Contents
6 Graph Sketching
87
6.1
Increasing Functions and Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
87
6.2
Intervals — Monotonically Increasing or Decreasing . . . . . . . . . . . . . . . . . . . . . . .
88
6.3
Extrema — Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89
6.4
Relative Maxima and Relative Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
90
6.5
The Second Derivative Test for Relative Extrema . . . . . . . . . . . . . . . . . . . . . . . . .
93
6.6
Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
94
6.7
Points of Inflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95
6.8
Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96
6.9
Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
6.10 Graph Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
6.11 Graphs of Trancendental Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
6.12 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6.13 Functions with Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2
CONTENTS
Chapter 6
Graph Sketching
6.1
Increasing Functions and Decreasing Functions
For problems numbered 1 to 5 show whether the function is increasing or decreasing at the indicated points.
1. (a)
f (x) = 2x2 − 1;
at x = 0, x = −3, and x =
1
2
(b)
f (x) = x3 − 3x2 + 1;
(c)
f (x) = |x| − 2;
at x = −2, x = −1, and x = 2
(d)
f (x) = |x − 1|;
at x = 0, x = 12 , and x = 2
at x = −1, x = 0, and x = 3
2. (a)
f (x) = x|x|;
(b)
f (x) =
x
x+1 ;
(c)
f (x) = 3x4 + 4x3 ;
(d)
f (x) =
(2x+1)
(x−2) ;
(e)
f (x) =
x2 +1
x ;
3. (a)
at x = −1, x = 1, x = 2, and x = 4
at x = −1, and x = 10
at x = −2, x = −1, x = 0, and x = 1
at x = 7
at x = 1
h(x) = cos x2 ;
at x =
(b)
g(x) = − sin 2x;
(c)
g(x) = tan x;
at x =
(d)
g(x) = x sin x;
at x
4. (a)
(b)
f (x) = xex ;
−π
4 ,
= π2 ,
π
4
and x =
x = 0, and x =
3π
4
π
4
and x = 0
at x = −10, x = −1, x = 0, and x = 1
at x = −10, x = −2, x = −1, and x = 0
at x = −1, x = 1, and x = 10
(c)
f (x) =
(d)
f (x) = e(x−1) ;
5. (a)
at x = 0, x =
f (x) = (x + 1)ex ;
ex
x ;
π
4
2
at x = −1, x = 1, and x = 2
h(x) = ln(1 − x);
at x = 2, and x = 0
(b)
h(x) = x ln x;
at x = 1, x = e−2 , and x = e3
(c)
h(x) = ln2 x;
at x = e−1 , x = 1, and x = e
(d)
h(x) = ln(sin x);
at x =
π
4,
x=
π
2,
and x =
3π
4
88
Graph Sketching
6.2
Intervals — Monotonically Increasing or Decreasing
For problems numbered 6 to 11 divide the domain of the function into a finite number of intervals on each
of which the function is strictly monotone. Indicate the intervals where the function is increasing and the
intervals where it is decreasing.
Example: f (x) = 6x4 − 20x3 − 6x2 + 72x + 12
Solution:
f 0 (x) = 12(2x3 − 5x2 − x + 6)
= 12(x + 1)(2x − 3)(x − 2)
(see Chapter 0 page 5 Theorem I)
f 0 (x) = 0 at x = −1, 32 or 2.
Because f is a continuous function we can conclude f (x) is strictly monotone on each of the intervals
(−∞, −1), (−1, 23 ), ( 32 , 2), (2, +∞).
Compute a value f 0 (−2) = −336 < 0 to conclude that f is decreasing on (−∞, −1) due to the fact f is
continuous on (−∞, −1) and −2 ∈ (−∞, −1).
f 0 (0) > 0 → increasing on −1, 23
f 0 47 < 0 → decreasing on 32 , 2
f 0 (3) > 0 → increasing on (2, +∞).
NOTE: To use this method you must check that f is continuous on the given interval and state this as part
of your solution.
6. (a)
(c)
f (x) = (x − 1)2 + 1
(b)
f (x) = x3 + 2
f (x) = |x − 2|
(d)
f (x) = x|x|
7. (a)
f (x) =
1
(x−1)(x−2)
(b)
f (x) = 4x3 − 3x
(c)
f (x) =
(x2 +1)
x2
(d)
f (x) =
f (x) = (x2 + 2x + 1) 2
(b)
f (x) =
f (x) = |(x − 1)(x − 2)(x − 3)|
(d)
f (x) =
f (x) = ex − x
(b)
f (x) = xe2x
(d)
f (x) = ex
f (x) = x2 ln x
(b)
f (x) =
f (x) = ln |x|
(d)
f (x) = ln(x2 + 1)
(b)
f (x) = | sin x|
(d)
f (x) = sin x + cos x
8. (a)
(c)
9. (a)
(c)
10. (a)
(c)
11. (a)
(c)
1
f (x) =
ex
x+1
s(x) = sin x −
f (x) = sin |x|
π
2
√
x−3
1
1
(x2 +4) 2
p
|x − 2|
2
−x−2
ln x
x
6.3 Extrema — Maxima and Minima
6.3
89
Extrema — Maxima and Minima
Examples
maximum
maximum
relative maximum
relative maximum
relative minimum
minimum
relative minimum
minimum
a
b
The maximum (global maximum) is the highest value a function attains on the given domain. The minimum
(global minimum) is the lowest value a function attains on the given domain. Some functions do not have a
maximum or a minimum.
For each of the following problems find the global maximum and the global minimum.
If it exists, state the place and value. If it does not exist, state why not. Domain is x ∈ (−∞, +∞) unless
otherwise stated.
12. f (x) = x2
13. f (x) = −x2
14. f (x) = |x|
15. f (x) = sin x on [0, 2π]
16. f (x) = cos x on
[0, 2π]
17. f (x) = 2x on x ≤ 2
18. f (x) = 2x on x < 2
19. f (x) = tan x
20. f (x) = csc x
21. f (x) = sec x
22. f (x) =
√
x2
on (−2, +2)
24. f (x) = |1 − x| on (−1, 3)
26. f (x) = |2 − x|,
28. f (x) =
for |x| > 4∗
1
x
23. f (x) = x2
on x > 2
25. f (x) = |x − 2| + 5 on (−3, 17]
27. f (x) = 5 − |x − 2| on x > 3
29. f (x) = x3 − 6x2 + 11x − 6
30. f (x) = (x − 1)(x − 2)(x − 3)
31. f (x) = | sin x|
32. f (x) = |(x − 1)(x − 2)(x − 3)|
33. f (x) =
34. f (x) =
36. f (x) =
∗ This
√ 1
25−x2
√
x
means domain of x is (−∞, −4) and (4, ∞)
√
25 − x2
35. f (x) = x2 + x1 ,
37. f (x) =
√
x 6= 0
x on x < 3
90
Graph Sketching
38. f (x) =
40. f (x) =
√
√
x † , f (x) < 9
39. f (x) =
25 − x2
41. f (x) =
on x < 3
√
√
25 − x2
sin x
42. f (x) = x2
on (−2, +2)
43. f (x) = x2
44. f (x) = x3
on [−2, +2]
45. f (x) = bxc
46. f (x) = bx2 c
on
(−2, +2)
48. f (x) = dxe
50. f (x) = dx2 e
on
[−2, +2]
54. f (x) =
on [−2, +2]
47. f (x) = bx2 c
on [−2, 2]
49. f (x) = dx2 e
on (−2, +2)
51. f (x) =
52. f (x) = (x − 5)2 + 2
on x < 10
1
x2 +9
53. f (x) = 2 − (x − 5)2



| sin x| on x > 0
3x on −3 < x ≤ 0
55. f (x) =

 2
x − 18 on x ≤ −3
1
9−4
 1

on x < −1


 x
x on x ∈ (−1, +1]
56. f (x) =



1

on x > +1
x
57. f (x) = | tan x|
58. f (x) = x2 , f (x) < 2
59. f (x) = x2 , f (x) > 2
60. f (x) = |2 − x|, f (x) > 4
61. f (x) = 2 − |x + 1|, f (x) > 0
62. f (x) = 2 − |x2 + 1|
63. f (x) = −x2 + 4 − 1
64. s = 128t − t2
65. y = x 3
66. f (θ) =
6.4
√
1 − cos2 θ
2
on
− π2 , + π2
67. y = (x − 5)3
on (−0, 10]
Relative Maxima and Relative Minima
Recall that a critical number for a function f is a number c such that either f 0 (c) = 0 or f 0 (c) does not exist.
√
√
function f is given by the formula x. The range is those x values which are less than 9. In other words for each
point on the graph, both statements must hold.
† The
6.4 Relative Maxima and Relative Minima
91
(ii)
(i)
f’(c) = 0
rel.max.
(iii)
f’(c) = 0
rel. min.
(v)
(iv)
f’(c) = 0
no max. or min.
f’(c) = 0
no max. or min.
(vi)
(point cusp)
f’(c) does not exist
rel. max.
(corner cusp)
f’(c) does not exist
rel.max.
Notice that sometimes a relative maximum or a relative minimum occurs when the derivative does not exist
(as in (v) and (vi)). At a point or a corner, the first derivative test still works.
For questions numbered 68 to 117
(a) Find the domain of the given function.
(b) Find the first derivative of the function.
(c) Find the critical numbers.
(d) On what intervals is f 0 (x) > 0, and on what intervals is f 0 (x) < 0?
(e) Use the first derivative test to give the relative maxima and the relative minima.
(f) If they exist, give global maximum, global minimum, and their location.
68. f (x) = 2x3 − 3x + 1
69. f (x) = 3x − 1
70. f (x) = (x − 1)2
71. f (x) = 1 − x − x2
72. f (x) = x3 + x2 − x − 1
73. f (x) = x2 ex
74. f (x) = x3 + x2 + x − 1
75. f (x) = x2 + x + 1
76. f (x) =
8
x2 +4
77. f (x) = sin πx
78. f (x) = sin2 πx
79. f (x) = x2 (x − 1)2
80. f (x) = x3 (x − 1)2
81. f (x) = ae−b
2
x2
,
a>0
92
Graph Sketching
82. f (x) =
x4
4
+
x3
3
−
x2
2
−x+1
1
2
83. f (x) = 1 + 2(x − 1) 3
1
1
84. f (x) = 1 + (x − 1) 3
85. f (x) = x 3 (x − 2) 3
86. f (t) = e−t sin t,
87. f (x) =
x2 −x−1
x2 −x+1
on [0, π]
88. f (x) =
x
x2 +1
89. f (x) =
1
x
90. f (x) =
x2
x2 +1
91. f (x) =
(2−x)3
2(1−x)
92. f (x) = ln
x4
(3x−4)2
93. f (x) = x + sin x
94. f (x) = x sin x
95. f (x) = sec x tan x
96. f (x) = x ln x
97. f (x) = eln(x
98. y = 4x + 9
99. y = −8x + 2
2
+3)
100. y = x3
101. y = 2x3 − 6x
102. y = 4x3 − 12x
103. y = x3 + 3x2 − 1
104. y = 2x3 − 3x2 − 36x + 4
105. y = x4 − 2x2
106. f (x) = x4 − 2x + 12
107. f (x) = x5
108. f (x) = x7
109. f (x) = x4
110. f (x) = x +
112. f (x) =
x−1
x+1
111. f (x) = 4x +
16
x
√
113. f (x) = x x − 1
2
1
115. f (x) = 3 x 3 − x 3
2
114. y = x 3
116. f (x) =
1
x
x2
x−1
2
117. f (x) = x 3 + x
For each of the questions numbered 118–131, sketch a graph of some function with the given properties.
(You do not have to give a formula for the function. Just give the picture.)
118. A function with a relative minimum at 2 but no global minimum.
119. A function with the domain [0, 2] and a maximum at 1.
120. A function with the domain (0, 2] with a relative maximum at 1 and a maximum at 2. Is f 0 (2) = 0?
121. A function with a relative minimum at 0, but with f 0 (0) not existing.
122. A differentiable function with neither a maximum nor a minimum.
123. A function with a relative maximum and a relative minimum but with no maximum or minimum.
124. A function whose range is all values less than 1 that does not have a maximum.
125. A function with a maximum at 5 but f 0 (5) does not exist.
6.5 The Second Derivative Test for Relative Extrema
93
126. A function which is defined everywhere but has no maximum, no minimum, no relative maximum and
no relative minimum.
127. A function with a maximum at 5, a minimum at 0, a relative maximum at 2 and a relative minimum
at 1.
128. A function with a domain of (0, 2) with a maximum and a minimum.
129. A function with a domain of (0, 2) with a relative maximum and a relative minimum.
130. A function with a domain of (0, 2) with no maximum, no minimum, no relative maximum and no
relative minimum.
131. A function whose domain is (−∞, ∞) and whose range is (π, 5]. Is there a maximum? Is there a
minimum?
6.5
The Second Derivative Test for Relative Extrema
For questions 132–171, use the second derivative test in finding the relative extrema of the indicated functions.
If the value of the second derivative is zero at a critical point, use the first derivative test. Then find the
extrema.
132. f (x) = 2x2 − 6x + 5
133. f (x) = −3x2 + 2x − 1
134. f (x) = x3 − 3x + 2
135. f (x) = x3 + x2 − x + 5
136. f (x) = x4 + 2x3
137. f (x) = x4 + 2x3 − 3x2 − 4x
138. f (x) = x2 +
140. f (x) =
1
x2
139. f (x) = 2x2 −
1
x2
√
141. f (x) = x x + 3
2x
x2 +1
142. f (x) = x3 − 23 x2 − 6x + 2
143. f (x) = x3 + x2 − x − 1
144. f (x) = x3 − 4x2 + 4x − 1
145. f (x) = x3 + 3x2 − 3x − 5
146. f (x) = x3 − x2 + x − 1
147. f (x) = x4 + 43 x3 − 4x2 −
148. f (x) = (x + 2)(x − 2)3
149. f (x) = x4 − 3x3 + 3x2
150. f (x) = x4 + 5x3 + 6x2
151. f (x) = x − 3 +
152. f (x) =
2
4x
x2 +4
2
x+1
5
153. f (x) = 5x 3 − x 3
√
154. f (x) = x 8 − x2 ,
|x| ≤
√
156. f (x) = x2 5 + x,
x ≥ −5
1
2
√
8
2
155. f (x) = x 3 (x + 2)−1
√
157. f (x) = x2 3 − x2 ,
158. f (x) = x 3 (x + 2)− 3
159. f (x) = x + sin x
160. f (x) = x4
161. f (x) =
162. f (x) = x + x1 ,
4
3
x 6= 0
x5
10
− x3 +
163. f (x) = x − cos x
9
10
|x| ≤
√
3
94
Graph Sketching
164. f (x) =
√1 ,
x−9
165. f (x) = x3 − 6x2
x>9
4
166. f (x) = 3x4 − 10x3 + 6x2 + 5
167. f (x) = x 3
1
2
x2
168. f (x) = 5x2 +
169. f (x) = x(2a − x) 2 ,
1
3
1
171. f (x) = x 2 (x − 18)− 2
170. f (x) = x(2a − x2 ) 2
6.6
a>0
Concavity
For problems numbered 172 to 189 decide:
(a) Is the given function increasing or decreasing at the indicated value of x? Why?
(b) Is the given function concave upward or concave downward for the indicated value of x? Why?
172. f (x) = 3x2 + 15x,
x = −3
173. f (x) = 4x2 − 28x + 7,
[decreasing, concave upward]
x = −10
174. f (x) = 8x3 + 9x2 − 18x + 15,
175. f (x) = x3 + x2 − x + 1,
176. y = x3 + 5x2 + 18x,
177. y = x2 + 3x + 14,
x = −2
x=7
179. y = −3x + 9,
x = −2
181. y =
x+1
x ,
√
x=0
x2 + 25,
182. y = x cos x,
2
185. y = xex ,
186. y = x2 ex ,
[not defined at x = 0]
x = −1
x=
183. y = tan x sec x,
184. y = e−x ,
x = −3
x=0
178. y = 4x − 14,
180. y =
x=0
π
4
x=
π
4
x = −1
x=0
x = −1
187. f (x) = ln |x + 4| + (x + 4),
188. f (x) = (x + 4)ex+4 ,
189. f (x) = ln |x + 2| +
x = −5
x = −4
1
x+2 ,
x = −1
6.7 Points of Inflection
6.7
95
Points of Inflection
In problems numbered 190 to 213 find:
(a) Intervals where the function is increasing and decreasing.
(b) Relative maximum and relative minimum.
(c) Intervals where function is concave up, concave down. Why?
(d) Points of inflection. Show why.
190. y = x3
[concave up x > 0; concave down x < 0; inflection at (0, 0)]
191. y = x10 + x8
√
192. y = 3 x
2
193. y = x 3
194. f (x) =
1
x
195. f (x) =
1
x2
196. g(x) =
1
x2 +1
[concave up x > 0, concave down x < 0]
197. h(x) = x3 + ax + b
[concave up, −1 + 2k < x < 2k; concave down 2k < x < 1 + 2k;
inflection points (k, 0), k ∈ Z]
198. y(x) = sin πx
199. T (x) = tan x
200. f (x) = x4
201. f (x) = xn , n an integer > 2
202. φ(x) = 5 − 2x + x2
[concave up everywhere]
1
203. ψ(x) = 3 + (x + 1) 3
2
204. λ(x) = 3 + (x + 1) 3
205. c(x) = −| sin x|
206. f (x) = ln x2
207. f (x) = xex
2
208. g(x) = ex
209. y =
x
x2 +1
210. y =
x
x2 −1
+1
211. f (x) = x + ex
212. f (x) = ln(sin2 x)
213. g(x) = sin x +
1
sin x
[concave up everywhere except x = nπ, n ∈ Z]
96
6.8
Graph Sketching
Asymptotes
For the functions numbered 214 to 241,
(a) Find all the asymptotes if there are any.
(b) Does the curve approach the asymptote
(i) from below?
(ii) from above?
(iii) neither?
(c) In the case of vertical asymptotes does the curve approach the asymptote at +∞ or −∞? Also check
the left and right hand sides of the asymptote separately.
x+1
(x+2)(x+1)(x−3)
214. (a)
y=
215. (a)
y =3+
216. (a)
y=
217. (a)
x
(x+2)(x−4)
(b)
y =2+
(b)
y =x−3+
1
(x−2)2
(b)
y=
1
(x−2)3
y=
x
|x−1|
(b)
y=
|x|
x−1
218. (a)
y=
x2 −4
x−2
(b)
y=
x+2
x2 −4
219. (a)
y=
x+2
x−2
(b)
y=
3x+2
x−2
220. (a)
y=
1
x2 −4
(b)
y=
1
x2 +4
221. (a)
y=
a
x−3
(b)
y=
ax+b
cx−d
222. (a)
y=
x+1
x3 +2x2 −x−2
(b)
y=
x−1
x2 +2x−3
223. (a)
y=
x2 +1
x
(b)
y=
x2 +x+1
x2
224. (a)
y=
x2 −2
x
(b)
y=
x3 −1
x
225. (a)
y=
(x−3)2
x
(b)
y=
x2
1−x2
226. (a)
y=
x2 +1
x2 −1
(b)
y=
√x
x−1
227. (a)
y=
√ x−1
9x2 +1
(b)
y =3−
228. (a)
(b)
229. (a)
2
x+1
√1
x2
√
x2 + 1.
(Hint: rationalize the numerator)
√
2
y = 2x − 4x + 3
y =x−
y=
(x−1)3
x2
(b)
y=
x2 (x−2)
(x+1)2
1
x+1
6.9 Symmetry
230. (a)
y=
231. (a)
y=
232. (a)
97
4
x2
3
−1
2 2
x2
(b)
y = 1+
(b)
y=
y = xex
(b)
y = 1 − ex
233. (a)
y = x2 ex
(b)
y = 2 − e−x
234. (a)
y = x 2 ex
(b)
y = 1 − e−x
235. (a)
y=
ex
x
(b)
y = e−x
236. (a)
y=
ex
x2
(b)
y=
xe−x +1
x
237. (a)
y = xe−x
(b)
y=
e−x +x2
x2
238. (a)
y = e−x sin x
(b)
y = e−x cos x
239. (a)
y=
ln x
x
(b)
y=
240. (a)
y =1+
(b)
y = e−x ln x
1
x
sin x
1
ln x
x2
1
x2
cos x
2
2
x ln x+1
x
241. Under certain conditions the size of a colony of bacteria is related to time by
x = 40000
2t + 1
t+1
(t ≥ 0)
where x is the number of bacteria and t is the time in hours. To what number does the
size of the colony tend over a long period of time? In how many hours will the colony have
achieved 95% of its ultimate size?
6.9
Symmetry
Symmetry is very easy to test for. It does not do anything spectacular for the graph as does the discovery
of asymptotes. However noticing symmetry can shorten your work appreciably.
242. An even function, f , is one for which f (−x) = f (x); and odd function is one for which
f (−x) = −f (x) for all values in its domain. What can we say concerning the symmetry of
the graph of an even function? Of an odd function?
243. Show that if every x in an equation is replaced by 2k − x and the resulting equation is
equivalent to the original, then its graph is symmetric about the line x = k.
For the equations numbered 244 to 263 test for symmetry and find the axis or point of symmetry if one
exists.
244. (a)
y 2 = a2 − x2
(b)
y = x3 + 1
245. (a)
y = x4
(b)
y=
1
x2
246. (a)
y 2 = x4
(b)
y=
x
x2 +1
247. (a)
y=
(b)
y =x+
1
x2 +2
1
x
98
Graph Sketching
5
248. (a)
y = x2
249. (a)
250. (a)
5
(b)
y = x3
y 3 + x3 = a3
(b)
y=
y=
−x
x2 −1
(b)
y = x|x|
251. (a)
y=
|x|
x
(b)
y = −x
252. (a)
y = sin x
(b)
y = cos x
253. (a)
y = tan x
(b)
y = sec x
254. (a)
y = csc x
(b)
y = cot x
255. (a)
y = sin x
(b)
y=
sin x
x
256. (a)
y = x cos x
(b)
y=
cos x
x
257. (a)
y = sin x cos x
(b)
y = x2 sin x
258. (a)
y = − cos x
(b)
y = sin2 x
259. (a)
y = 10x
(b)
y = e−x
260. (a)
y = 2xe−x
(b)
y = xex
261. (a)
y=
(b)
y=
262. (a)
y 2 = x2 − x3
(b)
y 2 = (x − 1)2 (x + 2)
263. (a)
y 2 = x2 (1 − x2 )
(b)
y = sin2 x + cos2 x
1
1
1
2
2
ex −e−x
2
8
1−x2
2
2
ex +e−x
2
264. If f (x) is an even function, is f 0 (x) even or odd?
265. If f (x) is an odd function, is f 0 (x) even or odd?
266. If f is an odd function and g is an even function, is f ◦ g even or odd?
267. If f is an odd function and g is an even function is f ◦ g even or odd? Is g ◦ f even or odd?
For each of the following graphs give the following information if the function has the stated property:
(a) Domain of the function.
(b) Symmetry.
(c) For what values of x is f (x) = 0? For what values of y is f (0) = y?
(d) On what intervals is f increasing? On what intervals is f decreasing?
(e) For what values of x is f (x) a relative maximum or a relative minimum?
(f) Estimate the value of x where the points of inflection occur. Is the slope of the tangent at the point of
inflection equal to zero, greater than zero or less than zero?
(g) On what intervals is f concave up and on what intervals is f concave down?
(h) Find all asymptotes.
6.9 Symmetry
268.
99
Y
Y
269.
X
X
Y
Y
270.
271.
X
X
272.
273.
Y
X
Y
X
100
Graph Sketching
274.
275.
Y
Y
X
276.
X
Y
X
6.10
Graph Sketching
In problems 277 to 303 sketch the graph of a function with the listed properties. State as much of the
following information as you can:
Where is the function increasing, decreasing, concave up, and concave down?
Where are the local maxima, local minima, and points of inflection?
What are the asymptotes and how does the function approach them (e.g., from above or below etc.)?
Use one page per problem and at least half the page for your sketch of the function.
Do not decide on a scale for your axes until you have come up with a picture of the function. You do not
have to use the same scale for both the x and y axes. The use of HB pencil and an eraser will make your
work easier.
6.10 Graph Sketching
277.
DOMAIN
R − {1}
101
f
f (0) = 0
f0
f 0 < 0 : −∞ < x < 1 or
f 00
f 00 > 0 : −∞ < x < 1
ASYMPTOTES
lim f (x) = 1
x→∞
f 00 < 0 : 1 < x < ∞
1<x<∞
lim f (x) = ∞
x→1+
lim f (x) = −∞
278.
0
R − {0}
f > 0 : −∞ < x < −1 or
1<x<∞
x→1−
00
f > 0 : −∞ < x < 0
f 00 < 0 : 0 < x < ∞
y =x+1
lim f (x) = ∞
x→0+
f 0 (1) = f 0 (−1) = 0
lim f (x) = −∞
x→0−
0
279.
R − {0}
f (2) = 0
f < 0 : −1 < x < 0 or
0<x<1
f 0 > 0 : −5 < x < 0 or
0<x<∞
f 00 > 0 : −9 < x < 0
f 00 (−9) = 0; f (−9) =
f 0 (−5) = 0; f (−5) = 1
f 00 < 0 : −∞ < x < −9 or
0
280.
R
f (−3) = 0
f (0) = 0
281.
R
282.
R
f (−3) = 0
f (−1) = 0
f (2) = 0
f (0) = −6
f (−5) = 0
f (0) = 0
f < 0 : −∞ < x < −5
f 0 > 0 : −2 < x < ∞
f 0 (−2) = 0; f (−2) = −5
f 0 < 0 : −∞ < x < −2
f 0 > 0 : −∞ < x < −2
f 0 > 0 : 23 < x < ∞
f 0 (−2) = f 0 ( 23 ) = 0
f 0 < 0 : −2 < x < 23
f 0 > 0 : −2 < x < ∞
0
f (−2) = 0; f (−2) = −3
0
283.
R
f (0) = 0
f (−3) = 3
284.
x 6= −2
0
0
f0 > 0 : 0 < x < 2
0
f (−3) = 4
f (3) = 4
f (0) = 0
f 0 < 0 : −2 < x < 0
f 0 < 0 : −∞ < x − 2
f 0 > 0 : −∞ < x < −3
0
0
0
f (−3) = f (3) = f (0) = 0
f 0 < 0 : −3 < x < 0
f0 < 0 : 3 < x < ∞
none
x→−∞
00
f <0:0<x<∞
f 00 > 0 : −5 < x < ∞
lim f (x) = 0
x→∞
00
00
lim f (x) = 0
x→−∞
00
f (−5) = f (5) = 0 = f (0)
f 00 < 0 : 0 < x < 5
f 00 < 0 : −∞ < x < −5
f 00 < 0 : −∞ < x < −2
00
f > 0 : −2 < x < 2
f 00 > 0 : −∞ < x < −4
f > 0 : −2 < x < 2
0
none
lim f (x) = ∞
f (0) = 0
00
f >0:0<x<3
lim f (x) = ∞
x→0−
lim f (x) = 6
f <0:2<x<∞
0
x→0+
x→∞
00
00
f >0:2<x<∞
1
2
f 00 > 0 : −∞ < x < 0
f > 0 : −5 < x < 0
f (−3) = f (3) = 0
f (0) = 2
f (0) =
0<x<∞
f 00 > 0 : −∞ < x < −1
f 00 (−1) = 0; f (−1) = −1
f 00 < 0 : −1 < x < ∞
f 00 > 0 : − 23 < x < ∞
f 00 (− 23 ) = 0
f 00 < 0 : −∞ < x < − 23
00
f < 0 : −∞ < x < −3
f 0 < 0 : 3 < x < −∞
x 6= 2
R
0
f (3) = 3
and
285.
f < 0 : −∞ < x < −2
f 0 > 0 : −3 < x < 3
y=2
lim f (x) = −∞
3
2
00
f >0:4<x<∞
f 00 (−4) = f 00 (−2) = f 00 (2) = f 00 (4) = 0
f 00 < 0 : −4 < x < −2
f 00 < 0 : 2 < x < 4
lim f (x) − 1
x→∞
lim f (x) = −1
x→−∞
x = −2
x=2
lim f (x) = 0
x→∞
lim f (x) = 0
x→−∞
102
Graph Sketching
DOMAIN
f
f (0) = 1
f (x) = f (−x)
286.
R
287.
R
f (0) = −30
f (−1) = 0
f (3) = 0
288.
x > −1
f (0) = 0
f0
f0 > 0 : 0 < x < ∞
f 0 (0) is undefined
f 0 < 0 : −∞ < x < 0
f 0 > 0 : −∞ < x < −1
f 0 > 0 : 34 < x < ∞
f 0 (−1) = f 0 ( 34 ) = 0
f 0 < 0 : −1 < x < 34
f0 > 0 : 1 < x <
0
f (2) = 0
f (1) =
f 0 ( 52 )
5
2
=0
0
289.
f < 0 : −1 < x < 1
f 0 < 0 : 52 < x < ∞
f0 > 0 : 0 < x < ∞
f (0) = −1
R
0
f (1) = 0
f 00
f 00 > 0 : −∞ < x < 0
f 00 > 0 : x < x < ∞
f 00 (0) is undefined
f 00 > 0 : − 13 < x < 32
f 00 > 0 : 3 < x < ∞
f 00 (− 13 ) = f 00 ( 32 ) = f 00 (3) = 0
f 00 < 0 : −∞ < x < − 13
f 00 < 0 : 32 < x < 3
f 00 > 0 : −1 < x < 2
00
f >0:3<x<∞
00
f (0) = 0
none
lim f (x) = 0
x→−1+
lim f (x) = 0
x→∞
00
f (2) = f (3) = 0
f 00 < 0 : 2 < x < 3
f 00 > 0 : −1 < x < 1
00
ASYMPTOTES
none
00
f (−1) = f (1) = 0
lim f (x) = 1
x→∞
lim f (x) = 1
x→−∞
290.
R
0
f (x) = f (−x)
f < 0 : −∞ < x < 0
f (0) = −5
f0 > 0 : 1 < x < ∞
0
f (3) = 0
f (1) = 0
f < 0 : −∞ < x < 1
R − {−1, 1}
f < 0 : −∞ < x < −1
f 00 < 0 : 1 < x < ∞
f 00 > 0 : − 13 < x < 2
f
0
291.
00
00
(− 13 )
= f (2) = 0
00
f 0 > 0 : −∞ < x < −2
f (2) = 5
f0 > 0 : 2 < x < ∞
f 00 > 0 : 0 < x < 1
0
f (2) = f (−2) = 0
0
f < 0 : −2 < x < −1
f 0 < 0 : −1 < x < 1
f0 < 0 : 1 < x < 2
00
f (0) = 0
00
f < 0 : −1 < x < 0
f 00 < 0 : 1 < x < ∞
292. (a) g 0 (x) exists for every x ∈ R
(b) g(x) = g(−x), x ∈ R
lim g(x) = 1
x→∞
(d) g(0) = 0
(e)
g 00 (x) > 0 on 0 < x < e and g 00 (x) < 0 on e < x < ∞
293. (a) f is symmetric about the y axis and the domain of f is R
(b)
(c)
lim f (x) = −∞ and lim f (x) = ∞
x→0
0
x→∞
f (x) > 0 on 0 < x < ∞
lim f (x) = −2
x→−∞
f (0) = 0
0
lim f (x) = 2
x→∞
− 13
f < 0 : −∞ < x <
f 00 < 0 : 2 < x < ∞
f 00 > 0 : −∞ < x < −1
f (−2) = −5
(c)
00
lim (f (x) − x) = 0
x→∞
lim f (x) = ∞
x→1+
lim f (x) = −∞
x→−1+
6.10 Graph Sketching
103
293. (d) f (1) = 0
(e)
f 00 (2) = 0
294. (a) f is symmetric about the origin and f (x) is defined for x 6= 0
(b)
(c)
lim f (x) = 0 and lim f 0 (x) = −1
x→∞
x→0+
f (x) > 0 on 0 < x < ∞
(d) f 0 (x) < 0 on 0 < x < ∞
295. (a) Domain of f is (−∞, −5) ∪ (−5, ∞)
(b) f (0) = 0
(c)
(d)
(e)
lim f (x) = −1 and lim f (x) = 0
x→∞
x→−∞
lim f (x) = ∞
lim f (x) = ∞ and
x→−5−
x→−5+
f 0 (x) > 0 on (−∞, −5) and f 0 (x) < 0 on (−5, ∞)
296. (a) Domain of f is R
(b) f (3) = 4
(c)
f 0 (x) = 32 , x ∈ R
297. (a) Domain of f is [0, 9]
(b) The point, (3, 2) is on the graph of f
(c)
f 0 (x) = − 13 , x ∈ [0, 9]
298. (a) Domain of f is R
(b) f (3) = 2 and f (7) =
(c)
14
3
f 0 (x) = c, x ∈ R and c is a constant.
299. (a) Domain of f is R
(b) f (−1) = 2 and f (3) = −6
(c)
f 00 (x) = 0, x ∈ R
300. (a) g(3) = 0
(b) g 0 (x) =
(c)
1
2
on (3, ∞)
g 0 (x) = − 21 on (−∞, 3)
104
Graph Sketching
√
301. (a) f 0 (0) = −2 and f ( 2) = 0
(b) f 0 (x) = 0 on (−∞, 0]
(c)
f 0 (x) = 2x on [0, ∞)
302. (a) f is a continuous function on R
(b) f (0) = 0 and f (−4) = 0
(c)
lim f (x) = ∞ and lim f (x) = −∞
x→∞
x→−∞
0
(d) f (x) > 0 on (−∞, −2) ∪ (0, ∞)
(e)
f 0 (x) < 0 on (−2, 0)
(f) f 00 (x) < 0 on (−∞, 23 ) and f 00 (x) > 0 on ( 32 , ∞)
(g)
lim f 0 (x) = −∞ and lim f 0 (x) = ∞
x→0−
x→0+
303. (a) f is symmetric about the y axis
(b)
(c)
lim f (x) = −3
x→∞
lim f (x) = −∞ and lim− f (x) = +∞
x→2+
x→2
(d) f (0) = 3
(e)
Example:
f 0 (0) = 0
Graph f (x) =
x+1
.
x2 + 2x
Write f (x) =
x+1
x2 +2x
=
x+1
x(x+2) .
Domain of f is R − {0, −2}.
Justification for following chart is:
x + 1 > 0 =⇒ x > −1, x + 2 > 0 =⇒ x > −2, and x > 0
−−−
−−−
−−−
f <0
+++
−−−
−−−
−2
f >0
+++
+++
−−−
−1
f <0
+++
+++
+++
0
f >0
(x + 2)
(x + 1)
x
1 + x1
x+1
= lim
= 0 implies x axis is a horizontal asymptote. The above chart shows how the
2
x→∞ x + 2x
x→∞ x + 2
function approaches the x axis.
lim
Vertical asymptotes (you must compute these limits!)
lim f (x) = −∞
x→−2−
lim f (x) = ∞
x→−2+
lim f (x) = −∞ lim+ f (x) = ∞
x→0−
x→0
6.10 Graph Sketching
105
Intercepts: x + 1 = 0 ⇒ x = −1 is the x intercept.
f 0 (x) =
x2 + 2x − (x + 1)(2x + 2)
−(x2 + 2x + 2)
(x + 1)2 + 1
=
=
−
<0
(x2 + 2x)2
(x2 + 2x)2
(x2 + 2x)2
for all real numbers; thus f is decreasing on its domain.
f 00 (x)
=
f 00 (x)
=
f 00 (x)
=
f 00 (x)
=
(2x + 2)(x2 + 2x)2 − 2(x2 + 2x + 2)(x2 + 2x)(2x + 2)
−(x2 + 2x)4
2
2
2x (x + 1)(x + 2) − 4x(x2 + 2x + 2)(x + 2)(x + 1)
−x4 (x + 2)4
2x(x + 1)(x + 2){x(x + 2) − 2(x2 + 2x + 2)}
−x4 (x + 2)4
−2(x + 1){(x + 1)2 + 3}
−x3 (x + 2)3
(x + 2)3 > 0 ⇒ x + 2 > 0 ⇒ x > −2, x3 > 0 ⇒ x > 0 and x + 1 > 0 ⇒ x > −1 provides information required
for this chart:
−−−
−−−
−−−
−2
f 00 < 0
concave
down
−−−
−−−
+++
−1
f 00 > 0
concave
up
−−−
+++
+++
0
f 00 < 0
concave
down
+++
+++
+++
x
(x + 1)
(x + 2)3
f 00 > 0
concave
up
Inflection point occurs at x = −1
f(x)
−2
Example:
−1
Graph f (x) = (x − 3)2/3 (x + 1).
Domain of f is R because cube root function is always defined.
Intercepts f (0) = (−3)2/3 = 91/3 , f (−1) = 0, and f (3) = 0.
Asymptotes:
lim f (x) = −∞ and lim f (x) = ∞.
x→−∞
There are no asymptotes.
x→∞
x
106
Graph Sketching
Increasing / decreasing:
f 0 (x) =
2(x + 1)
5x − 7
+ (x − 3)2/3 =
.
3(x − 3)1/3
3(x − 3)1/3
Since 5x − 7 > 0 =⇒ x > 7/5 and x − 3 > 0 =⇒ x > 3 we can construct the following chart:
−−−
−−−
f increasing r
(−∞, 7/5)
max/min
+ + ++
− − −−
7/5
decreasing r
(7/5, 3)
+++
+++
3
increasing
(3, ∞)
5x − 7
x−3
By the First Derivative Test we conclude f has a local maximum at x = 7/5 and a local minimum at 3
where f 0 is undefined.
Concavity:
f (x)
=
5 3(x − 3)1/3 − 3(5x − 7) 13 (x − 3)−2/3 (x − 3)2/3
9(x − 3)2/3
(x − 3)2/3
f 00 (x)
=
15(x − 3) − (5x − 7)
2(5x − 19)
=
9(x − 3)4/3
9(x − 3)4/3
00
f 00 (x) > 0 iff 5x − 19 > 0 =⇒ x >
Concave up on the interval
19
5 ,∞
19
5 .
.
Concave down on the interval −∞, 19
5 .
Inflection points:
Since f (19/5) is defined and concavity changes at the point (19/2, f (19/2)), it is the only inflection point
6.10 Graph Sketching
107
y
(19/5, f(19/5))
x
−1
7/5
3
√
Example: g(x) = 1/ 1 + x3 . Domain of g is (−1, ∞) because 1 + x3 > 0 is needed for square root to be
defined in the denominator of g
√
lim g(x) = 0 because 1 + x3 increases without bound.
x→∞
lim g(x)
x→−1+
=∞
because numerator remains nonzero as denominator
approaches zero through positive values.
2
√3x
2
−3x2
2 1+x
g 0 (x) = −
.
=
3
(1 + x )
2(1 + x3 )3/2
g is always decreasing on (−1, ∞) since g 0 (x) < 0 on domain of g. (1 + x3 ) > 0 and x2 ≥ 0 except at x = 0.
There are NO maximum or minimum points.
3 2x(1 + x3 )3/2 − 3x4 (1 + x3 )1/2
00
g (x) = −
2
(1 + x3 )3
00
g (x)
=
√
√
−3x 1 + x3 2(1 + x3 ) − 3x3
−3x 1 + x3 (2 − x3 )
=
.
2(1 + x3 )3
2(1 + x3 )3
−−−
+++
−−−
g 00 > 0
−−−
+++
+++
0
g 00 < 0
−−−−−
−−−−−
+++++
21/3
00
g >0
(−3)
(2 − x3 )
108
Graph Sketching
Conclusion: g is concave up on (−1, 0) and (21/3 , ∞) and concave down on (0, 21/3 ). There is a change of
concavity at x = 0 and x = 21/3 . Also g is defined at these two points so they qualify as inflection points.
g(x)
•
•
1
−1
•
•
1
23
x
For each function or relation given in problems numbered 304 to 373, sketch the graph.
Determine
(a) The domain.
(b) Any symmetry, if it exists. State why.
(c) The asymptotic behaviour. State why.
(d) The intervals where the function or relation is increasing, decreasing. State why.
(e) The relative maxima, relative minima, the maximum and the minimum if they exist. State why.
(f) The intervals where the function or relation is concave up, concave down. State why.
(g) The points of inflection. State why.
Include the above information on the graph.
304. y =
5
x−3
305. y =
306. y =
2
x+5
307. f (R) =
308. y =
2x+1
5−3x
309. y =
3x+1
x
311. y =
x2 +2
x2 +x−2
310. y = x2 + |x2 − 4|
x−1
x+1
E2 R
r+R
E > 0, r > 0 are constants
6.10 Graph Sketching
312. f (x) = |x2 + 2| + x2
314. f (x) =
316. y =
|x+a|
x+a
109
313. y =
x4
x4 +1
315. g(x) =
1
|x|
x2 +2x−3
x−1
317. h(x) = x +
318. y = cos θ +
π
4
319. y = sin2 x
320. f (θ) = | sin θ|
321. h(x) = 4 sin
322. f (θ) = 1 + cos θ
321. f (x) = tan
324. f (x) = x + 1 −
326. g(x) =
1
x2 −1
x2 −3x−4
x2 −3x
325. f (x) =
327. y =
x2
x4 −1
(x−4)
(x+1)(x−2)
329. y =
x3
x3 −2
330. y =
(x−1)2
x+3
331. y =
x2 −1
x−2
332. y =
x2 −3−4
x2 −3x
333. f (x) =
x2 −4x−5
x−5
x
2
x
π
x2 +x−2
x+2
328. y =
334. f (x) =
1
x
(x+1)2
(x+2)2 (x−1)
335. f (x) = −2x2 + 3x − 1
336. f 0 (x) and f 00 (x) for the preceding question
337. g(x) = x3 − 5x2 + 3x + 2
338. g 0 (x), g 00 (x), g 000 (x) for the preceding question
339. h(x) = x4 − 2x3 − 2x2 + 1
3
3
341. y = x 2 (x − 4)
342. y =
x2
x−2
4
344. y = x 3 (x − 4)
1
3
346. y = (x + 2)3 (x − 1)
343. y = x 3 (x2 − 4)
345. y = x 2 (x2 + 2)
347. y =
340. A(z) = z 4 − 6z 2 + 8z + 3
√x
x−9
349. P (R) =
350. f (x) =
352. f (x) =
E2 R
(r+R)2
348. y =
−
x4
2
+
x3
3
−1
E > 0, r > 0 are constraints
|x|
x2 +1
√
x5
5
1 + x3
354. f (x) = 2 tan x − sec x
351. f (x) = 2 sin 2x + cos 4x
353. f (x) = sin2 x +
355. f (x) =
x(x−1)
x2 −4
√
3 sin x
110
Graph Sketching
356. f (x) =
√
3
x (x + 4)
358. f (x) = x3 − 3x2 + 3x − 2
360. f (x) =
3
x2 −4
3
x
359. f (x) = − x+3
√
361. f (x) = x2 5 − x
3x−2
2x+3
362. f (x) = −3x2 + 2x − 1
364. f (x) =
357. f (x) =
x2 −2x−1
x+1
363. f (x) =
2x4 −1
x2
2
5
365. f (x) = 5x 3 − x 3
π
sin x, x <
2
366. f (x) =
 x + 1 − π, x > π
2
2


367. f (x) =
(
x2 + 1, x < −1
x + 1, x > −1
368. f (x) =
1
x2 −3
369. f (x) =
370. f (x) =
x−2
(x2 −4)(x−1)
371. f (x) = csc x1
372. f (x) =
(x−1)2
x+2
373. f (x) = x(x − |x|)
6.11
√x−2
x2 −4
Graphs of Trancendental Functions
For problems numbered 374 to 394 sketch the functions f and g on the same coordinate axes.
374. f (x) = ex
and g(x) = 2x
375. f (x) = ex
and g(x) = 3x
376. f (x) = ex
and g(x) = 10x
377. f (x) = ex
and g(x) = e−x
378. f (x) = 2x
and g(x) = 2−x
379. f (x) = 10x
and g(x) = 10−x
380. f (x) = 2x
and g(x) = log2 x
381. f (x) = 3x
and g(x) = log3 x
x
382. f (x) = 21
and g(x) = log 21 x
383. f (x) = 3|x|
and g(x) = 10|x|
384. f (x) = ln x and g(x) = log10 x
385. f (x) = ln x
and g(x) = ln(x + 1)
386. f (x) = ln x
and g(x) = ln(x − 1)
387. f (x) = 3x
388. f (x) = 10x
and g(x) = 3x−1
and g(x) = 10x+10
389. f (x) = ln(2x)
and g(x) = ln x
6.11 Graphs of Trancendental Functions
390. f (x) = ln
391. f (x) =
x
3
1
2x
111
and g(x) = ln x
x
and g(x) = 12
392. f (x) = log2 x and g(x) = log4 x
393. f (x) = log2 x and g(x) = log2 (x + 1)
x
394. f (x) = 13
and g(x) = log 31 x
For problems numbered 395 to 402, find the equation of the line tangent to the curve at the point x. Question
numbered 395 is solved as an example.
395. Curve is y = xe2x , x = 2.
Solution: y 0 = 2xe2x + e2x
At x = 2, y 0 = 4e4 + e4 = 5e4
At x = 2, y = 2e4
The tangent line has the equation y − y1 = m(x − x1 )
y − 2e4 = 5e4 (x − 2)
5e4 x − y = 8e4
or
Ans.
396. Curve is y = 2x , x = −1
397. Curve is y = ln x, x = e
398. Curve is y = x2 e−x , x = 1
[2x − y − e = 0]
399. Curve is y = x ln x, x = e
400. Curve is y = x−2 e2x , x = −1
401. Curve is y = x ln x2 , x = 3
402. Curve is y = (ln x)2 , x = e
For problems numbered 403 to 421, find:
(a) the intervals in which f is increasing, and those in which f is decreasing;
(b) the relative maxima and minima;
(c) the intervals where the curve is concave upward and where it is concave downward;
(d) the inflection points.
and sketch the graph.
403. f (x) = 102−x
404. f (x) = 10−(x+1)
405. f (x) = e−x
406. f (x) = ex
407. f (x) = xe−x
408. f (x) = ex
409. f (x) =
ex −e−x
2
2
3
410. f (x) =
ex +e−x
2
112
Graph Sketching
2
411. f (x) = e−x
412. f (x) = x2 e−x
413. f (x) = ln(x2 − 1)
414. f (x) = 2xe−( 2 )x
2
1
2
415. f (x) = x2 e−x
416. f (x) = ln(1 + x)
417. f (x) = x ln x
418. f (x) = 2x2 ln x
419. f (x) = 2x(ln x)2
420. f (x) = e−x sin x
421. f (x) = e−x cos x
Sketch the graphs of functions numbered 422 to 456 and provide the following information.
(a) The domain.
(b) Any symmetry, if it exists. State why.
(c) The asymptotic behaviour. State why.
(d) The intervals where the function or relation is increasing, decreasing. State why.
(e) The relative maxima, relative minima, the maximum and minimum if they exist. State why.
(f) The intervals where the function or relation is concave up, concave down. State why.
(g) The points of inflection. State why.
Include all the above information on the graph.
In cases where the function or the derivative takes an “indeterminate” form, find the limit.
422. y = xex
423. y = x2 ln x
424. y = x ln x
425. y =
426. y =
ex
x
ln x
x
1
427. y = e− x
2
428. y = xe−x
429. y = 2x(ln x)2
430. y =
ln x
x2
431. y =
e−x
x
432. y =
1−ln x
x
433. y =
1+ln x
x
434. y =
e−x
x2
435. y =
x
ln x
436. y =
ln x
x3
437. y =
ln2 x
x2
438. y =
x
ln2 x
439. y =
ex
x2 −3
440. y =
cos x
ex
441. y = e− x2
442. y = e|x|
1
443. y = | ln x|
6.12 Polynomials
113
√
x
444. y = ln |x|
445. y = e
446. y = x − ln x
447. y = x + ln x
448. y = x ln x − x
449. y 2 = ln x
450. y 2 = xe−x
451. y 2 = −xe−x
452. y 2 = e−x (8 − x2 )
453. y 2 = x ln x
454. y 2 = −x ln x
455. y 2 = − ln x
456. For y = xe−x , sketch the graph of y, y 0 and y 00 .
6.12
Polynomials
457. Find polynomials which have the following properties then draw their graphs making estimates whenever necessary.
(a)
no roots, one minimum
(b)
no roots, one maximum
(c)
one root, no extremum
(d)
one root, no extremum
(e)
one root, one maximum
(f)
one root, one minimum
(g)
one root, no extremum; one point of inflection
(h)
one root, two extremum
(i)
three roots all distinct
(j)
three roots all distinct and five extremum
(k)
two roots, two extremum
(l)
two roots, three extremum
(m)
no roots, three extremum
458. Same instructions as in question 458
(a)
a root at x = 1 and a maximum at x = 2
(b)
a root at x = 1 and a minimum at x = 2
(c)
roots at x = 2, 4 and a maximum at x = 3
(d)
roots at x = 2, 4 and a minimum at x = 3
(e)
a root at x = 1 and two extremum somewhere
(f)
a root at x = 0 and a point of inflection somewhere
459. graph the following polynomials estimating whenever necessary. Include numerical estimates (or exact
values if possible) of roots and extreme values.
114
Graph Sketching
(a)
x3 − x + 1
(b)
x3 + x + 1
(c)
x3 − 2x + 1
(d)
x3 + 2x2 + 4x + 3
(e)
x3 + 4x2 + 2x − 3
(f)
x3 + 2x + 3
(g)
x3 + 5x2 + 7x + 3
(h)
x4 + 3x3 + 4x2 + 3x + 1
(i)
x4 + x + 1
(j)
x4 + 3x2 + 1
(k)
x4 + 3x2 − 20
(l)
x4 + 4x3 + 4x2 + 2
(m)
x5 − x + 1
(n)
x5 + x + 1
(o)
x2 + 2x + 1
(p)
x3 + 3x2 + 3x + 1
(q)
x4 + 4x3 + 6x2 + 4x + 1
(r)
x5 + 5x4 + 10x3 + 10x2 + 5x + 1
6.13
Functions with Parameters
For the function or relation given in the following problems do the following:
(a) For each problem decide how to partition the range of the parameter. If c is the parameter the three
cases: c < 0, c = 0 and c > 0 may be appropriate. Other possibilities are c < −1, c = −1 or just one
case c ∈ R.
(b) For each case compute and describe all the important features of its graph. Draw a neat sketch of the
graph and label these features on your diagram:
f (x) = x2 − 2ax + a2
(b)
ax2 + x + 1
(c)
f (x) = x2 + x − a
(d)
x2 + ax + 3
(e)
f (x) = x2 + a2 x − 3
(f)
(x − a)2 + 1
461. (a)
A(x) = |x − a| − x
(b)
A(x) =
462. (a)
`(x) = (a2 − 1)x
(b)
`(x) = (a + 3)x − a
(b)
p(x) = (x − a)3 + (x − a)2 − χ + a
460. (a)
(c)
(x−a)2
|x−a|
`(x) = ax + (a − 4)
463. (a)
p(x) = (x − a)2 + 2(x − a) − 6
464. (a)
g(x) =
1
(a−1)x
(b)
g(x) =
465. (a)
f (x) =
4χ
1−ax2
(b)
f (x) = x +
(b)
g(x) =
466. g(x) =
467. (a)
(c)
(
1
x−a
1
x−a
x2 − a if x ≥ a
a − x2 if x < a
√
g(x) = 2 x − ax
√
g(x) = x2 − a2
468. (x − 1)2 + (y − 1)2 = 2 − `n(c)
469. x2 + (y − c)2 = c2
(d)
g(x) =
√
√
x−a
ax
−
1
x+a