MATH 147
Quiz Key #8
4/7/2016
2
Consider the function f ( x ) = x2x+1 .
(a) Find all vertical and horizontal asymptotes (if any) of f .
(b) Find the intervals on which f is increasing or decreasing and the local
extrema.
(c) Find the intervals on which f is concave upward or downward and the
inflection points.
(d) Sketch the graph of y = f ( x ) and clearly label the asymptote(s), local
extrema, and inflection points.
Solution: (a) To find vertical asymptotes, we check to see where the denominator is zero. In other words, we check wherex2 + 1 = 0. However, this occurs only at x = ±i, which are not real numbers. Hence, there are no vertical
asymptotes. To check horizontal asymptotes, we check the limits of f at ±∞.
1
2
x
1
x2
lim f ( x ) = lim 2
· = lim
= 1.
x →∞
x →∞ x + 1
x →∞ 1 + 1
1
2
x
2
x
1
x2
1
x2
· = lim
lim f ( x ) = lim 2
= 1.
x →−∞
x →−∞ x + 1
x →−∞ 1 + 1
1
x2
2
x
Hence the horizontal asymptote is y = 1.
(b) Now, we differentiate:
2x x2 + 1 − x2 · 2x
2x
0
=
.
f (x) =
2
2
2
2
( x + 1)
( x + 1)
Again, the denominator is never zero in the real numbers. Therefore, our only
critical value is when 2x = 0, which is at x = 0. Hence we make a sign chart as
follows:
(−∞, 0) (0, ∞)
2x
−
+
2
x2 + 1
+
+
f 0 (x)
−
+
Therefore, f is increasing on (0, ∞) and decreasing on (−∞, 0) . Furthermore,
f has no local maximum and a local minimum at (0, f (0)) = (0, 0) .
1
(c) Now, we find the second derivative:
f 00 ( x ) =
2 x2 + 1
∴ f (x) =
∴ f (x) =
− 2x ·
d
dx
4
h
x2 + 1
2 i
.
( x 2 + 1)
00
00
2
2 x2 + 1
2
− 4x x2 + 1 · 2x
4
( x 2 + 1)
2 x2 + 1
x2 + 1 − 4x2
( x 2 + 1)
4
=
.
2 1 − 3x2
( x 2 + 1)
3
.
Once again, the denominator is always positive, so it remains to check when
the numerator is zero.
1 − 3x2 = 0
3x2 = 1.
1
x2 = .
3
1
x = ±√ .
3
So we now use a sign chart to check concavity:
√1 , ∞
−∞, − √1
− √1 , √1
3
3
3
3
2 1 − 3x2
−
+
−
3
2
x +1
+
+
+
f 00 ( x )
−
+
−
Therefore, f is concave upward on − √1 , √1 and concave downward on
3
3
−∞, − √1 ∪ √1 , ∞ . Hence, f has points of inflection at − √1 , f − √1
3
3
3
3
1
1
√
√
and
,f
. Now, note that
3
3
2
1
− √1
3
1
3
= ,
= =
2
4
4
− √1
+1
3
f
1
−√
3
3
and f
1
√
3
= 2
√1
3
2
√1
3
Hence, f has points of inflection at − √1
3
2
1
3
= =
4
3
1
.
4
+1
, 14 and √1 , 14 .
3
(d)
3