For Thursday, read section 2.4.
“Paradox”:
“Solve
x
2
= 1.
We get
x
= +1 and
x
= –1.
So
1 =
x
= –1, implying 1 = –1.”
..?..
An equation is only true (or false) in a context
and in a
domain of validity
. Ignore the context and domain of
validity and you may get nonsense.
That may seem just silly, but later this week I’ll show you a
calculus counterpart of this paradox.
Section 2.2: The derivative as a function
Key facts (to be proved later):
Where
f
′
(
a
) > 0,
f
(
x
) is increasing on some neighborhood
of
a
;
where
f
′
(
a
) < 0,
f
(
x
) is decreasing on some neighborhood
of
a
;
where
f
′
(
a
) = 0, no conclusion can be drawn without more
information.
[Sketch
x
2
, –
x
2
,
x

x
, and –
x

x
.]
[Do problem 3 on page 92.]
Note that in each case, the function
f
(
x
) is either odd or
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even, and that when
f
(
x
) is odd
f
′
(
x
) is even and vice versa.
This is no accident:
Claim: The derivative of any even function is an odd
function.
Proof: Suppose
f
(
x
) is even, so that
f
(–
x
) =
f
(
x
) for all
x
.
Then for all
x
we have
f
′
(
x
) = lim
h
→
0
(
f
(
x
+
h
) –
f
(
x
))/
h
and
f
′
(–
x
) = lim
h
→
0
(
f
(–
x
+
h
) –
f
(–
x
))/
h
= lim
h
→
0
(
f
(
x
–
h
) –
f
(
x
))/
h
(since
f
(
x
) is even)
= lim
h
→
0
(
f
(
x
+
h
) –
f
(
x
))/(–
h
) (replace
h
by –
h
)
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 Fall '11
 Staff
 Calculus, Derivative, open interval

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