Graphics Reference
In-Depth Information
31
f
(
x
)
x
at
x
0.
32 For example,
x
x
at
x
0.
33
f
(
x
)
(cos
x
)(2
x
).
34
f
(
x
)
2
x
sin(1/
x
)
cos(1/
x
), when
x
0,
f
(0)
0.
f
is not continuous at 0, since
x
cos(1/
x
) is likethefunction in qn 20.
So
f
is not differentiable at 0.
35
f
has domain R.
f
has domain R
0
.
f
has domain R
0, 1
.
36 Apply qn 12 to
f
.
37 Consider limits from below and limits from above at 0.
(i) Yes. (ii) Yes. (iii) Yes. (iv) No.
38
f
-
is the derived function of
f
.
f
is the derived function of
f
.
39
f
(
x
)
f
(
a
)
x
a
1
x
a
1
2
a
as
x
a
.
So
f
(
a
)
1/(2
a
) and
g
(
f
(
a
))
2
a
.
41
g
(
f
(
a
)) ยท
f
(
a
)
1. But if
f
(
a
)
0, there is no possible value for
g
(
f
(
a
)).
Considr thegraphs of thefunctions
f
and
g
in qn 39 at 0.
42
(i)
g
(
f
(
x
))
g
(
f
(
a
))
x
a
f
(
x
)
f
(
x
)
f
(
a
)
x
a
f
(
a
)
f
(
a
)
1
.
f
(
x
)
Usethefact that
f
(
a
) exists and is non-zero, with qn 6.93, the
algebra of limits.
(ii) Because
f
is a bijection.
(iii) (
b
)
b
by definition. This may be rewritten (
f
(
a
))
f
(
a
).
By thecontinuity of
g
at
b
f
(
a
), (
g
(
f
(
a
)))
g
(
f
(
a
)). So (
a
)
a
.
(iv) From thelimit in (i),
g
(
f
(
a
))
g
(
f
(
a
))
1
f
(
a
)
,
f
(
a
)
f
(
a
)
so
g
(
b
)
g
(
b
)
1
f
(
a
)
.
b
b
(v) From thedefinition of (
b
), limits from aboveand blow
b
exist and
areequal, so thetwo-sided limit
g
(
y
)
g
(
b
)
y
b
1
f
(
a
)
.
lim
Thus
g
(
f
(
a
)) exists and equals 1/
f
(
a
).
