with base less than 1
1
3
and
1
10
are decreasing. The graph of
1
3
is the
reflection of that of 3 about the axis, and the graph of
1
10
is the reflection of
that of 10 about the axis. The graph of 10 increases more quickly than that of
3 for 0, and approaches 0 faster as → −∞.
8. Each of the graphs approaches ∞ as → −∞, and each approaches 0 as
→ ∞. The smaller the base, the faster the function grows as → −∞, and
the faster it approaches 0 as → ∞.
9. We start with the graph of = 3 (Figure 15) and shift
1 unit upward to get the graph of () = 3 + 1.
10. We start with the graph of =
1
2
(Figure 3) and stretch vertically by a factor of 2 to obtain the graph of = 2 1
2
. Then
we shift the graph 3 units downward to get the graph of () = 2 1
2
− 3.
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SECTION 1.4 EXPONENTIALFUNCTIONS
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45
11. We start with the graph of = (Figure 15) and reflect about the yaxis to get the graph of = − . Then we reflect the
graph about the xaxis to get the graph of = − − .
12. We start with the graph of = 4 (Figure 3) and shift
2 units to the left to get the graph of = 4 +2 .
13. We start with the graph of = (Figure 15) 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 =
1
2
−
and then reflect about the axis to get the graph
of = − 1
2
− . Finally, we shift the graph one unit upward to get the graph of = 1 − 1
2
− .
14. We start with the graph of = (Figure 15) and
reflect the portion of the graph in the first quadrant
about the axis to obtain the graph of = || .
15. (a) To find the equation of the graph that results from shifting the graph of = two 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 = two 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 xaxis, we multiply the original
function by −1 to get = − .
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46
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CHAPTER 1 FUNCTIONSAND MODELS
(d) To find the equation of the graph that results from reflecting the graph of = about the yaxis, 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 xaxis and then about the yaxis,
we first multiply the original function by −1 (to get = − ) and then replace with − in this equation to
get = − − .
16. (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) .
17. (a) The denominator is zero when 1 − 1−
2
= 0 ⇔ 1−
2
= 1 ⇔ 1 − 2 = 0 ⇔ = ±1. Thus,
the function () =
1 −
2
1 − 1− 2
has domain { | 6= ±1} = (−∞−1) ∪ (−11) ∪ (1∞).
(b) The denominator is never equal to zero, so the function () =
1 +
cos
has domain , or (−∞∞).
18. (a) The function () =
√ 10
− 100 has domain