Lang Short Calculus solutions

Serge Lang : Short Calculus

Solutions to exercices

1 Numbers and Functions

1.2 Inequalities

1. For |x| = x, we have x < 3. For |x| = x, we have x < 3 and 3 < x. So, for this inequality,we have : 3 < x < 3 or (3, 3).

2. For |2x + 1| = (2x + 1), we have 2x + 1 1 and x 0. For |2x + 1| = (2x + 1), we have2x 1 1 and x 1. So, for this inequality, we have : 1 x 0 or [1, 0].

3. For |x2 2| = (x2 2), we have x2 2 1 and x 3 or x 3 (sincex2 = |x| by

theorem 1, p. 10). For |x2 2| = (x2 2), we have x2 + 2 1 and x 1 or x 1. So,for this inequality, we have : 3 x 1 and 1 x 3 or [3,1] [1,3].

4.

5. We have (x + 1)(x 2) = 0 for x = 1 or x = 2. We verify the negativity of each intervalbetween (,1), (1, 2) and (2,). So, for this inequality, we have 1 < x and x < 2 or(1, 2).

6. We have (x 1)(x + 1) = 0 for x = 1 or x = 1. We verify the positivity of each intervalbetween (,1), (1, 1) and (1,). So, for this inequality, we have x < 1 and x > 1 or(,1) (1,).

7. We have (x 5)(x + 5) = 0 for x = 5 or x = 5. We verify the negativity of each intervalbetween (,5), (5, 5) and (5,). So, for this inequality, we have 5 < x and x < 5 or(5, 5).

8. We have x(x+ 1) = 0 for x = 0 or x = 1. We verify the negativity of each interval between(,1], [1, 0] and [0,). So, for this inequality, we have 1 x and x 0 or [1, 0].

9. We have x2(x 1) = 0 for x = 0 or x = 1. We verify the positivity of each interval between(, 0], [0, 1] and [1,). So, for this inequality, we have 0 and x 1 or [0] [1,).

10. We have (x 5)2(x+ 10) = 0 for x = 5 or x = 10. We verify the negativity of each intervalbetween (,10], [10, 5] and [5,). So, for this inequality, we have x 10 and x = 5or (,10] [5].

11. We have (x 5)4(x+ 10) = 0 for x = 5 or x = 10. We verify the negativity of each intervalbetween (,10], [10, 5] and [5,). So, for this inequality, we have x 10 and x = 5or (,10] [5].

12. We have (2x+ 1)6(x 1) = 0 for x = 12 or x = 1. We verify the positivity of each intervalbetween (,12 ], [12 , 1] and [1,). So, for this inequality, we have x = 12 or x 1 or[12 ] [1,).

1

13. We have (4x + 7)20(2x + 8) = 0 for x = 74 or x = 4. We verify the negativity of eachinterval between (,4), (4,74) and (74 ,). So, for this inequality, we have x < 4or (,4)

14. We have to show that |x+ y| |x| |y|.

Proof.

|x| = |x+ y y| = |(x+ y) + (y)| (Associativity) |x+ y|+ | y| (By theorem 3, p. 11) |x+ y|+ |y| (Absolute value property)

|x| |y| |x+ y|

15. We have to show that |x y| |x| |y|.

Proof.

|x| = |x y + y| = |(x y) + (y)| (Associativity) |x y|+ |y| (By theorem 3, p. 11)

|x| |y| |x y|

16. We have to show that |x y| |x|+ |y|.

Proof.

|x y| = |(x) + (y)| (Associativity) |x|+ | y| (By theorem 3, p. 11) |x|+ |y| (Absolute value property)

1.3 Functions

1. f(34) =43 , f(23) = 32

2. f(2x+ 1) = 12x+1 for x 6= 123. g(1) = 0, g(1) = 2, g(54) = 1084. f(z) = 2z z2, f(w) = 2w w2

5. The function 1x22 is defined for x 6=

2. We have f(5) = 123 .

6. f(x) = 3x is defined for all x. We have f(27) = 3.

2

7. (a) f(1) = 1

(b) f(2) = 1

(c) f(3) = 1(d) f(43) = 1

8. (a) f(12) = 1

(b) f(2) = 4

(c) f(4) = 0(d) f(43) = 0

9. (a) f(1) = 2(b) f(1) = 6(c) f(x+ 1) = 2(x+ 1) + (x+ 1)2 5 = x2 + 4x 2

10. f(x) = 4x is defined for x 0. We have f(16) = 2.

1.4 Powers

1. 23 = 8 and 32 = 9

2. 51 = 15 and (1)5 = 1

3. (12)4 = 116 and 4

12 = 2

4. (13)2 = 19 and 2

13

5. (12)4 = 116 and 412 = 12

6. 32 = 9 and 23 = 8

7. 31 = 13 and 13 = 18. 22 = 14 and 22 = 149. 14 = 1 and 41 = 14

10. (12)9 = 1512 and 912 = 13

2 Graphs and Curves

2.1 Coordinates

1.

3

2.

3. x is negative and y is positive.

4. x is negative and y is negative.

5.

6.

4

7.

2.2 Graphs

1.

2.

5

3.

4.

5.

6

6.

7.

8.

7

9.

10.

11.

8

12.

13.

14.

9

15.

16.

17.

10

18.

19.

20.

11

21.

22.

23.

12

24.

25.

26.

13

27.

28.

29.

14

30.

31.

32.

15

33.

34.

35.

16

36.

For other values of x, we define f(x) = x n if n < x n+ 1 for all n.

2.3 The straight line

1.

2.

17

3.

4.

5. We have y2 y1 = a(x2 x1), then a = 712(1) = 83 . With y = 83x+ b and taking (1, 1),we have (1) = 83(1) + b, then b = 53 . The equation of the line is : y = 83x 53 .

6. We have y2 y1 = a(x2 x1), then a = 112

43 = 32 . With y = 32x+ b and taking (4,1),we have (1) = 32(4) + b, then b = 5. The equation of the line is : y = 32x+ 5.

18

7. Here, x2 = x1 =

2, so this is not a straight line and the y-coordinate of any point can bearbitrary. Its a vertical line whose equation is x =

2.

8. We have y2 y1 = a(x2 x1), then a = 4(5)3(3) =93+3

. With y = 93+3

x + b and taking

(

3, 4), we have 4 = 93+3

(

3) + b, then b = 4 93

3+3. The equation of the line is :

y = 93+3

x+ 4 93

3+3.

9. With a = 4, we have y = 4x+ b. With (1, 1), we have 1 = 4(1) + b and b = 3. The equationof the line is : y = 4x 3.

10. With a = 2, we have y = 2x + b. With (12 , 1), we have 1 = 2(12) + b and b = 2. Theequation of the line is : y = 2x+ 2.

11. With a = 12 , we have y = 12x+ b. With (

2, 3), we have 3 = 12(

2) + b and b = 3 +22 .

The equation of the line is : y = 12x+ 3 +22 .

12. With a =

3, we have y =

3x+ b. With (1, 5), we have 5 = 2(1) + b and b = 5 +3.The equation of the line is : y =

3x+ 5 +

3.

13.

14.

19

15.

16.

17.

18.

20


11 = 14 .

20. We have y2 y1 = a(x2 x1), then a = 1112 1

4

= 8.

21. We have y2y1 = a(x2x1), then a = 1322 = 222 =

2(2+2)

(22)(2+2) =

22+42 =

2+2.

22. We have y2 y1 = a(x2 x1), then a = 2133) =1

33 =(3+3)

(33)(3+3) =3+3

93 =3+3

6 .

23. We have y2y1 = a(x2x1), then a = 312pi =22pi . With y =

22pix+b and taking (pi, 1),

we have 1 = 22pi (pi)+b, then b =

23pi2pi . The equation of the line is : y =

22pix+

23pi2pi =

22pix+

2pi22pi +

2pi2pi =

22pi (x pi) + 1.

24. We have y2 y1 = a(x2 x1), then a = pi212 . With y =pi212x + b and taking (1, pi), we

have pi = pi212(1) + b, then b =

pi(12)(pi2)12 =

2pi212 . The equation of the line is : y =

pi212x+

2pi212 =

pi212x

pi222

12 +22212 =

pi212x

2(pi2)12 +

2(12)12 =

pi212(x

2)+2.

25. We have y2 y1 = a(x2 x1), then a = 122(1) =32+1

. With y = 32+1

x + b and

taking (1, 2), we have 2 = 32+1

(1) + b, then b = 2212+1

. The equation of the line is :

y = 32+1

x+ 2212+1

= 32+1

x+ 32+1

+ 22+22+1

= 32+1

(x 1) + 2 = (x+ 1)( 32+1

) + 2.


2(1) =321 = 3 +

2. With y = (3 +

2)x+ b

and taking (2,3), we have 3 = (3 +2)(2) + b, then b = 3 + 22. The equation of theline is : y = (3 +

2)x+ 3 + 2

2 = (3 +

2)x+ 3 +

2 +

2 = (3 +

2)(x+ 1) =

2.

27. (a)

21

(b)

(c)

(d)

22

(e)

28. We have y = ax+ b and y = cx+ d with b 6= d.(a) We have to show that if the lines are parallel, they have no points in common.

Proof. Suppose there is a common point on the two lines; at that point, the lines havethe same y-coordinate. For x = 0, we have y = b = d which contradicts the fact thatc 6= d. For x 6= 0, we have a = ybx and c = ydx ; because the lines are parallel, theyhave the same slope, then a = c, and :

y bx

=y dx

y b = y db = d (Which contradicts the fact that b 6= d)

Therefore, the lines cannot have the same y-coordinate and we conclude that they haveno point in common.

(b) We have to show that, if the lines are not parallel, they have exactly one point incommon.

23

Proof. Because the lines are not parallel, a 6= c. When the two lines have the samey-coordinate, we have :

ax+ b = cx+ d

ax cx = d bx(a c) = d b

x =d ba c (With a 6= c)

Considering that a, b, c, d are contants, the x-coordinate has a unique value. We concludethat there is only one point in common between the lines.

29. (a) 3x+ 5 = 2x+ 1, then x = 4. By substitution in y = 3x+ 5, we find y = 7.(b) 3x 2 = x+ 4, then x = 32 . By substitution in y = 3x 2, we find y = 52 .(c) 2x+ 3 = x+ 2, then x = 13 . By substitution in y = 2x+ 3, we find y = 73 .(d) x+ 1 = 2x+ 7, then x = 6. By substitution in y = x+ 1, we find have y = 5.

2.4 Distance between two points

1. L =

(1 (3))2 + (4 (5))2 = 972. L =

(0 1)2 + (2 1))2 = 2

3. L =

(3 (1))2 + (2 4))2 = 524. L =

(1 1))2 + (2 (1))2 = 13

5. L =

(1 12))2 + (1 2)2 =

54 =

52

6. Let A(1, 2), B(4, 2), C(1,3) and D(x, y) be the four vertices of the rectangle. Since Aand B has the same y-coordinate, it means that AB is parallel to CD and C and D has thesame y-coordinate. Since A and C has the same x-coordinate, then B and D has the samex-coordinate. We conclude that D(4,3) is the fourth vertex.

7. We have AB = CD =

(4 (1))2 + (2 2)2 = 5. To check which segment is perpendicu-lar, we have to consider the slopes : aAB = aCD =

224(1)) = 0 so AB and CD are horizontal

lines; the slopes are not defined for aBD and aAC , so BD and AC are vertical lines that are per-pendicular to AB and CD. We can conclude thatBD = AC =

(3 2)2 + (1 (1))2 =

5.

8. Let A(2,2), B(3,2), C(3, 5) and D(x, y) be the four vertices of the rectangle. Since Aand B has the same y-coordinate, it means that AB is parallel to CD and C and D has thesame y-coordinate. Since B and C has the same x-coordinate, then A and D has the samex-coordinate. We can conclude that D(2, 5) is the fourth vertex.

9. We have AB = CD =

(2 (2))2 + (3 (2))2 = 5. To check which segment is per-pendicular, we have to consider the slopes : aAB = aCD =

2(2)3(2)) = 0 so AB and CD

are horizontal lines; the slopes are not defined for aBC or aAD, so BC and AD are ver-tical lines that are perpendicular to AB and CD. We can conclude that BC = AD =

(5 (2))2 + (3 3)2 = 7.

24

10. We have to show that, if the distance between numbers x and y is defined to be |x y|, thisis the same as the distance between the points (x, 0) and (y, 0) on the plane.

Proof. Let L be the distance between the points (x, 0) and (y, 0) on the plane. We have :

L =

(0 0)2 + (y x)2 =

(y x)2= |y x| (By theorem 1, p. 10)

11.* With d(x, y) as the distance between numbers x and y, we have to show that if x, y, z arenumbers, then d(x, z) d(x, y) + d(y, z) and d(x, y) = d(y, x).

Proof. Lets remind a square root property :a+ b a+b.

a+ b a+ b+ 2ab

(a+b)2

a+ b

(a+b)2

a+ b |a+

b| (Definition of absolute value)

a+ b a+b (Since a, b > 0)

Then we have :

d(x, y) =

(x1 x2)2 + (y1 y2)2 =

(y1 y2)2 + (x1 x2)2 = d(y, x)

d(x, z)2 = (x1 x2)2 + (z1 z2)2 (x1 x2)2 + (y1 y2)2 + (y1 y2)2 + (z1 z2)2 d(x, y)2 + d(y, z)2

d(x, z)2 d(x, y)2 + d(y, z)2

d(x, z) d(x, y)2 +

d(y, z)2 = d(x, y) + d(y, z) (Since

a+ b a+b)

2.6 The circle

1. (a) We have center (2,1) and radius 5.

25

(b) We have center (2,1) and radius 2.

(c) We have center (2,1) and radius 1.

(d) We have center (2,1) and radius 3.

26

2. (a) We have center (0, 1) and radius 3.

(b) We have center (0, 1) and radius 2.

(c) We have center (0, 1) and radius 5.

27

(d) We have center (0, 1) and radius 1.

3. (a) We have center (1, 0) and radius 1.

(b) We have center (1, 0) and radius 2.

28

(c) We have center (1, 0) and radius 3.

(d) We have center (1, 0) and radius 5.

29

4.

x2 + y2 2x+ 3y 10 = 0x2 2x+ 1 1 + y2 + 3y + 9

4 9

4 10 = 0 (Completing the square)

(x 1)2 + (y + 32

)2 = 10 + 1 +9

4

(x 1)2 + (y + 32

)2 =53

4

The center is (1,32) and the radius is532 .

5.

x2 + y2 + 2x 3y 15 = 0x2 + 2x+ 1 1 + y2 3y + 9

4 9

4 15 = 0 (Completing the square)

(x+ 1)2 + (y 32

)2 = 15 + 1 +9

4

(x+ 1)2 + (y 32

)2 =73

4

The center is (1, 32) and the radius is732 .

30

6.

x2 + y2 + x 2y 16 = 0x2 + x+

1

4 1

4+ y2 2y + 1 1 16 = 0 (Completing the square)

(x+1

2)2 + (y 1)2 = 16 + 1

4+ 1

(x+1

2)2 + (y 1)2 = 69

4

The center is (12 , 1) and the radius is692 .

7.

x2 + y2 x+ 2y 25 = 0x2 x+ 1

4 1

4+ y2 + 2y + 1 1 25 = 0 (Completing the square)

(x 12

)2 + (y + 1)2 = 25 +1

4+ 1

(x 12

)2 + (y + 1)2 =105

4

31

The center is (12 ,1) and the radius is1052 .

2.7 Dilations and the ellipse

1. We have center (0, 0). For x = 0, y = 4; for y = 0, x = 3, then the vertices of the ellipseare (0, 4), (0,4), (3, 0), (3, 0).

2. We have center (0, 0) For x = 0, y = 3; for y = 0, x = 2, then the vertices of the ellipseare : (0, 3), (0,3), (2, 0), (2, 0).

32



33

5. We have center (1,2) For x = 1, y2 + 4y+ 4 = 16, then (y+ 6)(y 2) = 0, and y = 6 and2; for y = 2, x2 2x+ 1 = 9, then (x 4)(x+ 2), then x = 4 and 2. The vertices of theellipse are : (1, 2), (1,6), (4,2), (2,2).

6. We have :

4x2 + 25y2 = 100

x2 +25

4y2 = 25

x2

25+y2

4= 1

So we have center (0, 0) For x = 0, y = 2; for y = 0, x = 5, then the vertices of the ellipseare : (0, 2), (0,2), (5, 0), (5, 0).

34

7. We have center (1,2) For x = 1, y2 + 4y + 4 = 4, then y(y + 4) = 0, and y = 0 and4; for y = 2, x2 + 2x+ 1 = 3. Solving the equation with the quadratic formula, we have :x = b

b24ac2a , then x =

3 1 and 3 1. The vertices of the ellipse are : (1, 0),

(1,4), (3 1,2), (3 1,2).

8. We have :

25x2 + 16y2 = 400

x2 +16

25y2 = 16

x2

16+y2

25= 1

So we have center (0, 0) For x = 0, y = 5; for y = 0, x = 5, then the vertices of the ellipseare : (0, 5), (0,5), (4, 0), (4, 0).

35

9. We have center (1,3) For x = 1, y2 + 6y + 9 = 4, then (y + 5)(y + 1) = 0, and y = 5 and1; for y = 3, x2 2x+ 1 = 1, then x(x 2), then x = 0 and 2. The vertices of the ellipseare : (1,1), (1,5), (2,3), (0,3).

2.8 The parabola

1.

2.

3.

4.

36

5. Its a circle.

x2 + y2 4x+ 2y 20 = 0x2 4x+ 4 4 + y2 + 2y + 1 1 20 = 0

(x 2)2 + (y + 1)2 = 20 + 4 + 1(x 2)2 + (y + 1)2 = 25

6. Its a circle.

x2 + y2 2y 8 = 0x2 y2 2y + 1 1 8 = 0

x2 + (y 1)2 = 8 + 1x2 + (y 1)2 = 9

7. Its a circle.

x2 + y2 + 2x 2 = 0x2 + 2x+ 1 1 + y2 2 = 0

(x+ 1)2 + y2 = 2 + 1

(x+ 1)2 + y2 = 3

8. Its a parabola.

y 2x2 x+ 3 = 0y

2= x2 +

1

2x 3

2y

2= x2 +

1

2x+

1

16 1

16 3

2y

2= (x+

1

4)2 1

16 3

2

y +25

8= 2(x+

1

4)2

9. Its a parabola.

y x2 4x 5 = 0y 5 = x2 + 4xy 5 = x2 + 4x+ 4 4y 1 = (x+ 2)2

10. Its a parabola.

y x2 + 2x+ 3 = 0y + 3 = x2 + 2x

y + 3 = x2 + 2x+ 1 1y + 3 = (x+ 1)2

y + 4 = (x+ 1)2

37

11. Its a circle.

x2 + y2 + 2x 4y = 3x2 + 2x+ y2 4y = 3

x2 + 2x+ 1 1 + y2 4y + 4 4 = 3(x+ 1)2 1 + (y 2)2 4 = 3

(x+ 1)2 + (y 2)2 = 2

12. Its a circle.

x2 + y2 4y 2y = 3x2 4y + y2 2y = 3

x2 4y + 4 4 + y2 2y + 1 1 = 3(x 2)2 4 + (y 1)2 1 = 3

(x 2)2 + (y 1)2 = 2

13. Its a parabola.

x 2y2 y + 3 = 0x = 2y2 + y 3x

2= y2 +

y

2 3

2x

2= y2 +

y

2+

1

16 1

16 3

2x

2= (y +

1

4)2 1

16 3

2x

2= (y +

1

4)2 1

16 3

2

x+25

8= 2(y +

1

4)2

14. Its a parabola.

x y2 4y = 5x 5 = y2 + 4yx 5 = y2 + 4y + 4 4x 5 = (y + 2)2 4x 1 = (y + 2)2

2.9 The hyperbola

1.

2.

3.

38

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

3 The derivative

3.1 The slope of a curve

1. We have y = 2x2 and look for the slope at (1, 2). We have f(1+h) = 2(1+h)2 = 2+4h+2h2;

so the closest point is (1+h, 2+4h+2h2). Finding the slope, we have 2+4h+2h22

1+h1 =4h+2h2

h =4 + 2h. When h approches 0, the slope is 4.

2. We have y = x2 + 1 and look for the slope at (1, 2). We have f(1 + h) = (1 + h)2 + 1 =2 2h + h2; so the closest point is (1 + h, 2 2h + h2). Finding the slope, we have22h+h221+h(1) =

2h+h2h = 2 + h. When h approches 0, the slope is 2.

3. We have y = 2x7 and look for the slope at (2,3), but we already know that a = 2 becauseits a straight line. Lets verify : we have f(2 + h) = 2(2 + h) 7; so the closest point is(2 + h, 2h 3). Finding the slope, we have 2h3(3)2+h2 = 2hh = 2. Whatever h, the slope is 2.

4. We have y = x3 and look for the slope at (12 ,18). We have f(

12+h) = (

12+h)

3 = 18+h3+ 3h4 +

3h2

2 ;

so the closest point is (12 +h,18 +h

3 + 3h4 +3h2

2 ). Finding the slope, we have18+h3+ 3h

4+ 3h

2

2 1

812+h 1

2

=

h2 + 34 +3h2 . When h approches 0, the slope is

34 .

5. We have y = 1x and look for the slope at (2,12). We have f(2+h) =

12+h ; so the closest point is

(2+h, 12+h). Finding the slope, we have1

2+h 1

2

2+h2 =1

2+h 1

2

h =2(2+h)2(2+h)

h =h4+h

h =1h(h4+h) = 14+h .

When h approches 0, the slope is 14 .

39

6. We have y = x2 + 2x and look for the slope at (1,1). We have f(1 + h) = (1 + h)2 +2(1 + h) = h2 1; so the closest point is (1 + h, h2 1). Finding the slope, we haveh21(1)h1(1) =

h2

h = h. When h approches 0, the slope is 0.

7. We have y = x2 and look for the slope at (2, 4). We have f(2+h) = (2+h)2 = h2 +4h+4; so

the closest point is (2+h, h2+4h+4). Finding the slope, we have h2+4h+442+h2 =

h2+4hh = h+4.

When h approches 0, the slope is 4.

8. We have y = x2 and look for the slope at (3, 9). We have f(3+h) = (3+h)2 = h2 +6h+9; so

the closest point is (3+h, h2+6h+9). Finding the slope, we have h2+6h+993+h3 =

h2+6hh = h+6.

When h approches 0, the slope is 6.

9. We have y = x3 and look for the slope at (1, 1). We have f(1 + h) = (1 + h)3 = h2 + h3 +2h+ 2h2 + 1 +h = h3 + 3h2 + 3h+ 1; so the closest point is (1 +h, h3 + 3h2 + 3h+ 1). Finding

the slope, we have h3+3h2+3h+11

1+h1 =h3+3h2+3h

h = h2 + 3h+ 3. When h approches 0, the slope

is 3.

10. We have y = x3 and look for the slope at (2, 8). We have f(2 + h) = (2 + h)3 = 2h2 + h3 +8h + 4h2 + 8 + 4h = h3 + 6h2 + 12h + 8; so the closest point is (2 + h, h3 + 6h2 + 12h + 8).

Finding the slope, we have h3+6h2+12h+88

2+h2 =h3+6h2+12h

h = h2 + 6h+ 12. When h approches

0, the slope is 12.

11. We have y = 2x+ 3 and look for the slope at (2, 7), since y=2(2)+3=7. We have f(2 + h) =2(2 + h) + 3 = 2h + 7; so the closest point is (2 + h, 2h + 7). Finding the slope, we have2h+772+h2 =

2hh = 2. Whatever h, the slope is 2.

12. We have y = 3x5 and look for the slope at (1,2), since y=3(1)-5=-2. We have f(1 +h) =3(1 + h) 5 = 3h 2; so the closest point is (1 + h, 3h 2). Finding the slope, we have3h2(2)

1+h1 =3hh = 3. Whatever h, the slope is 3.

13. We have y = ax + b and look for the slope at (x, ax + b). We have f(x + h) = a(x + h) +b = ax + ah + b; so the closest point is (x + h, ax + ah + b). Finding the slope, we haveax+ah+b(ax+b)

x+hx =ahh = a. Whatever h, the slope is a.

3.2 The derivative

1. We have f(x) = x2 + 1 and f(x + h) = (x + h)2 + 1 = x2 + 2xh + h2 + 1. Then f (x) =

limh0

f(x+ h) f(x)h

= limh0

x2 + 2xh+ h2 + 1 (x2 + 1)h

= limh0

(2x+ h) = 2x.

2. We have f(x) = x3 and f(x + h) = (x + h)3 = x3 + 3x2h + 3xh2 + h3. Then f (x) =

limh0

x3 + 3x2h+ 3xh2 + h3 x3h

= limh0

(3x2 + 3xh+ h2) = 3x2.

3. We have f(x) = 2x3 and f(x + h) = 2(x + h)3 = 2x3 + 6x2h + 6xh2 + 2h3. Then f (x) =

limh0

2x3 + 6x2h+ 6xh2 + 2h3 2x3h

= limh0

(6x2 + 6xh+ 2h2) = 6x2.

4. We have f(x) = 3x2 and f(x + h) = 3(x + h)2 = 3x2 + 6xh + 3h2. Then f (x) =

limh0

3x2 + 6xh+ 3h2 3x2h

= limh0

(6x+ 3h) = 6x.

40

5. We have f(x) = x2 5 and f(x + h) = (x + h)2 5 = x2 + 2xh + h2 5. Then f (x) =limh0

x2 + 2xh+ h2 5 (x2 5)h

= limh0

(2x+ h) = 2x.

6. We have f(x) = 2x2 + x and f(x+ h) = 2(x+ h)2 + x+ h = 2x2 + 4xh+ 2h2 + x+ h. Then

f (x) = limh0

2x2 + 4xh+ 2h2 + x+ h (2x2 + x)h

= limh0

(4x+ 2h+ 1) = 4x+ 1.

7. We have f(x) = 2x2 3x and f(x+ h) = 2(x+ h)2 3(x+ h) = 2x2 + 4xh+ 2h2 3x 3h.Then f (x) = lim

h02x2 + 4xh+ 2h2 3x 3h (2x2 3x)

h= lim

h0(4x+ 2h 3) = 4x 3.

8. We have f(x) = 12x3 + 2x and f(x+ h) = 12x

3 + 32x2h+ 32xh

2 + 12h3 + 2x+ 2h. Then f (x) =

limh0

12x

3 + 32x2h+ 32xh

2 + 12h3 + 2x+ 2h (12x3 + 2x)h

= limh0

(3

2x2+

3

2xh+

1

2h2+2) =

3

2x2+2.

9. We have f(x) = 1x+1 and f(x+h) =1

x+h+1 . Then f(x) = lim

h0

1x+h+1 1x+1

h= lim

h0

(x+1)(x+h+1)(x+h+1)(x+1)

h=

limh0

(x+ 1) (x+ h+ 1)h(x+ h+ 1)(x+ 1)

= limh0

1(x+ h+ 1)(x+ 1)

=1

(x+ 1)2.

10. We have f(x) = 2x+1 and f(x+h) =2

x+h+1 . Then f(x) = lim

h0

2x+h+1 2x+1

h= lim

h0

2(x+1)2(x+h+1)(x+h+1)(x+1)

h=

limh0

2(x+ 1) 2(x+ h+ 1)h(x+ h+ 1)(x+ 1)

= limh0

2(x+ h+ 1)(x+ 1)

=2

(x+ 1)2.

*11. For each case in Exercice 1 :

1. Finding the tangent line at (2, 5), the slope is 4 = y5x2 , and the equation of the tangentline is y = 4x 3.


3. Finding the tangent line at (2, 16), the slope is 24 = y16x2 , and the equation of thetangent line is y = 24x 32.

4. Finding the tangent line at (2, 12), the slope is 12 = y12x2 , and the equation of thetangent line is y = 12x 12.

5. Finding the tangent line at (2,1), the slope is 4 = y+1x2 , and the equation of the tangentline is y = 4x 9.




9. Finding the tangent line at (2, 13), the slope is 19 =y 1

3x2 , and the equation of the tangent

line is y = 19x+ 59 .10. Finding the tangent line at (2, 23), the slope is 29 =

y 23

x2 , and the equation of the tangentline is y = 29x+ 109 .

41

*12. We have f(x) = x if x 0 and f(x) = 2 if x > 0. We look for f (x) when x = 1and we have f (x) = lim

h0(x+ h) (x)

h= 1. For x = 0, we first look for the left

derivative : since 0 + h 0 for h 0, we have f(0 + h) = f(h) = h and f(0) = 0, thenf (x) = lim

h0f(0 + h) f(0)

h= lim

h0hh

= 1. For the right derivative : since 0 + h > 0 for

h > 0, we have f(0 + h) = 2 and f(0) = 0, then f (x) = limh0

f(0 + h) f(0)h

= limh0

2

h, which

is impossible, so there is no right derivative.

*13. We have f(x) = |x| + x. For h > 0, we take the right derivative for x = 0 : f(0 + h) =f(h) = h + h = 2h and f(0) = 0, and we have : f (x) = lim

h0f(0 + h) f(0)

h= lim

h02h

h= 2.

For h < 0, we take the left derivative for x = 0 : f(0 + h) = f(h) = 0 and f(0) = 0,

so : f (x) = limh0

f(0 + h) f(0)h

= 0. Since the right and the left derivatives are not

equal, the derivative is not defined for x = 0. If x > 0, then x + h > 0, so f (x) =

limh0

f(x+ h) f(x)h

= limh0

2x+ 2h 2xh

limh0

2h

h= 2; also, if x < 0, then x + h < 0, so

f (x) = limh0

f(x+ h) f(x)h

= limx0x h+ x+ h (x+ x)

h= 0. Then, for x 6= 0, the

left and right derivatives are equal for each value of x, so the derivative is defined for thesevalues.

*14. Let f(x) = 0 if x 1 and f(x) = x if x > 1. Finding the left derivative for x = 1, wehave h < 0, so (1 + h) 1; then we havef(1 + h) = 0 and f (1 + h) = 0. Finding the rightderivative, we have h > 0, so (1 + h) > 1; then we have f(1 + h) = 1 + h and f (1 + h) =

limh0

f(1 + h) f(1)h

= limh0

1 + h

h, which is impossible, so there is no right derivative; we

conclude that the derivative is not defined for x = 1. If x 6= 1, then if x > 1 and x + h > 1,so f (x) = lim

h0f(x+ h) f(x)

h= lim

h0x+ h x

hlimh0

h

h= 1; also, if x < 0, then x+ h < 0, so

f (x) = limh0

f(x+ h) f(x)h

= limx0

0

h= 0.

*15. (a) Let f(x) = x|x|. At x = 0, we find the right derivative : for h > 0, we have f(0 +h) = f(h) = h2 and f(0) = 0, so f (x) = lim

h0f(0 + h) f(0)

h= 0. We find the left

derivative : for h < 0, we have f(0 + h) = f(h) = h2 and f(0) = 0, so f (x) =limh0

f(0 + h) f(0)h

= 0. The derivative exists and is equal to 0.

(b) Let f(x) = x2|x|. At x = 0, we find the right derivative : for h > 0, we have f(0 +h) = f(h) = h3 and f(0) = 0, so f (x) = lim

h0f(0 + h) f(0)



f(0 + h) f(0)h


(c) Let f(x) = x3|x|. At x = 0, we find the right derivative : for h > 0, we have f(0 +h) = f(h) = h4 and f(0) = 0, so f (x) = lim

h0f(0 + h) f(0)



f(0 + h) f(0)h


42

3.3 Limits

1. Let f(x) = 2x2 + 3x, then :

f (x) = limh0

2(x+ h)2 + 3(x+ h) (2x2 + 3x)h

= limh0

2x2 + 4xh+ 2h2 + 3x+ 3h 2x2 3xh

= limh0

(4x+ 2h+ 3)

= limh0

4x+ limh0

2h+ limh0

3 (Property 1)

= 4x+ 3

2. Let f(x) = 12x+1 , then :

f (x) = limh0

12(x+h)+1 12x+1

h

= limh0

24x2 + 4x+ 4xh+ 2h+ 1

Taking the numerator, we have limh02 = 2, and taking de denominator, we have :

limh0

(4x2 + 4x+ 4xh+ 2h+ 1) = limh0

4x2 + limh0

4x+ limh0

4xh+ limh0

2h+ limh0

1 (Property 1)

= 4x2 + 4x+ 1

= (2x+ 1)2

By using Property 3, f (x) = 2(2x+1)2

.

3. Let f(x) = xx+1 , then :

f (x) = limh0

x+hx+h+1 xx+1

h

= limh0

1

x2 + 2x+ xh+ h+ 1

Taking the numerator, we have limh0

1 = 1, and taking de denominator, we have :

limh0

(4x2 + 4x+ 4xh+ 2h+ 1) = limh0

x2 + limh0

2x+ limh0

xh+ limh0

h+ limh0

1 (Property 1)

= x2 + 2x+ 1

= (x+ 1)2

By using Property 3, f (x) = 1(x+1)2

.

43

4. Let f(x) = x(x+ 1) = x2 + x, then :

f (x) = limh0

(x+ h)2 + x+ h (x2 + x)h

= limh0

x2 + 2xh+ h2 + x+ h x2 xh

= limh0

(2x+ h+ 1)

= limh0

2x+ limh0

h+ limh0

1 (Property 1)

= 2x+ 1

5. Let f(x) = x2x1 , then :

f (x) = limh0

x+h2x+2h1 x2x1

h

= limh0

14x2 + 4xh 4x 2h+ 1


limh0

(4x2 + 4xh 4x 2h+ 1) = limh0

4x2 + limh0

4xh limh0

4x limh0

2h+ limh0

1 (Property 1)

= 4x2 4x+ 1= (2x 1)2

By using Property 3, f (x) = 1(2x1)2 .

6. Let f(x) = 3x3, then :

f (x) = limh0

3(x+ h)3 3x3h

f (x) = limh0

9x2h+ 9xh2 + 3h3

h

f (x) = limh0

(9x2 + 9xh+ 3h2)

f (x) = limh0

9x2 + limh0

9xh+ limh0

3h2 (Property 1)

f (x) = 9x2

44

7. Let f(x) = x4, then :

f (x) = limh0

(x+ h)4 x4h

f (x) = limh0

x4 + 4x3h+ 6x2h2 + 4xh3 + h4 x4h

f (x) = limh0

4x3h+ 6x2h2 + 4xh3 + h4

h

f (x) = limh0

(4x3 + 6x2h+ 4xh2 + h3)

f (x) = limh0

4x3 + limh0

6x2h+ limh0

4xh2 + limh0

h3 (Property 1)

f (x) = 4x3

8. Let f(x) = x5, then :

f (x) = limh0

(x+ h)5 x5h

f (x) = limh0

x5 + 5x4h+ 10x3h2 + 10x2h3 + 5xh4 + h5 x5h

f (x) = limh0

(5x4 + 10x3h+ 10x2h2 + 5xh3 + h4)

f (x) = limh0

5x4 + limh0

10x3h+ limh0

10x2h2 + limh0

5xh3 + limh0

h4 (Property 1)

f (x) = 5x4

9. Let f(x) = 2x3, then :

f (x) = limh0

2(x+ h)3 2x3h

f (x) = limh0

2x3 + 6x2h+ 6xh2 + 2h3 2x3h

f (x) = limh0

(6x2 + 6xh+ 2h2)

f (x) = limh0

6x2 + limh0

6xh+ limh0

2h2 (Property 1)

f (x) = 6x2

10. Let f(x) = 12x3 + x, then :

f (x) = limh0

12(x+ h)

3 + (x+ h) 12x3 xh

f (x) = limh0

12x

3 + 32x2h+ 32xh

2 + 12h3 + x+ h 12x3 x

h

f (x) = limh0

(3

2x2 +

3

2xh+

1

2h2 + 1)

f (x) = limh0

3

2x2 + lim

h03

2xh+ lim

h01

2h2 + lim

h01 (Property 1)

f (x) =3

2x2 + 1

45

11. Let f(x) = 2x , then :

f (x) = limh0

2x+h 2x

h

= limh0

2x2 + xh


limh0

(x2 + xh) = limh0

x2 + limh0

xh = x2 (Property 1)

By using Property 3, f (x) = 2x2

.

12. Let f(x) = 3x , then :

f (x) = limh0

3x+h 3x

h

= limh0

3x2 + xh


limh0

(x2 + xh) = limh0

x2 + limh0

xh = x2 (Property 1)

By using Property 3, f (x) = 3x2

.

13. Let f(x) = 12x3 , then :

f (x) = limh0

12(x+h)3 12x3

h

= limh0

2h(2x+2h3)(2x3)

h

= limh0

2(2x+ 2h 3)(2x 3)


limh0

[(2x+ 2h 3)(2x 3)] = limh0

(2x+ 2h 3) limh0

(2x 3) (Property 2)= (2x 3)(2x 3)= (2x 3)2

By using Property 3, f (x) = 2(2x3)2 .

46

14. Let f(x) = 13x+1 , then :

f (x) = limh0

13(x+h)+1 13x+1

h

= limh0

3h(3x+3h+1)(3x+1)

h

= limh0

3(3x+ 3h+ 1)(3x+ 1)


limh0

[(3x+ 3h+ 1)(3x+ 1)] = limh0

(3x+ 3h+ 1) limh0

(3x+ 1) (Property 2)

= (3x+ 1)(3x+ 1)

= (3x+ 1)2

By using Property 3, f (x) = 3(3x+1)2

.

15. Let f(x) = 1x+5 , then :

f (x) = limh0

1x+h+5 1x+5

h

f (x) = limh0

h(x+h+5)(x+5)

h

f (x) = limh0

1(x+ h+ 5)(x+ 5)


limh0

[(x+ h+ 5)(x+ 5)] = limh0

(x+ h+ 5) limh0

(x+ 5) (Property 2)

= (x+ 5)(x+ 5)

= (x+ 5)2

By using Property 3, f (x) = 1(x+5)2

.

16. Let f(x) = 1x2 , then :

f (x) = limh0

1x+h2 1x2

h

f (x) = limh0

h(x+h2)(x2)

h

f (x) = limh0

1(x+ h 2)(x 2)

47


limh0

[(x+ h 2)(x 2)] = limh0

(x+ h 2) limh0

(x 2) (Property 2)= (x 2)(x 2)= (x 2)2

By using Property 3, f (x) = 1(x2)2 .

17. Let f(x) = 1x2

, then :

f (x) = limh0

1(x+h)2

1x2

h

f (x) = limh0

2xhh2(x+h)2x2

h

f (x) = limh0

2x h(x+ h)2x2


(2x h) = limh0

2x limh0

h = 2x, and taking dedenominator, we have :

limh0

[(x+ h)2x2] = limh0

(x+ h)2 limh0

x2 (Property 2)

= x2 x2= x4

By using Property 3, f (x) = 2xx4

= 2x3

.

18. Let f(x) = 1(x+1)2

, then :

f (x) = limh0

1(x+h+1)2

1(x+1)2

h

f (x) = limh0

2xhh22h(x+h+1)2(x+1)2

h

f (x) = limh0

2x h 2(x+ h+ 1)2(x+ 1)2


(2xh 2) = limh0

2x limh0

h limh0

2 = 2x 2, andtaking de denominator, we have :

limh0

[(x+ h+ 1)2(x+ 1)2] = limh0

(x+ h+ 1)2 limh0

(x+ 1)2 (Property 2)

= (x+ 1)2(x+ 1)2

= (x+ 1)4

By using Property 3, f (x) = 2x2(x+1)4

= 2(x+1)(x+1)4

= 2(x+1)3

.

48

3.4 Powers

1. We have :

(x+ h)4 = x(x+ h)(x+ h)(x+ h) + h(x+ h)(x+ h)(x+ h)

= x[x(x+ h)(x+ h) + h(x+ h)(x+ h)] + h[x(x+ h)(x+ h) + h(x+ h)(x+ h)]

= x[x(x2 + 2xh+ h2) + h(x2 + 2xh+ h2)] + h[x(x2 + 2xh+ h2) + h(x2 + 2xh+ h2)]

= x[x3 + 2x2h+ xh2 + x2h+ 2xh2 + h3] + h[x3 + 2x2h+ xh2 + x2h+ 2xh2 + h3]

= (x4 + 3x3h+ 3x2h2 + xh3) + (x3h+ 3x2h2 + 3xh3 + h4)

= x4 + 4x3h+ 6x2h2 + 4xh3 + h4

2. We have dfdx = nxn1. For f(x) = x4, then f (x) = 4x3.

3. (a) For f(x) = x23 , f (x) = 2

3x3.

(b) For f(x) = x32 , f (x) = 3

2x25

.

(c) For f(x) = x76 , f (x) = 76x

16 .

4. We have f(x) = x9, then f (x) = 9x8. The slope at f (1) = 9, so the equation at (1, 1) is9 = y1x1 , and y = 9x 8.

5. We have f(x) = x23 , then f (x) = 2

3x13

. The slope at f (8) = 13 , so the equation at (8, 4) is13 =

y4x8 , and y =

13x+

43 .

6. We have f(x) = x34 , then f (x) = 3

4x74

. The slope at f (16) = 329

. We have f(16) = 18 ,

so the equation at (16, 18) is 329 =y 1

8x16 , and y = 329x+ 732 .

7. We have f(x) =x = x

12 , then f (x) = 1

2x12

. The slope at f (3) = 123

=36 . We have

f(3) =

3, so the equation at (3,

3) is36 =

y3x3 , and y =

36 x+

32 .

8. (a) For f(x) = x14 , we have f (x) = 1

4x34

, so f (5) = 145 34

.

(b) For f(x) = 1x14

, we have f (x) = 14x

54

, so f (7) = 147 54

.

(c) For f(x) = x2, we have f (x) =

2x21, so f (10) =

2(10

21).

(d) For f(x) = xpi, we have f (x) = pixpi1, so f (7) = pi7pi1.

3.5 Sum, products, and quotients

1.

49

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