) 4 25 = 129612 25 = 6487 14. We start with the graph of = (05) (Figure 3) and shift it 2 units downward to obtain the graph of = (05) − 2. The horizontal asymptote of the final graph is = −2. 17. (a) To find the equation of the graph that results from shifting the graph of = 2 units downward, we subtract 2 from the original function to get = − 2. (b) To find the equation of the graph that results from shifting the graph of = 2 units to the right, we replace with − 2 in the original function to get = (−2) . (c) To find the equation of the graph that results from reflecting the graph of = about the -axis, we multiply the original function by −1 to get = −. (d) To find the equation of the graph that results from reflecting the graph of = about the -axis, we replace with −in the original function to get = −. (e) To find the equation of the graph that results from reflecting the graph of = about the -axis and then about the -axis, we first multiply the original function by −1 (to get = −) and then replace with −in this equation to get = −−. 21. Use = with the points (16) and (324). 6= 1 = 6 and 24 = 3 ⇒ 24 = 6 3 ⇒ 4= 2 ⇒ =2 [since 0] and = 6 2 =3. The function is ()=3 · 2 . 24. Suppose the month is February. Your payment on the 28th day would be 2 28−1 =2 27 = 134,217,728 cents, or $1,342,177.28. Clearly, the second method of payment results in a larger amount for any month. 25. 1 m = 100 cm, (100) = 100 2 cm = 10 000 cm = 100 m. (100) = 2 100 cm =2 100 (100 · 1000) km ≈ 127 × 10 25 km 10 25 km. 1
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c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.4 EXPONENTIAL FUNCTIONS ¤ 41
12. We start with the graph of = (05)
(Figure 3) and shift it 1 unit to the right to
obtain the graph of = (05)−1.
13. We start with the graph of = 10
(Figure 3) and shift it 2 units to the left to
obtain the graph of = 10+2.
14. We start with the graph of = (05)
(Figure 3) and shift it 2 units downward
to obtain the graph of = (05) − 2.
The horizontal asymptote of the final
graph is = −2.
15. We start with the graph of = (Figure 16) and reflect about the -axis to get the graph of = −. Then we compress
the graph vertically by a factor of 2 to obtain the graph of = 12− and then reflect about the -axis to get the graph of
= − 12−. Finally, we shift the graph upward one unit to get the graph of = 1− 1
2−.
16. We start with the graph of = (Figure 13) and reflect about the -axis to get the graph of = −. Then shift the graphupward one unit to get the graph of = 1− . Finally, we stretch the graph vertically by a factor of 2 to obtain the graph of
= 2(1− ).
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
42 ¤ CHAPTER 1 FUNCTIONS AND MODELS
17. (a) To find the equation of the graph that results from shifting the graph of = 2 units downward, we subtract 2 from the
original function to get = − 2.
(b) To find the equation of the graph that results from shifting the graph of = 2 units to the right, we replace with − 2
in the original function to get = (−2).
(c) To find the equation of the graph that results from reflecting the graph of = about the -axis, we multiply the original
function by −1 to get = −.
(d) To find the equation of the graph that results from reflecting the graph of = about the -axis, we replace with − inthe original function to get = −.
(e) To find the equation of the graph that results from reflecting the graph of = about the -axis and then about the
-axis, we first multiply the original function by−1 (to get = −) and then replace with − in this equation toget = −−.
18. (a) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)and then shifting this graph 2 · 4 = 8 units upward. So the equation is = − + 8.
(b) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is = −(−4).
(b) The denominator is never equal to zero, so the function () =1 +
coshas domain R, or (−∞∞).
20. (a) The function () =√
10 − 100 has domain | 10 − 100 ≥ 0
= | 10 ≥ 102
= { | ≥ 2} = [2∞).
(b) The sine and exponential functions have domain R, so () = sin( − 1) also has domain R.
21. Use = with the points (1 6) and (3 24). 6 = 1 = 6
and 24 = 3 ⇒ 24 =
6
3 ⇒
4 = 2 ⇒ = 2 [since 0] and = 62
= 3. The function is () = 3 · 2.
22. Given the -intercept (0 2), we have = = 2. Using the point2 2
9
gives us 2
9= 22 ⇒ 1
9= 2 ⇒ = 1
3
[since 0]. The function is () = 2
13
or () = 2(3)−.
23. If () = 5, then(+ )− ()
=
5+ − 5
=
55 − 5
=
55 − 1
= 5
5 − 1
.
24. Suppose the month is February. Your payment on the 28th day would be 228−1 = 227 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
25. 1 m = 100 cm, (100) = 1002 cm = 10000 cm = 100 m.
(100) = 2100 cm = 2100(100 · 1000) km ≈ 127× 1025 km 1025 km.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
42 ¤ CHAPTER 1 FUNCTIONS AND MODELS
17. (a) To find the equation of the graph that results from shifting the graph of = 2 units downward, we subtract 2 from the
original function to get = − 2.
(b) To find the equation of the graph that results from shifting the graph of = 2 units to the right, we replace with − 2
in the original function to get = (−2).
(c) To find the equation of the graph that results from reflecting the graph of = about the -axis, we multiply the original
function by −1 to get = −.
(d) To find the equation of the graph that results from reflecting the graph of = about the -axis, we replace with − inthe original function to get = −.
(e) To find the equation of the graph that results from reflecting the graph of = about the -axis and then about the
-axis, we first multiply the original function by−1 (to get = −) and then replace with − in this equation toget = −−.
18. (a) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)and then shifting this graph 2 · 4 = 8 units upward. So the equation is = − + 8.
(b) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is = −(−4).
(b) The denominator is never equal to zero, so the function () =1 +
coshas domain R, or (−∞∞).
20. (a) The function () =√
10 − 100 has domain | 10 − 100 ≥ 0
= | 10 ≥ 102
= { | ≥ 2} = [2∞).
(b) The sine and exponential functions have domain R, so () = sin( − 1) also has domain R.
21. Use = with the points (1 6) and (3 24). 6 = 1 = 6
and 24 = 3 ⇒ 24 =
6
3 ⇒
4 = 2 ⇒ = 2 [since 0] and = 62
= 3. The function is () = 3 · 2.
22. Given the -intercept (0 2), we have = = 2. Using the point2 2
9
gives us 2
9= 22 ⇒ 1
9= 2 ⇒ = 1
3
[since 0]. The function is () = 2
13
or () = 2(3)−.
23. If () = 5, then(+ )− ()
=
5+ − 5
=
55 − 5
=
55 − 1
= 5
5 − 1
.
24. Suppose the month is February. Your payment on the 28th day would be 228−1 = 227 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
25. 1 m = 100 cm, (100) = 1002 cm = 10000 cm = 100 m.
(100) = 2100 cm = 2100(100 · 1000) km ≈ 127× 1025 km 1025 km.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
42 ¤ CHAPTER 1 FUNCTIONS AND MODELS
17. (a) To find the equation of the graph that results from shifting the graph of = 2 units downward, we subtract 2 from the
original function to get = − 2.
(b) To find the equation of the graph that results from shifting the graph of = 2 units to the right, we replace with − 2
in the original function to get = (−2).
(c) To find the equation of the graph that results from reflecting the graph of = about the -axis, we multiply the original
function by −1 to get = −.
(d) To find the equation of the graph that results from reflecting the graph of = about the -axis, we replace with − inthe original function to get = −.
(e) To find the equation of the graph that results from reflecting the graph of = about the -axis and then about the
-axis, we first multiply the original function by−1 (to get = −) and then replace with − in this equation toget = −−.
18. (a) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)and then shifting this graph 2 · 4 = 8 units upward. So the equation is = − + 8.
(b) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is = −(−4).
(b) The denominator is never equal to zero, so the function () =1 +
coshas domain R, or (−∞∞).
20. (a) The function () =√
10 − 100 has domain | 10 − 100 ≥ 0
= | 10 ≥ 102
= { | ≥ 2} = [2∞).
(b) The sine and exponential functions have domain R, so () = sin( − 1) also has domain R.
21. Use = with the points (1 6) and (3 24). 6 = 1 = 6
and 24 = 3 ⇒ 24 =
6
3 ⇒
4 = 2 ⇒ = 2 [since 0] and = 62
= 3. The function is () = 3 · 2.
22. Given the -intercept (0 2), we have = = 2. Using the point2 2
9
gives us 2
9= 22 ⇒ 1
9= 2 ⇒ = 1
3
[since 0]. The function is () = 2
13
or () = 2(3)−.
23. If () = 5, then(+ )− ()
=
5+ − 5
=
55 − 5
=
55 − 1
= 5
5 − 1
.
24. Suppose the month is February. Your payment on the 28th day would be 228−1 = 227 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
25. 1 m = 100 cm, (100) = 1002 cm = 10000 cm = 100 m.
(100) = 2100 cm = 2100(100 · 1000) km ≈ 127× 1025 km 1025 km.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
42 ¤ CHAPTER 1 FUNCTIONS AND MODELS
17. (a) To find the equation of the graph that results from shifting the graph of = 2 units downward, we subtract 2 from the
original function to get = − 2.
(b) To find the equation of the graph that results from shifting the graph of = 2 units to the right, we replace with − 2
in the original function to get = (−2).
(c) To find the equation of the graph that results from reflecting the graph of = about the -axis, we multiply the original
function by −1 to get = −.
(d) To find the equation of the graph that results from reflecting the graph of = about the -axis, we replace with − inthe original function to get = −.
(e) To find the equation of the graph that results from reflecting the graph of = about the -axis and then about the
-axis, we first multiply the original function by−1 (to get = −) and then replace with − in this equation toget = −−.
18. (a) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)and then shifting this graph 2 · 4 = 8 units upward. So the equation is = − + 8.
(b) This reflection consists of first reflecting the graph about the -axis (giving the graph with equation = −)
and then shifting this graph 2 · 2 = 4 units to the right. So the equation is = −(−4).
(b) The denominator is never equal to zero, so the function () =1 +
coshas domain R, or (−∞∞).
20. (a) The function () =√
10 − 100 has domain | 10 − 100 ≥ 0
= | 10 ≥ 102
= { | ≥ 2} = [2∞).
(b) The sine and exponential functions have domain R, so () = sin( − 1) also has domain R.
21. Use = with the points (1 6) and (3 24). 6 = 1 = 6
and 24 = 3 ⇒ 24 =
6
3 ⇒
4 = 2 ⇒ = 2 [since 0] and = 62
= 3. The function is () = 3 · 2.
22. Given the -intercept (0 2), we have = = 2. Using the point2 2
9
gives us 2
9= 22 ⇒ 1
9= 2 ⇒ = 1
3
[since 0]. The function is () = 2
13
or () = 2(3)−.
23. If () = 5, then(+ )− ()
=
5+ − 5
=
55 − 5
=
55 − 1
= 5
5 − 1
.
24. Suppose the month is February. Your payment on the 28th day would be 228−1 = 227 = 134,217,728 cents, or
$1,342,177.28. Clearly, the second method of payment results in a larger amount for any month.
25. 1 m = 100 cm, (100) = 1002 cm = 10000 cm = 100 m.
(100) = 2100 cm = 2100(100 · 1000) km ≈ 127× 1025 km 1025 km.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.