Section 4.2 The Natural Exponential Function
It is known that
(
)x
1
1+
→ 2.7182818284590452353602874713526624977572470936...
x
as x → ±∞. We denote this number by e.
(
)x
1
1+
x
Value
x
1
21
2
10
1.110
2.593742460
100
1.01100
2.704813829
1000
1000
1.001
2.716923932
10000 1.000110000 2.718145927
DEFINITION: The natural exponential function is f (x) = ex .
PROPERTIES OF THE NATURAL EXPONENTIAL FUNCTION: The exponential function
f (x) = ex is a continuous function with domain R and range (0, ∞). Thus ex > 0 for all x. Also
ex → 0 as x → −∞
and
ex → ∞ as x → ∞
So the x-axis is a horizontal asymptote of f (x) = ex .
EXAMPLE: Sketch the graph of each function.
(a) f (x) = e−x
(b) g(x) = 3e0.5x
Solution:
(a) We start with the graph of y = ex and reflect in the y-axis to obtain the graph of y = e−x .
(b) We calculate several values, plot the resulting points, then connect the points with a smooth
curve.
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Continuous Compounded Interest
If $100 is invested at 2% interest, compounded annually, then after 1 year the investment
is worth
$100(1 + 0.02) = $102
If $100 is invested at 2% interest, compounded semiannually, then after 1 year the investment is worth
(
)
0.02
$100 1 +
= $101 (after first 6 months)
2
(
)
0.02
$101 1 +
= $102.01 (after 1 year)
2
The same result can be obtained in a more elegant way:
(
)
0.02
$100 1 +
= $101 (after first 6 months)
2
(
(
)(
)
)2
0.02
0.02
0.02
$100 1 +
1+
= $100 1 +
= $102.01 (after 1 year)
2
2
2
If $100 is invested at 2% interest, compounded quarterly, then after 1 year the investment
is worth
(
)
0.02
$100 1 +
= $100.5 (after first 3 months)
4
(
)
0.02
$100.5 1 +
= $101.0025 (after first 6 months)
4
(
)
0.02
$101.0025 1 +
= $101.5075125 (after first 9 months)
4
(
)
0.02
$101.5075125 1 +
= $102.0150500625 (after 1 year)
4
As before, the same result can be obtained in a more elegant way:
)
(
0.02
= $100.5 (after first 3 months)
$100 1 +
4
(
)(
)
(
)2
0.02
0.02
0.02
$100 1 +
1+
= $100 1 +
= $101.0025 (after first 6 months)
4
4
4
)2 (
)
(
)3
(
0.02
0.02
0.02
$100 1 +
1+
= $100 1 +
= $101.5075125 (after first 9 months)
4
4
4
(
(
)3 (
)
)4
0.02
0.02
0.02
1+
= $100 1 +
= $102.0150500625 (after 1 year)
$100 1 +
4
4
4
In general, if we invest A0 dollars at interest r, compounded n times a year, then after 1 year
the investment is worth
(
r )n
A0 1 +
dollars
n
Moreover, after t years the investment is worth
(
r )nt
A(t) = A0 1 +
(6)
n
2
QUESTION: What happens if n → ∞?
Answer: We have
(
A(t) = lim A0
n→∞
[
= A0
[(
]rt
[
]rt
(
r )nt
r )n/r
r )n/r
1+
= lim A0 1 +
= A0 lim 1 +
n→∞
n→∞
n
n
n
(
lim
n→∞
1
1+
n/r
)n/r ]rt
[
= A0
(
lim
m→∞
1
1+
m
)m ]rt
= A0 ert
so
A(t) = A0 ert
(7)
EXAMPLE: If $100 is invested at 2% interest, compounded continuously, then after 1 year
the investment is worth
A(1) = $100e0.02·1 ≈ $102.02
EXAMPLE: If $200, 000 is borrowed at 5.5% interest, find the amounts due at the end of 30
years if the interest compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) continuously.
Solution:
(i) By (6) we have
(
A(30) = A0
(
)1·30
0.055
r )n·30
= $200, 000 1 +
1+
≈ $996, 790.26
n
1
(ii) By (6) we have
(
A(30) = A0
(
)4·30
r )n·30
0.055
1+
= $200, 000 1 +
≈ $1, 029, 755.36
n
4
which gives ≈ $32, 965.10 difference between (ii) and (i).
(iii) By (6) we have
(
A(30) = A0
)12·30
(
r )n·30
0.055
1+
= $200, 000 1 +
≈ $1, 037, 477.57
n
12
which gives ≈ $7, 722.21 difference between (iii) and (ii).
(iv) By (7) we have
A(30) = A0 er·30 = $200, 000e0.055·30 ≈ $1, 041, 395.97
which gives ≈ $3, 918.38 difference between (iv) and (iii).
EXAMPLE: If $200, 000 is borrowed at 5.6% interest, find the amounts due at the end of 30
years if the interest compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) continuously.
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EXAMPLE: If $200, 000 is borrowed at 5.6% interest, find the amounts due at the end of 30
years if the interest compounded (i) annually, (ii) quarterly, (iii) monthly, (iv) continuously.
Solution:
(i) By (6) we have
(
A(30) = A0
(
)1·30
r )n·30
0.056
1+
= $200, 000 1 +
≈ $1, 025, 528.05
n
1
which gives ≈ $28, 737.79 difference between 5.6% and 5.5%.
(ii) By (6) we have
(
A(30) = A0
(
)4·30
r )n·30
0.056
1+
= $200, 000 1 +
≈ $1, 060, 680.53
n
4
which gives ≈ $35, 152.48 difference between (ii) and (i).
(iii) By (6) we have
(
A(30) = A0
(
)12·30
0.056
r )n·30
= $200, 000 1 +
≈ $1, 068, 925.95
1+
n
12
which gives ≈ $8, 245.43 difference between (iii) and (ii).
(iv) By (7) we have
A(30) = A0 er·30 = $200, 000e0.056·30 ≈ $1, 073, 111.19
which gives ≈ $4, 185.24 difference between (iv) and (iii).
4