Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
Exponential Functions
DEFINITION: An exponential function is a function of the form
f (x) = ax
where a is a positive constant.
1 x
x
10 1 −3
-3
= 103 = 1000
10 1 −2
-2
= 102 = 100
10 −1
1
-1
= 101 = 10
10 0
1
0
=1
10 1 1
1
1
= 10
= 0.1
10 1 2
1
2
= 102 = 0.01
10 3
1
3
= 1013 = 0.001
10
x
-3
-2
-1
0
1
2
3
10x
1
10−3 = 1013 = 1000
= 0.001
1
1
−2
10 = 102 = 100 = 0.01
1
10−1 = 10
= 0.1
0
10 = 1
101 = 10
102 = 100
103 = 1000
BASIC ALGEBRAIC PROPERTIES:
1. an = a
| · a ·{z. . . · a} if n is a positive integer.
n
factors
0
2. a = 1.
1
.
an
√
√
= q ap = ( q a)p .
3. a−n =
4. ap/q
THEOREM: If a > 0 and a 6= 1, then f (x) = ax is a continuous function with domain R and
range (0, ∞). In particular, ax > 0 for all x. If a, b > 0 and x, y ∈ R, then
1. ax+y = ax ay
2. ax−y =
ax
ay
3. (ax )y = axy
4. (ab)x = ax bx
BASIC CALCULUS PROPERTIES:
1. If a > 1, then lim ax = ∞ and lim ax = 0.
x→∞
x→−∞
2. If 0 < a < 1, then lim ax = 0 and lim ax = ∞.
x→∞
x→−∞
EXAMPLES:
1. Find lim (5x + 2x + 4).
x→−∞
Solution: Since lim 5x = 0 and lim 2x = 0, we have
x→−∞
x→−∞
lim (5x + 2x + 4) = lim 5x + lim 2x + lim 4 = 0 + 0 + 4 = 4
x→−∞
x→−∞
x→−∞
2. Find lim (0.1x + 0.7x ) and lim (0.1x + 0.7−x ).
x→∞
x→∞
1
x→−∞
Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
2. Find lim (0.1x + 0.7x ) and lim (0.1x + 0.7−x ).
x→∞
x→∞
Solution: Since lim 0.1x = 0 and lim 0.7x = 0, we have
x→∞
x→∞
lim (0.1x + 0.7x ) = lim 0.1x + lim 0.7x = 0 + 0 = 0
x→∞
x→∞
x→∞
Similarly, since lim 0.1x = 0 and lim 0.7−x = ∞, we have
x→∞
x→∞
lim (0.1x + 0.7−x ) = ∞
x→∞
The Number e and the Natural Exponential Function
It is known that
lim (1 + x)
1/x
x→0
x
1
= 2.7182818284590452353602874713526624977572470936...
= lim 1 +
x→±∞
x
We denote this number by e.
x
1
x
1+
x
1
1
2
10
1.110
100
1.01100
1000
1.0011000
10000 1.000110000
Value
2
2.593742460
2.704813829
2.716923932
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
lim ex = 0
lim ex = ∞
x→−∞
x→∞
So the x-axis is a horizontal asymptote of f (x) = ex .
EXAMPLES:
1. Find lim e−x
x→∞
3 −x+5
.
Solution: Put t = −x3 − x + 5. We know that t → −∞ as x → ∞. Therefore
lim e−x
x→∞
3 −x+5
= lim et = 0
t→−∞
2. Find lim− e2/(x−3) and lim+ e2/(x−3) .
x→3
x→3
2
Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
2. Find lim− e2/(x−3) and lim+ e2/(x−3) .
x→3
x→3
Solution: Put t = 2/(x − 3). We know that t → −∞ as x → 3− and t → ∞ as x → 3+ .
Therefore
lim− e2/(x−3) = lim et = 0
t→−∞
x→3
and
lim+ e2/(x−3) = lim et = ∞
t→∞
x→3
49x − 7x + 8
.
x→∞ 5 · 72x − 1
3. Find lim
Solution: Put t = 7x . We know that t → ∞ as x → ∞. Therefore
(7x )2 − 7x + 8
t2 − t + 8
49x − 7x + 8
=
lim
=
lim
= lim
lim
x→∞ 5 · (7x )2 − 1
t→∞ 5t2 − 1
t→∞
x→∞ 5 · 72x − 1
t2 −t+8
t2
5t2 −1
t2
= lim
1 − 1t +
t→∞
5 − t12
= lim
4. Find lim √
x→∞
ex
e2x + 1
.
5. Find lim e−2x sin x.
x→∞
3
t→∞
8
t2
=
t2
t2
−
5t2
t2
t
t2
+
−
1
t2
8
t2
1−0+0
1
=
5−0
5
Section 3.1 Exponential Functions
4. Find lim √
x→∞
ex
e2x + 1
2010 Kiryl Tsishchanka
.
Solution: Put t = ex . We know that t → ∞ as x → ∞. Therefore
ex
t
= lim √
lim √
= lim
x→∞
e2x + 1 t→∞ t2 + 1 t→∞
√t
t2
√
2
t +1
√
t2
t
= lim q t
t→∞
t2 +1
t2
= lim q
t→∞
t2
t2
1
+
1
t2
1
= lim q
t→∞
1+
5. Find lim e−2x sin x.
x→∞
Solution: Put t = −2x. We know that t → −∞ as x → ∞. Therefore
t
−2x
t
lim e
sin x = lim e sin −
x→∞
t→−∞
2
Note that
t
−1 ≤ sin −
2
≤1
Multiplying all three parts of this inequality by et , we get
t
t
t
−e ≤ e sin −
≤ et
2
Since
lim
t→−∞
it follows that
−et = lim et = 0
t→−∞
t
lim e sin −
t→−∞
2
t
=0
by the Squeeze Theorem. Thus
lim e−2x sin x = 0
x→∞
4
1
t2
=√
1
=1
1+0
Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
Appendix
EXAMPLE: Graph the following functions:
(a) f (x) = 2x−1
=⇒
y = 2x
f (x) = 2x−1 (horizontal shift)
(b) g(x) = 2x − 1
=⇒
y = 2x
g(x) = 2x − 1 (vertical shift)
(c) h(x) = −2x
=⇒
y = 2x
h(x) = −2x (reflection)
(d) p(x) = 2−x
=⇒
y = 2x
p(x) = 2−x (reflection)
x
1
1
REMARK: An other way to graph p(x) is to rewrite it as p(x) = x =
, which gives the
2
2
same result by the Figure on page 1.
5
Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
EXAMPLE: Graph f (x) = −2−x .
Solution: Note that
f (x) = −2
−x
1
=− x =−
2
x
1
2
Therefore
=⇒
x
1
y=
2
x
1
f (x) = −
(reflection)
2
x+1
1
+ 2.
EXAMPLE: Graph f (x) =
3
Solution: We have
=⇒
x
1
y=
3
x+1
1
y=
(horizontal shift)
3
⇓
x+1
1
+ 2 (vertical shift)
f (x) =
3
6
Section 3.1 Exponential Functions
2010 Kiryl Tsishchanka
EXAMPLE: Graph f (x) = −e1−x + 3.
Solution: We have
=⇒
y = 2x
y = −2x (reflection)
⇓
y = −2x+1 (horizontal shift)
⇓
y = −2x+1 + 3 (vertical shift)
⇓
f (x) = −2−x+1 + 3 (reflection)
7