1 Definition and Properties of the Exp Function - UH
[Pages:10]Lecture 4Section 7.4 The Exponential Function Section 7.5
Arbitrary Powers; Other Bases
Jiwen He
1 Definition and Properties of the Exp Function
1.1 Definition of the Exp Function
Number e
Definition 1. The number e is defined by ln e = 1
i.e., the unique number at which ln x = 1. Remark Let L(x) = ln x and E(x) = ex for x rational. Then
L E(x) = ln ex = x ln e = x, i.e., E(x) is the inverse of L(x). ex: Inverse of ln x
1
Definition 2. The exp function E(x) = ex is the inverse of the log function
L(x) = ln x:
L E(x) = ln ex = x, x.
Properties ? ln x is the inverse of ex: x > 0, E L = eln x = x. ? x > 0, y = ln x ey = x.
? graph(ex) is the reflection of graph(ln x) by line y = x.
? range(E) = domain(L) = (0, ), domain(E) = range(L) = (-, ).
? lim ex = 0 lim ln x = -, lim ex = lim ln x =
x-
.
x0+
x
x
1.2 Properties of the Exp Function
Algebraic Property Lemma 3. ? ex+y = ex ? ey.
?
e-x =
1 ex .
?
ex-y =
ex ey .
? erx = (ex)r, r rational.
2
Proof ln ex+y = x + y = ln ex + ln ey = ln (ex ? ey) .
Since ln x is one-to-one, then ex+y = ex ? ey.
1 = e0 = ex+(-x) = ex ? e-x
e-x
=
1 ex .
ex-y
= ex+(-y)
= ex ? e-y
= ex ?
1 ey
=
ex ey .
m
m
? For r = m N, emx = ex + ? ? ? + x = ex ? ? ? ex = (ex)m.
?
For
r
=
1 n
,
nN
and
n = 0,
ex
=
e
n n
x
=
e
1 n
x
n
e
1 n
x
=
(ex
)
1 n
.
?
For r
e
1 n
x
rational, let
m
=
(ex
)
1 n
r
m
= =
m n
,
m,
n
(ex
)
m n
=
N and (ex )r .
n
=
0.
Then
erx
=
e
m n
x
=
Derivatives
Lemma 4.
? d ex = ex dx
ex dx = ex + C.
?
dm dxm
ex
=
ex
>
0
E(x) = ex is concave up, increasing, and positive.
Proof Since E(x) = ex is the inverse of L(x) = ln x, then with y = ex,
d ex = E (x) = 1 = 1
dx
L (y) (ln y)
=
1
1
=
y
=
ex.
y
First, for m = 1, it is true. Next, assume that it is true for k, then
dk+1 dxk+1
ex
=
d dx
dk dxk
ex
= d (ex) = ex. dx
By the axiom of induction, it is true for all positive integer m.
3
1.3 Another Definition of the Exp Function
ex: as the series
xk k=0 k!
Definition 5. (Section 11.5)
ex =
xk
x2 x3 = 1+x+ + +???
k!
2! 3!
k=0
n xk
= lim
, x R.
n
k!
k=0
(k! = 1 ? 2 ? ? ? k)
Number e
1
11
1
n1
? e=
=1+ + +
+ ? ? ? = lim
.
k!
1 1?2 1?2?3
n
k!
k=0
k=0
? e 2.71828182845904523536 . . .
4
Limit:
limx
ex xn
Theorem 6.
ex
lim
x
xn
=
,
n N.
Proof. ? Recall that
ex =
xk
x
=1+ +
x2
+
x3
+??? .
k!
1 1?2 1?2?3
k=0
? For large x > 0,
ex > xp p!
ex xp-n
xn >
. p!
?
For
p > n,
lim xp-n
x
= ,
then
ex
lim
x
xn
= .
Quiz Quiz
1. domain of ln 1 + x2 :
(a) x > 1, (b) x > -1, (c) any x.
2. domain of ln x 4 + x2 : (a) x = 0, (b) x > 0, (c) any x.
2 Differentiation and Graphing
2.1 Chain Rule
Differentiation: Chain Rule
Lemma 7. d eu = eu du .
dx
dx
Proof By the chain rule,
d eu = d (eu) du = eu du dx du dx dx
Examples 8.
? d ekx = ekx ? k = kekx. dx
?
d
e x =e x?
d
x
=
ex
?
1
ex
y=
dx
dx
2x 2x
? d e-x2 = e-x2 d -x2 = e-x2 (-2x) = -2xe-x2 .
dx
dx
5
Examples: Chain Rule Examples 9. ? d e4 ln x.
dx ? d esin 2x.
dx ? d ln cos e2x .
dx
Solution Simplify it before the differentiation:
e4 ln x = eln x 4 = x4 d e4 ln x = d x4 = 4x3.
dx
dx
By the chain rule,
d esin 2x = esin 2x d sin 2x = esin 2x ? 2 cos 2x
dx
dx
By the chain rule,
d ln
dx
cos e2x
1 = cos e2x ?
- sin e2x
? d e2x = -2e2x tan e2x. dx
2.2 Graphing
Graph
of
f (x)
=
e-
x2 2
6
Example 10. Let
f (x)
=
e-
x2 2
.
Determine the symmetry of graph and
asymptotes. On what intervals does f increase? Decrease? Find the extrem
values of f .Determine the concavity and inflection points.
Solution
7
Since
f (-x)
=
e-
(-x)2 2
=
e-
x2 2
= f (x) and
lim
e-
(-x)2 2
= 0, the graph is
x?
symmetry w.r.t. the y-axis, and the x-axis is a horizontal asymptote.
? We have
f
(x)
=
e-
x2 2
(-x)
=
-xe-
x2 2
.
? Thus f on (-, 0) and on (0, ).
? At x = 0, f (x) = 0. Thus is the (only) local and absolute maximum.
f (0) = e0 = 1
?
From f
(x)
=
-xe-
x2 2
,
we
have
f
(x)
=
-e-
x2 2
+
x2e-
x2 2
=
(x2
-
1)e-
x2 2
.
? At x = ?1, f (x) = 0. Then, the graph is concave up on (-, -1) and (1, ); the graph is concave down on (-1, 1).
? The points are points of inflection.
(?1,
f
(?1))
=
(?1,
e-
1 2
)
Quiz (cont.) Quiz (cont.)
d 3. (ln |x|) =? :
dx
1 (a) ,
x
1 (b) ,
|x|
1 (c) - .
x
4. x-1 dx =? : (a) ln x + C, (b) ln |x| + C, (c) x-1 + C.
3 Integration
3.1 u-Substitution
Integration: u-Substitution Theorem 11.
eg(x)g (x) dx = eg(x) + C.
Proof. Let u = g(x), thus du = g (x)dx, then
eg(x)g (x) dx = eudu = eu + C = eg(x) + C.
Example 12. Calculate
xe-
x2 2
dx.
Let
u
=
-
x2 2
,
thus
du
=
-xdx,
then
xe-
x2 2
dx
=
-
eudu
=
-eu
+C
=
-e-
x2 2
+ C.
8
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