Analysis of Functions: Rational Functions, Cusps,
and Vertical Tangents
Solutions To Selected Problems
Calculus 9th Edition Anton, Bivens, Davis
Matthew Staley
October 18, 2011
1. Give a graph of the rational function and label the coordinates of the stationary
points and inflection points. Show the horizontal and vertical asymptotes and
label them their equations. Label points, if any, where the graph crosses a
horizontal asymptote.
(a) f (x) =
x
x2 −4
i. Symmetry:
−x
(−x)2 − 4
x
=− 2
x −4
= −f (x)
→ Origin Symmetry
f (−x) =
ii. x, y Intercepts:
f (0) =
02
0
=0
−4
f (x) = 0
x
=0
2
x −4
x=0
iii. Vertical asymptotes: Occur for f (x) when denominator is zero.
x2 − 4 = 0
(x + 2)(x − 2) = 0
x = 2, x = −2
iv. Sign of f (x): Check points x = −3, −1, 1, 3
−3
9−4
−1
f (−1) =
1−4
1
f (1) =
1−4
3
f (3) =
9−4
f (−3) =
−3
<0
5
−1
1
=
= >0
−3
3
1
=− <0
3
3
= >0
5
=
1
v. End Behavior:
lim
x→∞
lim
x→−∞
1
x
0
x
=
lim
=0
=
4
2
x − 4 x→∞ 1 − x2
1−0
1
x
0
x
=
lim
=0
=
4
x2 − 4 x→−∞ 1 − x2
1−0
vi. Derivatives:
f (x) =
x2
x
−4
(x2 − 4) − x(2x)
(x2 − 4)2
x2 − 4 − 2x2
=
(x2 − 4)2
−(x2 + 4)
= 2
<0
Always Decreasing and 6= 0
(x − 4)2
f 0 (x) =
(x2 − 4)2 (2x) − (x2 + 4)2(x2 − 4)(2x)
(x2 − 4)4
(x2 − 4)[(x2 − 4)(4x) − 4x(x2 + 2)]
=−
(x2 − 4)4
4x3 − 16x − 4x3 − 8x
=−
(x2 − 4)3
−24x
=− 2
(x − 4)3
24x
= 2
= 0 when x = 0
(x − 4)3
f 00 (x) = −
Interval
Sign of f 0 (x)
x < −2
−
−2 < x < 0
−
0<x<2
−
x>2
−
2
Sign of f 00 (x)
−
+
−
+
vii. Conclusions and Graph:
3
(b) f (x) =
x2
x2 +4
i. Symmetry:
(−x)2
(−x)2 + 4
x2
= 2
= f (x)
x +4
f (−x) =
→ Symmetric w.r.t. x-axis
ii. x, y Intercepts:
f (0) =
02
=0
02 + 4
f (x) = 0
x2
=0
x+ 4
x2 = 0
x=0
iii. Vertical asymptotes: By inspection we see that f (x) is continuous for
all values of x, so there are no vertical asymptotes on (−∞, +∞)
iv. Sign of f (x): f (x) > 0 on (−∞, +∞)
Also note that since x2 < x2 + 4, then
2
0 < x2x+4 < 1.
x2
x2 +4
< 1, so we know that
v. End Behavior:
lim
x2
1
1
= lim
=
=1
4
2
x + 4 x→∞ 1 + x2
1+0
lim
x2
1
1
= lim
=
=1
4
2
x + 4 x→−∞ 1 + x2
1+0
x→∞
x→−∞
4
vi. Derivatives:
f (x) =
x2
x2 + 4
(x2 + 4)2x − x2 (2x)
(x2 + 4)2
2x3 + 8x − 2x3
=
(x2 + 4)2
8x
= 2
(x + 4)2
f 0 (x) =
f 0 (x) = 0 when x = 0
(x2 + 4)2 (1) − x(2(x2 + 4)2x)
(x2 + 4)4
(x2 + 4)(x2 + 4 − 4x2 )
=8
(x2 + 4)4
4 − 3x2
=8 2
(x + 4)3
3x2 − 4
= −8 2
(x + 4)3
f 00 (x) = 8
3(0)2 − 4
f (0) = −8
(0 + 4)3
−4
1
= −8 3 = 8 2 > 0 → Concave up at x = 0 by 2nd derivative test
4
4
00
f 00 (x) = 0 when 3x2 − 4 = 0
3x2 = 4
4
x2 =
3
2
x = ±√
3
5
f
2
√
3
=
=
4
3
2
√2
3
2
√2
3
4
3
+4
+4
=
4
3
16
3
=
1
4
=
16
4
By symmetry, we have f − √23 = f √23 = 14 .
vii. Conclusions and Graph: Here is a sign chart for our function.
Sign of f (x)
Sign of f 0 (x)
Sign of f 00 (x)
√
x < −2/ 3
+
−
−
√
−2/ 3 < x < 0
+
−
+
√
0 < x < 2/ 3
+
+
+
+
−
Interval
√
x > 2/ 3
+
Inflection points occur at − √23 , 41 and
and tends towards y = 1 as x → ±∞.
6
√2 , 1
3 4
. f (x) is always positive
f (x) =
x2
x2 +4
− √23 , 14
7
√2 , 1
3 4
2. Show that y = x + 3 is an oblique asymptote of the graph of f (x) =
Sketch the graph of y = f (x) showing the asymptotic behavior.
x2
.
x−3
(a) First we long divide:
x+3
x−3
x2
− x2 + 3x
3x
− 3x + 9
9
(b)
f (x) = x + 3 +
f (x) − (x + 3) =
9
x2
9
x2
9
=0
x→∞ x2
f (x) → x + 3 x → ∞
lim (f (x) − (x + 3)) = lim
x→∞
Thus we have shown that y = x+3 is an oblique asymptote to the function
x2
. To graph the rest of the function we need to due some more analysis.
x−3
i. Symmetry:
f (−x) =
(−x)2
x2
=−
−x − 3
x+3
There is no symmetry.
ii. x, y Intercepts:
f (0) = 0
f (x) = 0 when x = 0
iii. Vertical asymptotes: x = 3
8
iv. Sign of f (x):
4
<0
−1
16
f (4) =
>0
1
f (2) =
v. End Behavior: This was already done above.
vi. Derivatives: Take f 0 and f 00 of original f (x).
f (x) =
x2
x−3
(x − 3)(2x) − x2 (1)
(x − 3)2
x2 − 6x
=
(x − 3)2
x(x − 6)
=
(x − 3)2
f 0 (x) =
f 0 (x) = 0 when x = 0, x = 6
(x − 3)2 (2x − 6) − (x2 − 6x)(2(x − 3)(1))
(x − 3)4
(x − 3)[(x − 3)(2x − 6) − 2x2 − 12x]
=
(x − 3)4
2x2 − 12x + 18 − 2x2 − 12x
=
(x − 3)3
18
=
(x − 3)3
f 00 (x) =
f 00 (0) =
18
< 0 Concave Down and Rel. Max
−33
f 00 (6) =
18
> 0 Concave Up and Rel. Min
33
9
vii. Graph:
10
3
3. Sketch the graph of the rational function f (x) = (x−2)
and label the coordix2
nates of the stationary points and inflection points. Show all possible asymptotes and label them with their equations.
Symmetry: f (−x) =
(−x−2)3
(−x)2
=
−(x+2)3
x2
→ There is no symmetry.
x, y Intercepts: f (x) = 0 only when x = 2. There is no y intercept as f (x) is
undefined at x = 0
Vertical asymptotes: x = 0
Sign of f (x): It is easy to see that for x < 2, f (x) < 0 and for x > 2, f (x) > 0
End Behavior: Rewrite f (x) =
(x−2)3
x2
=
x3 −6x2 +12x−8
x2
=x−6+
Now we can see that limx→±∞ (f (x) − (x − 6)) = limx→±∞
So f (x) → (x − 6) as x → ±∞
Thus there is an oblique asymptote y = x − 6
11
12
x
12
x
−
−
8
x2
8
x2
=0
Derivatives:
f (x) =
(x − 2)3
x2
x2 (3(x − 2)2 − (x − 2)3 (2x))
x4
2
x(x − 2) (3x − 2(x − 2))
=
x4
(x − 2)2 (x + 4)
=
x3
f 0 (x) =
f 0 (x) = 0 when x = 2, x = −4
f (2) = 0
(−6)3
(−2 · 3)3
−23 33
33
27
f (−4) =
=
=
=
−
=
−
(−4)2
(−22 )2
24
2
2
f 00 (x) =
x3 [2(x − 2)(x + 4) + (x − 2)2 (1)] − 3x2 (x − 2)2 (x + 4)
x6
=
x3 (x − 2)[2(x + 4) + (x − 2)] − 3x2 (x − 2)2 (x + 4)
x6
=
x2 (x − 2)[[x(2x + 8 + x − 2)] − 3(x − 2)(x + 4)]
x6
=
(x − 2)[x(3x + 6) − 3(x2 + 2x − 8)]
x4
=
(x − 2)[3x2 + 6x − 3x2 − 6x + 24]
x4
=
24(x − 2)
x4
f 00 (x) = 0 when x = 2
12
Conclusions Note that f 00 (4) = 24(−2)
< 0, so by the second derivative test,
44
f (x) is concave down at x = −4 and thus there is a relative max at x = −4
The second derivative test is inconclusive at x = 2 as f 00 (2) = 0. Make a
sign chart to determine other possible inflections points.
Interval
Sign of f (x)
Sign of f 0 (x)
Sign of f 00 (x)
x < −4
−
+
−
−4 < x < 0
−
−
−
0<x<2
−
+
−
x>2
+
+
+
We see that there is only one inflection point at (2, 0), which
is also a stationary
27
point. The other stationary point occurs at −4, − 2 .
Remember that there is an oblique asymptote y = x − 6, and a vertical asymptote x = 0
13
Graph of f (x) =
(x−2)3
x2
14
4. Give a graph of the function f (x) = 4x1/3 − x4/3 and identify the locations of
all relative extrema and inflection points.
f (x) = 4x1/3 − x4/3 = x1/3 (4 − x)
We see that f (x) = 0 for x = 0 and x = 4, and that there are no vertical
asymptotes. Here is a simple sign chart of f (x) to help us determine where is
above or below the x axis:
x < 0, f (x) < 0
0 < x < 4, f (x) > 0
x > 4, f (x) < 0
Now take the first and second derivatives to determine critical points and inflection points.
4
4
f 0 (x) = x−2/3 − x1/3
3
3
4
= x−2/3 (1 − x)
3
=
4 1
√ (1 − x)
3 3 x2
f 0 (x) is undefined at x = 0
f 0 (x) = 0 when x = 1, with f (1) = 11/3 (4 − 1) = 3. Stationary point at (1, 3).
Sign chart for f 0 (x):
x < 0 f 0 (x) > 0
0 < x < 1 f 0 (x) > 0
x > 1 f 0 (x) < 0
15
Now the second derivative:
8
4
f 00 (x) = − x−5/3 − x−2/3
9
9
4
= − x−5/3 (2 + x)
9
4 1
=− √
(2 + x)
9 3 x5
f 00 (x) is undefined also at x = 0
√
3
1/3
f 00 (x) = 0 when x = −2, with
f
(−2)
=
(−2)
(4
−
(−2))
=
−6
2.
√
Inflection point at (−2, −6 3 2)
1
Since f 00 (1) = − 49 √
3 5 (2 + 1) < 0, the stationary point is Concave down, and a
1
Relative Max by the second derivative test.
Sign chart for f 00 (x):
√
3
x < −6 2 f 00 (x) < 0
√
3
−6 2 < x < 0 f 00 (x) > 0
x > 0 f 00 (x) < 0
Now it would be good to check the end behavior before sketching the graph.
lim x1/3 (4 − x) → −∞
x→∞
lim x1/3 (4 − x) → −∞
x→−∞
16
Graph of f (x) = 4x1/3 − x4/3
17