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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
19. We have y2 y1 = a(x2 x1), then a = 112
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.
26. We have y2 y1 = a(x2 x1), then a = 32
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
3. We have center (0, 0) For x = 0, y = 3; for y = 0, x = 5, then the vertices of the ellipseare : (0, 3), (0,3), (5, 0), (5, 0).
4. We have center (0, 0) For x = 0, y = 5; for y = 0, x = 2, then the vertices of the ellipseare : (0, 5), (0,5), (2, 0), (2, 0).
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.
2. Finding the tangent line at (2, 8), the slope is 12 = y8x2 , and the equation of the tangentline is y = 12x 16.
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.
6. Finding the tangent line at (2, 10), the slope is 9 = y10x2 , and the equation of the tangentline is y = 9x 8.
7. Finding the tangent line at (2, 2), the slope is 5 = y2x2 , and the equation of the tangentline is y = 5x 8.
8. Finding the tangent line at (2, 8), the slope is 8 = y8x2 , and the equation of the tangentline is y = 8x 8.
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)
h= 0. We find the left
derivative : for h < 0, we have f(0 + h) = f(h) = h3 and f(0) = 0, so f (x) =limh0
f(0 + h) f(0)h
= 0. The derivative exists and is equal to 0.
(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)
h= 0. We find the left
derivative : for h < 0, we have f(0 + h) = f(h) = h4 and f(0) = 0, so f (x) =limh0
f(0 + h) f(0)h
= 0. The derivative exists and is equal to 0.
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
Taking the numerator, we have limh01 = 1, and taking de denominator, we have :
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
Taking the numerator, we have limh02 = 2, and taking de denominator, we have :
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
Taking the numerator, we have limh03 = 3, and taking de denominator, we have :
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)
Taking the numerator, we have limh02 = 2, and taking de denominator, we have :
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)
Taking the numerator, we have limh03 = 3, and taking de denominator, we have :
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)
Taking the numerator, we have limh01 = 1, and taking de denominator, we have :
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
Taking the numerator, we have limh01 = 1, and taking de denominator, we have :
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
Taking the numerator, we have limh0
(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
Taking the numerator, we have limh0
(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