Problem Sheet #26CEGEP CHAMPLAIN - ST. LAWRENCE
201-NYA-05: Differential Calculus
Patrice Camire
Curve Sketching
1. Complete the following statements about the graph of the function y = f(x).
(a) Ifdy
dx> 0, then the graph of y = f(x) is ...
(b) Ifdy
dx< 0, then the graph of y = f(x) is ...
(c) Ifdy
dx= 0, then the graph of y = f(x) is ...
(d) Ifd 2y
dx2> 0, then the graph of y = f(x) is ...
(e) Ifd 2y
dx2< 0, then the graph of y = f(x) is ...
Sketch the graph of the following functions indicating the y-intercept, all zeros, critical points,inflection points, vertical and horizontal asymptotes, local/global maximums and minimums. Makesure to accurately portray concavity and to justify fully all key features of the graph.
Essential problems:
2. y = −x2 + 3x− 2
3. y = x3 − 3x
4. y = x4 − 4x3
5. y = 3x5 − 5x3 + 2
6. y =(x− 2)2
x2
7. y = xex
8. y =x2 − 3
x3
9. y =2x2
x2 − 1
10. y =x3 − 8
x3 + 8
11. y =x2
(x+ 1)3
12. y = 5x2/3 − x5/3
Additional problems:
13. y = x2 − 4x+ 1
14. y = x3 − 3x2 + 5
15. y =ex
ex + 1
16. y =4(1 − x)
x2
17. y =x2
x2 + 3
18. y =x
5+
5
x
19. y =x
x2 + 4
20. y = ln(x2 + 9)
21. y = x2/3(x− 1)1/3
Answers
1. (a) increasing.
(b) decreasing.
(c) flat.
(d) concave up.
(e) concave down.
Essential problems:
2. y = −x2 + 3x− 2 = −(x− 1)(x− 2) y′ = −2x+ 3 y′′ = −2
y-intercept: y(0) = −2
zeros: x = 1, 2
vertical asymptotes: none
critical points: y′ = 0 : x = 3/2
possible inflection points: none
classification of critical points:
x = 3/2y′(3/2) = 0y′′(3/2) < 0
•
table of values:x 0 1 2 3/2
y −2 0 0 1/4
horizontal asymptotes: limx→−∞
y = limx→−∞
x2(
−1 +3
x− 2
x2
)
= −∞
limx→∞
y = limx→∞
x2(
−1 +3
x− 2
x2
)
= −∞
−1
−2
−3
1
1 2 3−1−2x
y
•
3. y = x3 − 3x = x(x2 − 3) y′ = 3x2 − 3 = 3(x− 1)(x+ 1) y′′ = 6x
y-intercept: y(0) = 0
zeros: x = 0,±√3 (
√3 ≈ 1.732)
vertical asymptotes: none
critical points: y′ = 0 : x = −1, 1
possible inflection points: y′′ = 0 : x = 0
classification of critical points:
x = −1 x = 1y′(−1) = 0 y′(1) = 0y′′(−1) < 0 y′′(1) > 0
••
table of values:x 0 −
√3
√3 −1 1
y 0 0 0 2 −2
horizontal asymptotes: limx→−∞
y = limx→−∞
x3(
1− 3
x2
)
= −∞
limx→∞
y = limx→∞
x3(
1− 3
x2
)
= ∞
−1
−2
−3
1
2
1 2−1−2−3x
y
•
•
•
4. y = x4 − 4x3 = x3(x− 4) y′ = 4x3 − 12x2 = 4x2(x− 3) y′′ = 12x2 − 24x = 12x(x− 2)
y-intercept: y(0) = 0
zeros: x = 0, 4
vertical asymptotes: none
critical points: y′ = 0 : x = 0, 3
possible inflection points: y′′ = 0 : x = 0, 2
classification of critical points:
x = 0 x = 3y′(0) = 0 and y′′(0) = 0 y′(3) = 0y′′(−0.1) > 0 and y′′(0.1) < 0 y′′(3) > 0
••
table of values:x 0 2 3 4
y 0 −16 −27 0
horizontal asymptotes: limx→−∞
y = limx→−∞
x4(
1− 4
x
)
= ∞
limx→∞
y = limx→∞
x4(
1− 4
x
)
= ∞
−9
−18
−27
9
18
27
1 2 3 4−1−2x
y
•
•
•
5. y = 3x5 − 5x3 + 2 y′ = 15x2(x2 − 1) y′′ = 30x(2x2 − 1)
y-intercept: y(0) = 2
zeros: x = 1 (the other one is not easily found)
vertical asymptotes: none
critical points: y′ = 0 : x = 0,−1, 1
possible inflection points: y′′ = 0 : x = 0,−√2
2,
√2
2(√2/2 ≈ 0.707)
classification of critical points:
x = −1 x = 0 x = 1y′(−1) = 0 y′(0) = 0 and y′′(0) = 0 y′(1) = 0y′′(−1) < 0 y′′(−0.1) > 0 and y′′(0.1) < 0 y′′(1) > 0
• ••
table of values:x −1 −
√2/2 0
√2/2 1
y 4 ≈ 3.237 2 ≈ 0.763 0
horizontal asymptotes: limx→−∞
y = limx→−∞
x5(
3− 5
x2+
2
x5
)
= −∞
limx→∞
y = limx→∞
x5(
3− 5
x2+
2
x5
)
= ∞
−1
−2
1
2
3
4
1−1−2x
y
•
•
•
•
•
6. y =(x− 2)2
x2y′ =
4(x− 2)
x3y′′ =
8(3− x)
x4
y-intercept: y(0) = 4/0 is undefined
zeros: x = 2
vertical asymptotes: x = 0, since limx→0
(x− 2)2
x2
[
=4
0+case
]
= ∞
critical points: y′ = 0 : x = 2 and y′ = ∅ : x = 0
possible inflection points: y′′ = 0 : x = 3 and y′′ = ∅ : x = 0
classification of critical points:
x = 0 x = 2y′(0) = ∅ and y′′(0) = ∅ y′(2) = 0y′′(−0.1) > 0 and y′′(0.1) > 0 y′′(2) > 0
•
table of values:x −1 2 3 0− 0+
y 9 0 1/9 = 0.1 ∞ ∞
horizontal asymptotes: limx→−∞
y = limx→−∞
(
1− 2
x
)2
= 1
limx→∞
y = limx→∞
(
1− 2
x
)2
= 1
additional work: To show that x = 3 is an inflection point, we evaluate the second derivative atx = 4: y′′(4) < 0. This shows that the graph is concave down to the right of x = 3.
−1
1
2
3
4
5
6
1 2 3 4 5 6 7 8−1−2−3−4−5−6−7x
y
• •
7. y = xex y′ = (x+ 1)ex y′′ = (x+ 2)ex
y-intercept: y(0) = 0
zeros: x = 0
vertical asymptotes: none
critical points: y′ = 0 : x = −1
possible inflection points: y′′ = 0 : x = −2
classification of critical points:
x = −1y′(−1) = 0y′′(−1) > 0
•
table of values:x 0 −1 −2
y 0 −e−1 ≈ −0.37 −2e−2 ≈ −0.27
horizontal asymptotes:
limx→−∞
y = limx→−∞
xex [−∞ · 0 case] = limx→−∞
x
e−x
[−∞∞ case
]
L.H.= lim
x→−∞
1
−e−x
[
1
−∞ case
]
= 0
limx→∞
y = limx→∞
xex = ∞
additional work: To show that x = −2 is an inflection point, we evaluate the second derivative atx = −3: y′′(−3) < 0. This shows that the graph is concave down to the left of x = −2.
−1
1
2
3
4
5
6
7
1 2−1−2−3−4−5x
y
••
8. y =x2 − 3
x3y′ =
9− x2
x4y′′ =
2(x2 − 18)
x5
y-intercept: y(0) = −3/0 is undefined
zeros: x = ±√3 ≈ ±1.732
vertical asymptotes: x = 0, since limx→0−
x2 − 3
x3
[−3
0−case
]
= ∞ and limx→0+
x2 − 3
x3
[−3
0+case
]
= −∞
critical points: y′ = 0 : x = −3, 3 and y′ = ∅ : x = 0
possible inflection points: y′′ = 0 : x = −√18,
√18 and y′′ = ∅ : x = 0 (
√18 ≈ 4.243)
classification of critical points:
x = −3 x = 0 x = 3y′(−3) = 0 y′(0) = ∅ and y′′(0) = ∅ y′(3) = 0y′′(−3) > 0 y′′(−0.1) > 0 and y′′(0.1) < 0 y′′(3) < 0
••
table of values:x −
√3
√3 −3 3 −
√18
√18 0− 0+
y 0 0 −2/9 = −0.2 2/9 = 0.2 ≈ −0.196 ≈ 0.196 ∞ −∞
horizontal asymptotes: limx→−∞
y = limx→−∞
(
1
x− 3
x3
)
= 0 limx→∞
y = limx→∞
(
1
x− 3
x3
)
= 0
additional work: To show that x = −√18 is an inflection point, we evaluate the second derivative
at x = −5: y′′(−5) < 0. This shows that the graph is concave down to the left of x = −√18.
To show that x =√18 is an inflection point, we evaluate the second derivative at x = 5: y′′(5) > 0.
This shows that the graph is concave up to the right of x =√18.
−1
−2
−3
−4
−5
1
2
3
4
5
1 2 3 4 5−1−2−3−4−5x
y
••
••
9. y =2x2
x2 − 1=
2x2
(x− 1)(x+ 1)y′ =
−4x
(x2 − 1)2y′′ =
4(3x2 + 1)
(x2 − 1)3
y-intercept: y(0) = 0
zeros: x = 0
vertical asymptotes:
x = −1 : limx→−1−
2x2
(x− 1)(x+ 1)
[
2
0+case
]
= ∞ and limx→−1+
2x2
(x− 1)(x + 1)
[
2
0−case
]
= −∞
x = 1 : limx→1−
2x2
(x− 1)(x+ 1)
[
2
0−case
]
= −∞ and limx→1+
2x2
(x− 1)(x+ 1)
[
2
0+case
]
= ∞
critical points: y′ = 0 : x = 0 and y′ = ∅ : x = −1, 1
possible inflection points: y′′ = ∅ : x = −1, 1
classification of critical points:
x = −1 x = 0 x = 1y′(−1) = ∅ and y′′(−1) = ∅ y′(0) = 0 y′(1) = ∅ and y′′(1) = ∅y′′(−1.1) > 0 and y′′(−0.9) < 0 y′′(0) < 0 y′′(0.9) < 0 and y′′(1.1) > 0
•
table of values:x 0 −2 2 −1− −1+ 1− 1+
y 0 8/3 = 2.6 8/3 = 2.6 ∞ −∞ −∞ ∞
horizontal asymptotes: limx→−∞
y = limx→−∞
2x2
x2 − 1·
1
x2
1
x2
= limx→−∞
2
1− 1
x2
= 2
limx→∞
y = limx→∞
2x2
x2 − 1·
1
x2
1
x2
= limx→∞
2
1− 1
x2
= 2
−1
−2
−3
−4
−5
1
2
3
4
5
1 2 3 4 5−1−2−3−4−5x
y
•
10. y =x3 − 8
x3 + 8y′ =
48x2
(x3 + 8)2y′′ =
192x(4 − x3)
(x3 + 8)3
y-intercept: y(0) = −1 zeros: x = 2
vertical asymptotes:
x = −2 : limx→−2−
x3 − 8
x3 + 8
[−16
0−case
]
= ∞ and limx→−2+
x3 − 8
x3 + 8
[−16
0+case
]
= −∞
critical points: y′ = 0 : x = 0 and y′ = ∅ : x = −2
possible inflection points: y′′ = 0 : x = 0,3√4 and y′′ = ∅ : x = −2 ( 3
√4 ≈ 1.587)
classification of critical points:
x = −2 x = 0y′(−2) = ∅ and y′′(−2) = ∅ y′(0) = 0 and y′′(0) = 0y′′(−2.1) > 0 and y′′(−1.9) < 0 y′′(−0.1) < 0 and y′′(0.1) > 0
•
table of values:x −3 0 3
√4 2 −2− −2+
y 35/19 ≈ 1.842 −1 −0.3 0 ∞ −∞
horizontal asymptotes: limx→−∞
y = limx→−∞
x3 − 8
x3 + 8·
1
x3
1
x3
= limx→−∞
1− 8
x3
1 + 8
x3
= 1
limx→∞
y = limx→∞
x3 − 8
x3 + 8·
1
x3
1
x3
= limx→∞
1− 8
x3
1 + 8
x3
= 1
additional work: To show that x = 3√4 is an inflection point, we evaluate the second derivative at
x = 2: y′′(2) < 0. This shows that the graph is concave down to the right of x = 3√4.
−1
−2
−3
−4
−5
1
2
3
4
5
1 2 3 4 5−1−2−3−4−5x
y
••
11. y =x2
(x+ 1)3y′ =
x(2− x)
(x+ 1)4y′′ =
2(x2 − 4x+ 1)
(x+ 1)5
y-intercept: y(0) = 0 zeros: x = 0
vertical asymptotes:
x = −1 : limx→−1−
x2
(x+ 1)3
[
1
0−case
]
= −∞ and limx→−1+
x2
(x+ 1)3
[
1
0+case
]
= ∞
critical points: y′ = 0 : x = 0, 2 and y′ = ∅ : x = −1
possible inflection points: y′′ = 0 : x = 2±√3 ≈ 0.268, 3.732 and y′′ = ∅ : x = −1
classification of critical points:
x = −1 x = 0 x = 2y′(−1) = ∅ and y′′(−1) = ∅ y′(0) = 0 y′(2) = 0y′′(−1.1) < 0 and y′′(−0.9) > 0 y′′(0) > 0 y′′(2) < 0
••
table of values:x −2 0 2−
√3 2 2 +
√3 −1− −1+
y −4 0 ≈ 0.0352 ≈ 0.148 ≈ 0.1314 −∞ ∞
horizontal asymptotes: limx→−∞
y = limx→−∞
x2
(x+ 1)3·
1
x3
1
x3
= limx→−∞
1
x(
1 + 1
x
)3= 0
limx→∞
y = limx→∞
x2
(x+ 1)3·
1
x3
1
x3
= limx→∞
1
x(
1 + 1
x
)3= 0
additional work: To show that x = 2+√3 is an inflection point, we evaluate the second derivative
at x = 4: y′′(4) > 0. This shows that the graph is concave up to the right of x = 2 +√3.
−1
−2
−3
1
2
3
1 2 3 4 5−1−2−3−4−5x
y
• • • •
12. y = 5x2/3 − x5/3 = (5− x)x2/3 y′ =5(2 − x)
3x1/3y′′ =
−10(x+ 1)
9x4/3
y-intercept: y(0) = 0
zeros: x = 0, 5
vertical asymptotes: none
critical points: y′ = 0 : x = 2 and y′ = ∅ : x = 0
possible inflection points: y′′ = 0 : x = −1 and y′′ = ∅ : x = 0
classification of critical points: (A line of slope 0 is horizontal and a line of slope ±∞ is vertical.)
x = 0 x = 2y′(0−) = −∞ and y′(0+) = ∞ y′(2) = 0y′′(−0.1) < 0 and y′′(0.1) < 0 y′′(2) < 0
••
table of values:x −1 0 2 5
y 6 0 ≈ 4.762 0
horizontal asymptotes: limx→−∞
y = limx→−∞
(5− x)x2/3 = ∞limx→∞
y = limx→∞
(5− x)x2/3 = −∞
additional work: To show that x = −1 is an inflection point, we evaluate the second derivative atx = −2: y′′(−2) > 0. This shows that the graph is concave up to the left of x = −1.
−1
−2
1
2
3
4
5
6
7
8
9
1 2 3 4 5−1−2−3x
y
•
•
•
Additional problems:
13. y = x2 − 4x+ 1 y′ = 2(x− 2) y′′ = 2
y-intercept: y(0) = 1
zeros: x = 2±√3 ≈ 0.268, 3.732
vertical asymptotes: none
critical points: y′ = 0 : x = 2
possible inflection points: none
classification of critical points:
x = 2y′(2) = 0y′′(2) > 0
•
table of values:x 2 0 ≈ 0.268 ≈ 3.732
y −3 1 0 0
horizontal asymptotes: limx→−∞
y = limx→−∞
x2(
1− 4
x+
1
x2
)
= ∞
limx→∞
y = limx→∞
x2(
1− 4
x+
1
x2
)
= ∞
−1
−2
−3
−4
1
2
3
4
5
1 2 3 4−1−2x
y
•
14. y = x3 − 3x2 + 5 y′ = 3x2 − 6x = 3x(x− 2) y′′ = 6x− 6 = 6(x− 1)
y-intercept: y(0) = 5
zeros: not easily found
vertical asymptotes: none
critical points: y′ = 0 : x = 0, 2
possible inflection points: y′′ = 0 : x = 1
classification of critical points:
x = 0 x = 2y′(0) = 0 y′(2) = 0y′′(0) < 0 y′′(2) > 0
••
table of values:x 0 1 2
y 5 3 1
horizontal asymptotes: limx→−∞
y = limx→−∞
x3(
1− 3
x+
5
x3
)
= −∞
limx→∞
y = limx→∞
x3(
1− 3
x+
5
x3
)
= ∞
−1
−2
1
2
3
4
5
1 2 3−1−2−3x
y
•
•
•
15. y =ex
ex + 1y′ =
ex
(ex + 1)2y′′ =
ex(1− ex)
(ex + 1)3
y-intercept: y(0) = 1/2
zeros: none
vertical asymptotes: none
critical points: none
Since y′ > 0 for all values of x, then the graph is always increasing.
possible inflection points: y′′ = 0 : x = 0
table of values:x 0
y 1/2
horizontal asymptotes: limx→−∞
y = limx→−∞
ex
ex + 1
[
=0+
1case
]
= 0
limx→∞
y = limx→∞
ex
ex + 1
1
ex
1
ex= lim
x→∞
1
1 + 1
ex= 1
additional work: To show that x = 0 is an inflection point, we evaluate the second derivative leftand right: y′′(−0.1) > 0 and y′′(0.1) < 0. This shows that the graph is concave up to the left ofx = 0 and concave down to the right of x = 0.
−1
1
1 2 3 4−1−2−3−4x
y
•
16. y =4(1 − x)
x2y′ =
4(x− 2)
x3y′′ =
8(3− x)
x4
y-intercept: y(0) = 4/0 is undefined
zeros: x = 1
vertical asymptotes: x = 0, since limx→0
4(1 − x)
x2
[
=4
0+case
]
= ∞
critical points: y′ = 0 : x = 2 and y′ = ∅ : x = 0
possible inflection points: y′′ = 0 : x = 3 and y′′ = ∅ : x = 0
classification of critical points:
x = 0 x = 2y′(0) = ∅ and y′′(0) = ∅ y′(2) = 0y′′(−0.1) > 0 and y′′(0.1) > 0 y′′(2) > 0
•
table of values:x 1 2 3 −1 0− 0+
y 0 −1 −8/9 = 0.8 8 ∞ ∞
horizontal asymptotes: limx→−∞
y = limx→−∞
4
(
1
x2− 1
x
)
= 0
limx→∞
y = limx→∞
4
(
1
x2− 1
x
)
= 0
additional work: To show that x = 3 is an inflection point, we evaluate the second derivative atx = 4: y′′(4) < 0. This shows that the graph is concave down to the right of x = 3.
−1
−2
1
2
3
4
5
6
1 2 3 4 5 6 7 8−1−2−3−4−5−6−7x
y
• •
17. y =x2
x2 + 3y′ =
6x
(x2 + 3)2y′′ =
18(1 − x2)
(x2 + 3)3=
18(1 − x)(1 + x)
(x2 + 3)3
y-intercept: y(0) = 0
zeros: x = 0
vertical asymptotes: none
critical points: y′ = 0 : x = 0
possible inflection points: y′′ = 0 : x = −1, 1
classification of critical points:
x = 0y′(0) = 0y′′(0) > 0
•
table of values:x 0 −1 1
y 0 1/4 = 0.25 1/4 = 0.25
horizontal asymptotes: limx→−∞
y = limx→−∞
x2
x2 + 3·
1
x2
1
x2
= limx→−∞
1
1 + 3
x2
= 1
limx→∞
y = limx→∞
x2
x2 + 3·
1
x2
1
x2
= limx→∞
1
1 + 3
x2
= 1
additional work: To show that x = −1 is an inflection point, we evaluate the second derivative atx = −2: y′′(−2) < 0. This shows that the graph is concave down to the left of x = −1.
To show that x = 1 is an inflection point, we evaluate the second derivative at x = 2: y′′(2) < 0.This shows that the graph is concave down to the right of x = 1.
−1
1
1 2 3 4 5−1−2−3−4−5x
y
•••
18. y =x
5+
5
xy′ =
x2 − 25
5x2y′′ =
10
x3
y-intercept: y(0) is undefined
zeros: none, since y =x2 + 25
5x
vertical asymptotes:
x = 0 : limx→0−
x
5+
5
x
[
5
0−case
]
= −∞ and limx→0+
x
5+
5
x
[
5
0+case
]
= ∞
critical points: y′ = 0 : x = −5, 5 and y′ = ∅ : x = 0
possible inflection points: y′′ = ∅ : x = 0
classification of critical points:
x = −5 x = 0 x = 5y′(−5) = 0 y′(0) = ∅ and y′′(0) = ∅ y′(5) = 0y′′(−5) < 0 y′′(−0.1) < 0 and y′′(0.1) > 0 y′′(5) > 0
••
table of values:x −5 5 0− 0+
y −2 2 −∞ ∞
horizontal asymptotes: limx→−∞
y = limx→−∞
x
5+
5
x= −∞
limx→∞
y = limx→∞
x
5+
5
x= ∞
−1
−2
−3
−4
−5
−6
−7
1
2
3
4
5
6
7
2 4 6 8 10−2−4−6−8−10x
y
•
•
19. y =x
x2 + 4y′ =
4− x2
(x2 + 4)2y′′ =
2x(x2 − 12)
(x2 + 4)3
y-intercept: y(0) = 0
zeros: x = 0
vertical asymptotes: none
critical points: y′ = 0 : x = −2, 2
possible inflection points: y′′ = 0 : x = 0,−√12,
√12 (
√12 ≈ 3.464)
classification of critical points:
x = −2 x = 2y′(−2) = 0 y′(2) = 0y′′(−2) > 0 y′′(2) < 0
••
table of values:x −
√12 −2 0 2
√12
y ≈ −0.217 −0.25 0 0.25 ≈ 0.217
horizontal asymptotes: limx→−∞
y = limx→−∞
x
x2 + 4
1
x2
1
x2
= limx→−∞
1
x
1 + 4
x2
= 0
limx→∞
y = limx→∞
x
x2 + 4
1
x2
1
x2
= limx→∞
1
x
1 + 4
x2
= 0
additional work: To show that x = −√12 is an inflection point, we evaluate the second derivative
at x = −4: y′′(−4) < 0. This shows that the graph is concave down to the left of x = −√12.
To show that x =√12 is an inflection point, we evaluate the second derivative at x = 4: y′′(4) > 0.
This shows that the graph is concave up to the right of x =√12.
−0.5
0.5
1 2 3 4 5 6−1−2−3−4−5−6x
y
• •
•
• •
20. y = ln(x2 + 9) y′ =2x
x2 + 9y′′ =
−2(x2 − 9)
(x2 + 9)2
y-intercept: y(0) = ln(9) ≈ 2.2
zeros: none
vertical asymptotes: none
critical points: y′ = 0 : x = 0
possible inflection points: y′′ = 0 : x = −3, 3
classification of critical points:
x = 0y′(0) = 0y′′(0) > 0
•
table of values:x −3 0 3
y ≈ 2.89 ≈ 2.2 ≈ 2.89
horizontal asymptotes: limx→−∞
y = limx→−∞
ln(x2 + 9) = ∞limx→∞
y = limx→∞
ln(x2 + 9) = ∞
additional work: To show that x = −3 is an inflection point, we evaluate the second derivative atx = −4: y′′(−4) < 0. This shows that the graph is concave down to the left of x = −3.
To show that x = 3 is an inflection point, we evaluate the second derivative at x = 4: y′′(4) < 0.This shows that the graph is concave down to the right of x = 3.
1
2
3
4
1 2 3 4 5 6−1−2−3−4−5−6x
y
•
•
•
21. y = x2/3(x− 1)1/3 y′ =3x− 2
3x1/3(x− 1)2/3y′′ =
−2
9x4/3(x− 1)5/3
y-intercept: y(0) = 0
zeros: x = 0, 1
vertical asymptotes: none
critical points: y′ = 0 : x = 2/3 and y′ = ∅ : x = 0, 1
possible inflection points: y′′ = ∅ : x = 0, 1
classification of critical points: (A line of slope 0 is horizontal and a line of slope ±∞ is vertical.)
x = 0 x = 2/3 x = 1y′(0−) = ∞ and y′(0+) = −∞ y′(2/3) = 0 y′(1−) = ∞ and y′(1+) = ∞y′′(−0.1) > 0 and y′′(0.1) > 0 y′′(2/3) > 0 y′′(0.9) > 0 and y′′(1.1) < 0
•
••
table of values:x 0 2/3 1
y 0 ≈ −0.53 0
horizontal asymptotes: limx→−∞
y = limx→−∞
x2/3(x− 1)1/3 = −∞limx→∞
y = limx→∞
x2/3(x− 1)1/3 = ∞
−1
−2
−3
1
2
3
1 2 3−1−2x
y
•
•
•