Math 137
Assignment No. 8  Solutions
Fall 2011
1. Required to prove the identity
arcsin
parenleftbigg
x
−
1
x
+ 1
parenrightbigg
= 2 arctan
√
x
−
π
2
.
(1)
Because of the appearance of the square root, we consider only nonnegative values of
x
, i.e.,
x
≥
0. Define
the function
f
(
x
) = arcsin
parenleftbigg
x
−
1
x
+ 1
parenrightbigg
−
2 arctan
√
x
+
π
2
,
x
≥
0
.
If we can show that
f
(
x
) = 0 for all
x
≥
0 then the identity in (1) follows. Note that
f
(0) = arcsin(
−
1)
−
2 arctan(0) +
π
2
=
−
π
2
+ 0 +
π
2
= 0
.
We now compute
f
′
(
x
) by means of the Chain Rule, using the following derivatives,
d
du
arcsin
u
=
1
√
1
−
u
2
,
d
du
arctan
u
=
1
1 +
u
2
.
We find, after a little algebra, that
d
dx
arcsin
x
−
1
x
+ 1
=
x
+ 1
2
√
x
·
2
(
x
+ 1)
2
=
1
√
x
(
x
+ 1)
d
dx
arctan
√
x
=
1
1 +
x
·
1
2
√
x
=
1
2
√
x
(
x
+ 1)
.
This implies that
f
′
(
x
) = 0 for all
x >
0. We can now employ Theorem 8 of Stewart’s text (which is based
on the Mean Value Theorem), Section 4.2, Page 287, to conclude that
f
(
x
) =
f
(0) = 0 for all
x
≥
0, thus
proving the identity in (1).
For those who prefer to prove the desired result without recourse to Theorem 8 of the textbook, we may
proceed as follows: For any
x >
0, define the closed interval
I
= [0
, x
]. The function
f
is continuous on [0
, x
]
and we have shown that
f
′
(
t
) = 0 for all
t
∈
(0
, x
). As such, the hypotheses of the Mean Value Theorem are
satisfied and we may conclude that
f
(
x
)
−
f
(0)
x
−
0
=
f
(
x
)
x
=
f
′
(
c
)
for some
c
∈
(0
, x
)
.
But
f
′
(
c
) = 0 which implies that
f
(
x
) = 0.
1
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2. This question is concerned with the function
f
(
x
) = 3
x
2
/
3
−
x
.
(a) Required to find intercepts, intervals where
f
(
x
)
>
0 and intervals where
f
(
x
)
<
0.
It helps if we examine the first term in the definition of
f
(
x
), which we’ll call
g
(
x
) = 3
x
2
/
3
. Note that
we may express
g
(
x
) as 3(
x
1
/
3
)
2
. The graph of the function
x
1
/
3
may be obtained from the graph of
x
3
by reflecting the latter about the line
y
=
x
. By “squaring” the resulting graph (which essentially
means reflecting any negative values about the
x
axis to make them positive) and multiplying by 3 we
obtain the graph of
g
(
x
) which is expected to have a cusp at
x
= 0. The graph is shown below.
5
4
3
2
1
0
1
2
3
4
5
0
1
2
3
4
5
6
7
8
9
The above analysis also shows that
f
(
x
) is defined for all
x
∈
R
.
Now rewrite
f
(
x
) as
f
(
x
) =
x
2
/
3
(3
−
x
1
/
3
)
,
which implies that
f
(
x
) = 0 when
x
= 0 or
x
= 27.
And since
x
2
/
3
>
0 for all
x
negationslash
= 0, it follows that
f
(
x
)
>
0 when
x
1
/
3
<
3.
The latter inequality is
satisfied for all
x <
27 (the cube root of a negative number is a negative number) so if we exclude the
point
x
= 0, we have the result that
f
(
x
)
>
0 over the two intervals (i) (
−∞
,
0) and (ii) (0
,
27).
(b) Regarding asymptotes, note that we may express
f
(
x
) as follows,
f
(
x
) =
x
(3
x
−
1
/
3
−
1)
.
As
x
→ ±∞
, the term 3
x
−
1
/
3
→
0, implying that 3
x
−
1
/
3
−
1
≈ −
1 for

x

sufficiently large. As such,
we may conclude that
f
(
x
)
≈ −
x
for

x

sufficiently large. As such, there is no constant
L
such that
lim
x
→∞
f
(
x
) =
L
or lim
x
→−∞
f
(
x
) =
L
. Therefore,
f
(
x
) has no horizontal asymptotes.
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 Fall '08
 ANDREWCHILDS
 Math, Algebra, Derivative, lim, Mathematical analysis, Limit of a function, Convex function

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