Unformatted text preview: 9
The Derivative Limits Onesided Limits and Continuity The Derivative Basic Rules of Differentiation The Product and Quotient Rules:
The HigherOrder Derivatives The Chain Rule Differentiation of Exponential and
Differentiation
Logarithmic Functions
Logarithmic Marginal Functions in Economics 9.1
9.1
Limits
y
400 f ( x) 300
200
100
10 20 30 40 50 60 x Introduction to Calculus
Introduction Historically, the development of calculus by Isaac Newton
Historically,
Isaac and Gottfried W. Leibniz resulted from the investigation
Gottfried
of the following problems:
of
1. Finding the tangent line to a curve at a given point on the
Finding
tangent
curve:
curve:
y
T t Introduction to Calculus
Introduction Historically, the development of calculus by Isaac Newton
Historically,
Isaac and Gottfried W. Leibniz resulted from the investigation
Gottfried
of the following problems:
of
1. Finding the area of a planar region bounded by an
Finding
area
arbitrary curve.
arbitrary
y R t Introduction to Calculus
Introduction The study of the tangentline problem led to the
The
tangentline creation of differential calculus, which relies on the
differential
which
concept of the derivative of a function.
derivative The study of the area problem led to the creation of
The
area
integral calculus, which relies on the concept of the
integral
which
antiderivative, or integral, of a function.
antiderivative or integral Example: A Speeding Maglev
Example: From data obtained in a test run conducted on a
From
test prototype of maglev, which moves along a straight
prototype
which
monorail track, engineers have determined that the
position of the maglev (in feet) from the origin at time t
position
from
time
is given by
is
s = f(t) = 4t2
(0 ≤ t ≤ 30)
30) f is called the position function of the maglev.
position The position of the maglev at time t = 0, 1, 2, 3, … , 10 is
0,
The position f(0) = 0
(0) f(1) = 4
(1) f(2) = 16
(2) f(3) = 36 … f(10) = 400
(3) But what if we want to find the velocity of the maglev at
But
velocity any given point in time?
point Example: A Speeding Maglev
Example: Say we want to find the velocity of the maglev at t = 2.
Say
velocity We may compute the average velocity of the maglev
We
average over an interval of time, such as [2, 4] as follows:
interval
such [2,
Distance covered f (4) − f (2)
=
Time elapsed
4−2
4(42 ) − 4(22 )
=
2
64 − 16
=
2
= 24 or 24 feet/second.
or 24 This is not the velocity of the maglev at exactly t = 2,
This not
exactly
but it is a useful approximation.
approximation Example: A Speeding Maglev
Example: We can find a better approximation by choosing a
We
better smaller interval to compute the speed, say [2, 3].
smaller
[2, More generally, let t > 2. Then, the average velocity of
let
Then,
average
the maglev over the time interval [2, t] is given by
time
Distance covered f (t ) − f (2)
=
Time elapsed
t−2
4(t 2 ) − 4(22 )
=
t−2
4(t 2 − 4)
=
t−2 Example: A Speeding Maglev
Example:
4(t 2 − 4)
Average velocity =
t−2 By choosing the values of t closer and closer to 2, we
By
closer
we
obtain average velocities of the maglev over smaller and
average
smaller time intervals.
smaller The smaller the time interval, the closer the average
The smaller
time
the closer
velocity becomes to the instantaneous velocity of the train
instantaneous
at t = 2, as the table below demonstrates:
t 2.5 2.1 2.01 2.001 2.0001 Average Velocity 18 16.4 16.04 16.004 16.0004 The closer t gets to 2, the closer the average velocity gets
The closer
the closer
average to 16 feet/second.
16 Thus, the instantaneous velocity at t = 2 seems to be
Thus,
instantaneous
16 feet/second.
16 Intuitive Definition of a Limit
Intuitive Consider the function g, which gives the average velocity
Consider
g,
average of the maglev:
of 4(t 2 − 4)
g (t ) =
t −2 Suppose we want to find the value that g(t) approaches
Suppose
approaches
as t approaches 2.
approaches
✦ We take values of t approaching 2 from the right
We
approaching from
(as we did before), and we find that g(t) approaches
approaches
16:
16
t 2.5 2.1 2.01 2.001 2.0001 g (t ) 18 16.4 16.04 16.004 16.0004 ✦ Similarly, we take values of t approaching 2 from the left,
Similarly,
approaching from
and we find that g(t) also approaches 16:
approaches 16
t 1.5 1.9 1.99 1.999 1.9999 g(t) 14 15.6 15.96 15.996 15.9996 Intuitive Definition of a Limit
Intuitive We have found that as t approaches 2 from either side,
We
approaches from g(t) approaches 16.
approaches 16 In this situation, we say that the limit of g(t) as t
In
the
approaches 2 is 16.
approaches
16 This is written as
4(t 2 − 4)
lim g (t ) = lim
= 16
t →2
t →2
t −2 Observe that t = 2 is not in the domain of g(t) .
Observe
not But this does not matter, since t = 2 does not play any
But
does
since role in computing this limit.
role Limit of a Function
Limit The function f has a limit L as x approaches a, written
The function
limit
approaches lim f ( x ) = L
x→a
if the value of f(x) can be made as close to the number
if
close
L as we please by taking x values sufficiently close to
values sufficiently
(but not equal to) a.
not Examples
Examples Let f(x) = x3. Evaluate
Let
Evaluate lim f ( x ).
x→2 Solution
Solution You can see in the graph
You
that f(x) can be as close
can
close
to 8 as we please by taking
x sufficiently close to 2.
sufficiently Therefore,
Therefore, f( x ) = x y
8 3 6
4
2 lim x 3 = 8
x→2 –2 –1 1
–2 2 3 x Examples
Examples
x + 2
1 Let g ( x ) = Let if x ≠ 1
if x = 1 Evaluate lim g ( x ).
x→1 Solution
Solution You can see in the graph that
You
g(x) can be as close
can
close
to 3 as we please by taking
x sufficiently close to 1.
sufficiently Therefore,
Therefore, y g(x) 5
3
1 lim g ( x ) = 3
x→1 –2 –1 1 2 3 x Examples
Examples Let f ( x ) =
Let 1
x2 Evaluate lim f ( x ). Solution
Solution The graph shows us that as x
The
approaches 0 from either
approaches
side, f(x) increases without
side
bound and thus does not
bound
not
approach any specific real
number.
number. Thus, the limit of f(x) does
Thus,
not exist as x approaches 0.
not
approaches x→0 y
5 f ( x) = –2 –1 1 2 1
x2 x Theorem 1
Theorem
Properties of Limits lim f ( x ) = L
x→a Suppose
Then,
1. lim [
x→a and lim g ( x ) = M x→a r f ( x )] = lim f ( x ) = Lr x →a r r, a real number 2. lim cf ( x ) = c lim f ( x ) = cL
2. x→a
x →a
3. lim
x →a [ f ( x ) ± g ( x )] = lim
x→a 4. lim
4. x→a [ f ( x ) g ( x )] = lim x→a 5.
5. f ( x ) ± lim g ( x ) = L ± M
x→a f ( x ) lim g ( x ) = LM x→a f ( x ) lim f ( x ) L
lim
= x→a
=
x →a g ( x )
lim g ( x ) M
x→a c, a real number Provided that M ≠ 0
Provided Examples
Examples Use theorem 1 to evaluate the following limits:
Use theorem
3 lim x = lim x = 23 = 8
x→2
3 x→2 3/2 lim 5x = 5 lim x = 5(4)3/2 = 40
x→4
3/2 x→4 4 lim (5x − 2) = 5 lim x − lim 2 = 5(1)4 − 2 = 3
x→1
4 x→1 x→1 Examples
Examples Use theorem 1 to evaluate the following limits:
Use theorem lim 2x
x →3 x + 7 = 2 lim x x→3 3 2 3 lim x 2 + 7 = 2(3)3 (3)2 + 7 = 216
x→3 lim (2 x 2 + 1) 2(2)2 + 1 9
2x + 1
= x→2
=
= =3
lim
x→2
lim ( x + 1)
2 +1
3
x +1
x→2
2 Indeterminate Forms
Indeterminate
4( x 2 − 4) Let’s consider lim
x →2 x − 2
which we evaluated earlier for the maglev example by
which
maglev
looking at values for x near x = 2.
near If we attempt to evaluate this expression by applying
If
Property 5 of limits, we get
Property lim 4( x 2 − 4) 0
4( x − 4) x→2
lim
=
=
x →2 x − 2
lim x − 2
0
x →2
2 In this case we say that the limit of the quotient f(x)/g(x) as
In
quotient x approaches 2 has the indeterminate form 0/0.
approaches
0/0 This expression does not provide us with a solution to our
This
not
solution
problem.
problem. Strategy for Evaluating Indeterminate Forms
Strategy 1. Replace the given function with an appropriate one that takes on the same values as the
same
original function everywhere except at x = a.
everywhere except
2. Evaluate the limit of this function as x
approaches a.
approaches Examples
Examples
4( x 2 − 4) Evaluate lim
x →2 x − 2 Solution As we’ve seen, here we have an indeterminate form 0/0.
As
indeterminate
0/0 We can rewrite 4( x − 2)( x + 2)
We 4( x 2 − 4)
=
= 4( x + 2)
x−2
x−2
x≠2 Thus, we can x 2 − 4)
4( say that lim
x →2 x−2 = lim 4( x + 2) = 16
x →2 Note that 16 is the same value we obtained for the maglev
Note
16 example through approximation.
example
approximation Examples
Examples
4( x 2 − 4) Evaluate lim
x →2 x − 2 Solution Notice in the graphs below that the two functions yield the
Notice
same graphs, except for the value x = 2:
same
except
4( x 2 − 4)
f ( x) =
x−2 y
20 20 16 8 4
–1 12 8 –2 16 12 –3 g ( x ) = 4( x + 2) y 4
1 2 3 x –3 –2 –1 1 2 3 x Examples
Examples Evaluate 1+ h −1
lim
h→0
h Solution As we’ve seen, here we have an indeterminate form 0/0.
As
indeterminate
0/0 We can rewrite (with the constraint that h ≠ 0):
We
constraint
1+ h −1
1+ h −1 1+ h +1
h
1
=
⋅
=
=
h
h
1 + h + 1 h( 1 + h + 1)
1+ h +1 Thus, we can say that 1+ h −1
1
1
1
lim
= lim
=
=
h→0
h→0 1 + h + 1
h
1 +1 2 Limits at Infinity
Limits There are occasions when we want to know whether f(x)
There approaches a unique number as x increases without
approaches unique
bound.
bound In the graph below, as x increases without bound, f(x)
In
without
approaches the number 400.
400 We call the line y = 400
We
y
a horizontal asymptote.
horizontal
400 In this case, we can say
In
f ( x)
that
that
300
lim f ( x ) = 400
x→∞ and we call this a limit
and
limit
of a function at infinity. 200
100 10 20 30 40 50 60 x Example
Example
2x2 Consider the function f ( x ) =
1 + x2 Determine what happens to f(x) as x gets larger and larger.
Determine
larger
Solution We can pick a sequence of values of x and substitute them
We
sequence in the function to obtain the following values:
in
x 1 2 5 10 100 1000 f(x) 1 1.6 1.92 1.98 1.9998 1.999998 As x gets larger and larger, f(x) gets closer and closer to 2.
As
larger
closer Thus, we can say that
2x2 lim x→∞ 1 + x 2 =2 Limit of a Function at Infinity
Limit The function f has the limit L as x increases without
The
limit bound (as x approaches infinity), written
bound
approaches lim f ( x ) = L x→∞ if f(x) can be made arbitrarily close to L by taking x
if
close
large enough.
large Similarly, the function f has the limit M as x decreases
Similarly,
limit
without bound (as x approaches negative infinity),
without
approaches
),
written
written
lim f ( x ) = M
x→−∞ if f(x) can be made arbitrarily close to M by taking x
if
close
large enough in absolute value.
large Examples
Examples −1 Let f ( x ) = Let
1 Evaluate if x < 0
if x ≥ 0 lim f ( x ) lim f ( x ) and
x→∞ x→−∞ Solution Graphing f(x) reveals that
Graphing y lim f ( x ) = 1 x→∞ lim f ( x ) = −1
x→−∞ f ( x) 1
–3 3
–1 x Examples
Examples Let g ( x ) =
Let Evaluate 1
x2 lim g ( x ) and lim g ( x ) x→∞ x→−∞ Solution Graphing g(x) reveals that
Graphing y 1
g ( x) = 2
x lim g ( x ) = 0
x→∞
lim g ( x ) = 0
x→−∞
–3 –2 –1 1 2 3 x Theorem 2
Theorem
Properties of Limits All properties of limits listed in Theorem 1 are valid when a is replaced by ∞ or –∞ . In addition, we have the following properties for
limits to infinity: For all n > 0,
For 1
lim
=0
x→∞ x n and 1
provided that n is defined.
x 1
lim
=0
x→−∞ x n Examples
Examples
x2 − x + 3 Evaluate lim
Evaluate x→∞
2 x3 + 1 Solution
Solution The limits of both the numerator and denominator do not
The limits
numerator
denominator
exist as x approaches infinity, so property 5 is not
exist
property
not
applicable.
applicable. We can find the solution instead by dividing numerator
We
dividing
and denominator by x3:
and
by
11
3
− 2+ 3
( x 2 − x + 3) / x 3
xx
x = 0−0+0 = 0 = 0
lim
= lim
x→∞ (2 x 3 + 1) / x 3
x→∞
1
2+0
2
2+ 3
x Examples
Examples
3x 2 + 8 x − 4 Evaluate lim
Evaluate x→∞
2 x2 + 4 x − 5 Solution
Solution Again, we see that property 5 does not apply.
Again,
property
not So we divide numerator and denominator by x2:
So
divide
and
by
84
−2
(3x + 8 x − 4) / x
x x = 3+ 0−0 = 3
lim 2
= lim
x→∞ (2 x + 4 x − 5) / x 2
x→∞
45
2+ − 2 2+0−0 2
xx
2 2 3+ Examples
Examples
2 x 3 − 3x 2 + 1 Evaluate lim
Evaluate x→∞ 2
x + 2x + 4 Solution
Solution Again, we see that property 5 does not apply.
Again,
property
not But dividing numerator and denominator by x2 does not
But dividing
and
by
not
1
help in this case:
help
2x − 3 + 2
3
2
2 lim
x→∞ (2 x − 3x + 1) / x
x
= lim
x→∞
24
( x 2 + 2 x + 4) / x 2
1+ + 2
xx In other words, the limit does not exist.
In
the We indicate this by writing 2
3 lim
x→∞ 2 x − 3x + 1
=∞
2
x + 2x + 4 9.2
9.2
OneSided Limits and Continuity y a b c d x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 Its graph shows that
Its f does not have a
does not
limit as x approaches
limit
zero, because
zero because
approaching from each
side results in different
values.
values y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 If we restrict x to be greater
If
restrict
greater than zero (to the right of zero),
than
we see that f(x) approaches 1 as
approaches
close as we please as x
approaches 0.
approaches In this case we say that the
In
righthand limit of f as x
righthand
approaches 0 is 1, written
approaches lim f ( x ) = 1 x→0+ y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided
x −1
x +1 Consider the function f ( x ) = if x < 0
if x ≥ 0 Similarly, if we restrict x to be
Similarly,
restrict less than zero (to the left of zero),
less
we see that f(x) approaches –1
approaches –1
as close as we please as
x approaches 0.
approaches In this case we say that the
In
lefthand limit of f as x
lefthand
approaches 0 is – 1, written
approaches lim f ( x ) = −1 x→0− y f ( x)
1
–1 1
–1 x OneSided Limits
OneSided The function f has the righthand limit L as x
The
righthand approaches from the right, written
approaches lim f ( x ) = L x→a + if the values of f(x) can be made as close to L as we
if
please by taking x sufficiently close to (but not equal
to) a and to the right of a.
to Similarly, the function f has the lefthand limit L as x
Similarly,
lefthand
approaches from the left, written
approaches lim f ( x ) = L x→a − if the values of f(x) can be made as close to L as we
if
please by taking x sufficiently close to (but not equal
to) a and to the left of a.
to Theorem 3
Theorem
Properties of Limits The connection between onesided limit and the twosided limit defined earlier is given by the following theorem. Let f be a function that is defined for all values of x
Let close to x = a with the possible exception of a itself.
Then
Then lim f ( x ) = L
x →a if and only if lim f ( x ) = lim f ( x ) = L x→a − x→a + Examples Show that lim f ( x ) exists by studying the onesided
Show
onesided
x →0 limits of f as x approaches 0:
limits
approaches
x f ( x) = −x Solution
Solution For x > 0, we find
For if x > 0
if x ≤ 0
y lim f ( x ) = 0 x→0+ And for x ≤ 0, we find
And lim f ( x ) = 0 1 x→0− Thus,
Thus, lim f ( x ) = 0
x →0 f ( x) 2 –2 –1 1 2 x Examples
Examples Show that lim g ( x ) does not exist.
Show
x →0 −1
g ( x) = 1 if x < 0
if x ≥ 0 Solution For x < 0, we find
For y lim g ( x) = −1 x →0− And for x ≥ 0, we find
And lim g ( x) = 1 g ( x)
1 x →0+ Thus,
Thus, x lim g ( x) does
x →0 not exist.
not –1 Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y a x This function is discontinuous at the following points:
This
discontinuous ✦ At x = a, f is not defined (x = a is not in the domain of f ).
At
not
not
). Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y a b x This function is discontinuous at the following points:
This
discontinuous ✦ At x = b, f(b) is not equal to the limit of f(x) as x approaches b.
At
not
of
as approaches Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y b c x This function is discontinuous at the following points:
This
discontinuous ✦ At x = c, the function does not have a limit, since the lefthand
At
does
since
lefthand
and righthand limits are not equal.
righthand
not Continuous Functions
Continuous Loosely speaking, a function is continuous at a given point if
Loosely
function its graph at that point has no holes, gaps, jumps, or breaks.
no holes gaps jumps or breaks Consider, for example, the graph of f
Consider,
y c d x This function is discontinuous at the following points:
This
discontinuous ✦ At x = d, the limit of the function does not exist, resulting in a
At
limit
resulting
break in the graph.
break Continuity of a Function at a Number
Continuity A function f is continuous at a number x = a
function
continuous if the following conditions are satisfied:
conditions
1. f(a) is defined.
2. lim f ( x ) exists.
x→a
3. lim f ( x ) = f (a )
3. x→a If f is not continuous at x = a, then f is said to
If
then be discontinuous at x = a.
discontinuous Also, f is continuous on an interval if f is
Also,
continuous
continuous at every number in the interval.
continuous Examples
Examples Find the values of x for which the function is continuous:
Find
continuous f ( x) = x + 2 Solution The function f is continuous everywhere because the three
The
continuous
conditions for continuity are satisfied for all values of x.
y
5 f ( x) = x + 2 4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous x2 − 4
g ( x) =
x−2 Solution The function g is discontinuous at x = 2 because g is not
The
defined at that number. It is continuous everywhere else.
defined
y x2 − 4
g ( x) =
x−2 5
4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous x + 2
h( x ) = 1 if x ≠ 2
if x = 2 Solution The function h is continuous everywhere except at x = 2
The
continuous
except
where it is discontinuous because
because
h (2) = 1 ≠ lim h ( x ) = 4
x→2 5 y y = h( x ) 4
3
2 1
– 2 –1 1 2 x Examples
Examples Find the values of x for which the function is continuous:
Find
continuous −1 if x < 0
F ( x) = 1 if x ≥ 0 Solution The function F is discontinuous at x = 0 because the limit
The
at
of F fails to exist as x approaches 0. It is continuous
fails
approaches It
everywhere else.
y y = F ( x) 1 x
–1 Examples
Examples Find the values of x for which the function is continuous:
Find
continuous 1
if x > 0 G( x) = x −1 i f x ≤ 0 Solution The function G is discontinuous at x = 0 because the limit
The
at
of G fails to exist as x approaches 0. It is continuous
fails
approaches It
everywhere else.
y y = G ( x)
x
–1 Properties of Continuous Functions
Properties
1. The constant function f(x) = c is continuous everywhere.
The constant
continuous
2. The identity function f(x) = x is continuous everywhere.
The identity
continuous
If f and g are continuous at x = a, then
and are
at
1. [f(x)]n, where n is a real number, is continuous at
where
continuous
x = a whenever it is defined at that number.
2. f ± g is continuous at x = a.
continuous
3. fg is continuous at x = a.
fg continuous
4. f /g is continuous if g(a) ≠ 0.
/g continuous Properties of Continuous Functions
Properties Using these properties, we can obtain the following
Using additional properties.
additional 1. A polynomial function y = P(x) is continuous at every
polynomial
value of x.
2. A rational function R(x) = p(x)/q(x) is continuous at
rational
every value of x where q(x) ≠ 0. Examples
Examples Find the values of x for which the function is continuous.
Find
values
continuous f ( x ) = 3x 3 + 2 x 2 − x + 10 Solution The function f is a polynomial function of degree 3, so f(x)
The
polynomial
so
is continuous for all value...
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 Spring '11
 michael
 Derivatives, Derivative, lim g

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