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170
CHAPTER 2
.
.
Differentiation
226
intersections.) Sketch a graph where
F
±
(1)
<
1 and there are no
solutions to the equation
F
(
x
)
=
x
between 0 and 1 (although
x
=
1i
sa solution). Solutions have a connection with the
probability of the extinction of animals or family names. Sup
pose you and your descendants have children according to the
following probabilities:
f
0
=
0
.
2is the probability of having
no children,
f
1
=
0
.
3is the probability of having exactly one
child, and
f
2
=
0
.
5is the probability of having two children.
Deﬁne
F
(
x
)
=
0
.
2
+
0
.
3
x
+
0
.
5
x
2
and show that
F
±
(1)
>
1.
Find the solution of
F
(
x
)
=
x
between
x
=
0 and
x
=
1; this
number is the probability that your “line” will go extinct some
time into the future. Find nonzero values of
f
0
,
f
1
and
f
2
such
that the corresponding
F
(
x
) satisﬁes
F
±
(1)
<
1 and hence the
probability of your line going extinct is 1.
4.
The
symmetric difference quotient
of a function
f
centered at
x
=
a
has the form
f
(
a
+
h
)
−
f
(
a
−
h
)
2
h
.If
f
(
x
)
=
x
2
+
1
and
a
=
1, illustrate the symmetric difference quotient as a
slope of a secant line for
h
=
1 and
h
=
0
.
5. Based on your pic
ture, conjecture the limit of the symmetric difference quotient
as
h
approaches 0. Then compute the limit and compare to the
derivative
f
±
(1) found in example 1.1. For
h
=
1
,
h
=
0
.
5 and
h
=
0
.
1, compare the actual values of the symmetric difference
quotient and the usual difference quotient
f
(
a
+
h
)
−
f
(
a
)
h
.
In general, which difference quotient provides a better esti
mate of the derivative? Next, compare the values of the dif
ference quotients with
h
=
0
.
5 and
h
=−
0
.
5tothe deriva
tive
f
±
(1). Explain graphically why one is smaller and one is
larger. Compare the average of these two difference quotients
to the symmetric difference quotient with
h
=
0
.
5. Use this
result to explain why the symmetric difference quotient might
provide a better estimate of the derivative. Next, compute sev
eral symmetric difference quotients of
f
(
x
)
=
±
4i
f
x
<
2
2
x
if
x
≥
2
centered at
a
=
2. Recall that in example 2.7 we showed that
the derivative
f
±
(2) does not exist. Given this, discuss one
major problem with using the symmetric difference quotient
to approximate derivatives. Finally, show that if
f
±
(
a
)exists,
then lim
h
→
0
f
(
a
+
h
)
−
f
(
a
−
h
)
2
h
=
f
±
(
a
).
2.3
COMPUTATION OF DERIVATIVES: THE POWER RULE
Youhavenow computed numerous derivatives using the limit deﬁnition. In fact, you may
have computed enough that you have started taking some shortcuts. In exploratory exercise 1
in section 2.2, we asked you to compile a list of derivatives of basic functions and to
generalize. We continue that process in this section.
The Power Rule
We ﬁrst revisit the limit deﬁnition of derivative to compute two very simple derivatives.
Forany constant
c
,
d
dx
c
=
0
.
(3.1)
y
x
a
c
y
±
c
FIGURE 2.20
A horizontal line
Notice that (3.1) says that for any constant
c
, the horizontal line
y
=
c
has a tangent
line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line
(see Figure 2.20).
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 Spring '10
 Lagerstrom

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