Download 5.1-Exponential_Functions

Exponential functions
LRT
03/24/2017
The exponential functions discussed in section 5.1 are
exponential functions with a constant base, b.
The exponential function with base b is the function which
assigns to x the number denoted bx . You know some values
of this function already.
1
For example b2 = b · b, b 2 =
√
1
b, b−1 = .
b
If b is to be the base, then b must be positive since there is
no square root if b < 0 and no b−1 if b = 0.
Using the Laws of Exponents we can determine many other
values.
The laws are
Let x and y be any real numbers and let a and b be any
positive real numbers. Then,
1. bx · by = bx+y
bx
2.
= bx−y
y
b
y
3.
bx = bx·y
4. (a · b)x = ax · bx
a x ax
5.
= x
b
b
6.
b0 = 1
There are many consequence of these Laws. In particular,
from 1 we see that for any positive integer n, bn = b| ·{z
· · }b.
n times
From 5 and 2 we see that for any negative integer n,
√
1
1
1
bn = · · · . If n is a positive integer, b n = n b. If,
|b {z b}
|n| times
m
additionally, m is any integer b n =
√
m
n
b .
The takeaway from this is that we already know the value
of bx for any rational number x. The function bx is the
unique continuous function which has these values at the
rational numbers.
The book requires b 6= 1 but 1x is defined for all x: the
value is 1.
You need b= 1 if you want the Laws to hold as stated.
a x ax
2x
= x with a = b = 2. It says x = 1x .
Look at 5.
b
b
2
The left-hand side is defined, but if you don’t have 1x you
don’t have the right-hand side!
y = 2−x
y = 2x
y = 1x
These graphs reflect the behavior of exponential functions
with any base.
The exponential function y = bx for b > 0 has the following
properties.
1. Its domain is (−∞, ∞).
2. If b 6= 1 its range is (0, ∞). For all b, bx > 0.
3. Its graph passes through (0, 1).
4. It is continuous on (−∞, ∞).
5. If b > 1 it is increasing; if b < 1 it is decreasing.
6. If b 6= 1 it is concave up.
7. If b > 1 lim bx = 0; lim bx = ∞.
x→−∞
x→∞
If b < 1 lim bx = ∞; lim bx = 0.
x→−∞
x→∞
If b 6= 1 0 is a horizontal asymptote in one direction
but not the other.
(1) says that you can
plug any number you
like into bx . (2) says
that if b 6= 1 you can always solve the equation
bx = a for any positive
number a.
(5) says the solution is unique.
(2) also says you can never solve the equation bx = 0 or
bx = a for a < 0.
Since bx is continuous, the technique we discussed on day 1
for solve inequalities works if the inequalities involve
exponential functions, provided you can solve equations
involving exponential function.Some useful special cases of
equalities and inequalities:
I If b 6= 1 and bx = by , x = y.
I If b > 1 then x < y if and only if bx < by .
I If b < 1 then x < y if and only if bx > by .
Example
2
Solve 2x = 4x . First make the bases the same. The base of
the exponential function on the left is 2 but on the right it
is 4. Exponential functions with different bases are difficult
x
to manipulate so write 4 = 22 and so 4x = 22 = 22x so
2
we need to solve 2x = 22x . Divide both sides by 22x so
2
2x
2
= 1 or 2x −2x = 1. Hence the exponent must be 0 so
2x
2
x2 − 2x = 0 and x(x − 2) = 0 so x = 0 or x = 2.
2
2
Where is 2x > 4x ? Let g(x) = 2x − 4x and solve g(x) = 0.
We just did this and saw x = 0 or x = 2.
2
0
1
> 0; g(1) = −2 < 0 and
4
g(3) = 29 − (22 )3 = 29 − 26 > 0 so
g(−1) = 2 −
-
+
•
−1
2
+
•
0
so 2x > 4x on (−∞, 0) ∪ (2, ∞).
1
•
2
3
There is a number defined by
m
1
e = lim 1 +
m→∞
m
Its value is about e ≈ 2.7182818284590452353602875.
Suppose you know that some function f has the form
f (t) = Aekt for some (currently unknown) constants A and
k. Suppose you know f (0) = 10 and f (2) = 40. Find a
formula for f (t).
Since f (0) = 10, 10 = Aek·0 = Ae0 = A.
Since f (2) = 40, 40 = 10ek·2 = 10e2k .
√
40
Hence e2k =
= 4. ek = 4 = 2.
10
t
ekt = ek = 2t .
f (t) = 10 · 2t
#23
The picture referred to below is the picture for problem
#23 in the book.
To solve a problem like this it is necessary to write all
exponentials to a common base.
In this case the two bases are 8 and
1
.
32
You need to see 8 = 23 and 32 = 25 to proceed.
Then 8x = (23 )x = 23x and
x−2
x−2
1
= 32−1
= 32(−1)(x−2) = (25 )2−x = 210−5x .
32
Solve 23x = 210−5x , 3x = 10 − 5x, 8x = 10, x =
5
.
4