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

•

•

•