1. The document discusses the concept of tangent lines and slope. It provides 5 examples of calculating the slope of a function at different points to derive the equation of the tangent line. 2. The slopes are calculated by taking the limit as h approaches 0 of the change in y over the change in x. 3. The slopes found were 2, 0, -1/2, 4, and 1/2, leading to tangent lines of y=2x-3, y=-2, y=-x/2+1, y=4x+2, and y=x/2+1 respectively.

1. 3
2
1
0
−3 −2 −1 0 1 2 3
x
Chapter 2 y
−1
−2
−3
Differentiation 3. Slope is
f (−2 + h) − f (−2)
lim
h→0 h
2
(−2 + h) − 3(−2 + h) − (10)
= lim
h→0 h
−7h + h2
= lim = −7.
h→0 h
Tangent line is y = −7(x + 2) + 10
2.1 Tangent Line and 120
Velocity 100
80
60
1. Slope is
40
f (1 + h) − f (1)
lim 20
h→0 h
2 0
(1 + h) − 2 − (−1) −10 −8 −6 −4 −2 0 2 4 6 8 10
= lim x
−20
h→0 h
h2 + 2h −40
= lim −60
h→0 h
= lim (h + 2) = 2.
h→0
Tangent line is y = 2(x−1)−1 or y = 2x−3. 4. Slope is
f (1 + h) − f (1)
lim
5.0 h→0 h
(1 + 3h + 3h2 + h3 ) + (1 + h) − 2
= lim
x 2.5 h→0 h
−3 −2 −1 0 1 2 3 4h + 3h2 + h3
0.0 = lim = lim 4 + 3h + h2 = 4.
h→0 h h→0
−2.5
Tangent line is y = 4(x − 1) + 2.
30
−5.0 25
−7.5 20
y 15
10
5
2. Slope is 0
f (0 + h) − f (0) −1 0 1 2 3
lim x
h→0 h −5
h2
= lim = 0.
h→0 h
Tangent line is y = −2. 5. Slope is
78

2. 2.1. TANGENT LINE AND VELOCITY 79
f (1 + h) − f (1) (h + 1) − 1
lim = lim √
h→0 h h→0 h( h + 1 + 1)
2 2
(1+h)+1 − 1+1 1 1
= lim = lim √ = .
h→0 h h→0 h+1+1 2
2 2−(2+h) 1 1
2+h −1 2+h Tangent line is y = (x+2)+1 or y = x+2.
= lim = lim 2 2
h→0 h h→0 h 4.0
−h 3.2
2+h −1 1
= lim =− .
= lim 2.4
h→0 h h→0 2 + h 2
1 1.6
Tangent line is y = − (x − 1) + 1 or
2 0.8
x 3
y=− + . 0.0
2 2 −4 −2
x −0.8
0 2 4
4.0
−1.6
3.2 y
−2.4
2.4
−3.2
1.6
−4.0
0.8
0.0 8. Slope is
−5 −4 −3 −2 −1
−0.8
0 1 2 3 4 5
f (1 + h) − f (x)
x
lim
−1.6 h→0 h √
y
−2.4 (1 + h) + 3 − 1 + 3
= lim
−3.2 h→0 √ h
−4.0 h+4−2
= lim
h→0 √ h √
6. Slope is h+4−2 h+4+2
= lim ·√
f (0 + h) − f (0) h→0 h h+4+2
lim h+4−4 1
h→0 h = lim ·√
h h h+4+2
−0 h→0
= lim h−1 1 1
h→0 h = lim √ = .
1 h→0 h+4+2 4
= lim = −1 1
h→0 h − 1 Tangent line is y = (x − 1) + 2.
Tangent line is y = −x. 4
4.0
4
3.2
2.4
3
1.6
0.8
2
0.0
−2 −1 0 1 2
x −0.8
1
−1.6
y
−2.4
−3.2
−2.5 0.0 2.5 5.0 7.5 10.0
−4.0
9. f (x) = x3 − x
7. Slope is No. Points (x, y) Slope
f (−2 + h) − f (−2)
lim (a) (1,0) and (2,6) 6
h→0 h (b) (2,6) and (3,24) 18
(−2 + h) + 3 − 1 (c) (1.5,1.875) and (2,6) 8.25
= lim
h→0 √ h (d) (2,6) and (2.5,13.125) 14.25
h+1−1 (e) (1.9,4.959) and (2,6) 10.41
= lim
h→0 √ h √ (f) (2,6) and (2.1,7.161) 11.61
h+1−1 h+1+1
= lim ·√ (g) Slope seems to be approximately 11.
h→0 h h+1+1

3. 80 CHAPTER 2. DIFFERENTIATION
2
10. f (x) = x2 + 1 −4.9(2 + h) + 5 − (−14.6)
= lim
No. Points (x, y) Slope h→0 h
(a) (1,1.414) and (2,2.236) 0.504 −4.9(4 + 4h + h2 ) + 5 − (−14.6)
= lim
(b) (2,2.236) and (3,3.162) 0.926 h→0 h
(c) (1.5,1.803) and (2,2.236) 0.867 −19.6h − 4.9h2
= lim
(d) (2,2.236) and (2.5,2.269) 0.913 h→0 h
(e) (1.9,2.147) and (2,2.236) 0.89 h (−19.6 − 4.9h)
= lim = −19.6
(f) (2,2.236) and (2.1,2.325) 0.899 h→0 h
(g) Slope seems to be approximately 0.89. 16. (a) Velocity at time t = 0 is,
s(0 + h) − s(0)
x−1 lim
11. f (x) = h→0 h
x+1 4h − 4.9h2
No. Points (x, y) Slope = lim
h→0 h
(a) (1,0) and (2,0.33) 0.33 h (4 − 4.9h)
= lim
(b) (2,0.33) and (3,0.5) 0.17 h→0 h
(c) (1.5,0.2) and (2,0.33) 0.26 = 4 − lim 4.9h = 4.
h→0
(d) (2,0.33) and (2.5,0.43) 0.2 (b) Velocity at time t = 1 is,
(e) (1.9,0.31) and (2,0.33) 0.2 s(1 + h) − s(1)
(f) (2,0.33) and (2.1,0.35) 0.2 lim
h→0 h
2
(g) Slope seems to be approximately 0.2. 4(1 + h) − 4.9(1 + h) − (−0.9)
= lim
h→0 h
12. f (x) = ex 4 + 4h − 4.9 − 9.8h − 4.9h2 + 0.9
No. Points (x, y) Slope = lim
h→0 h
(a) (1,2.718) and (2,7.389) 4.671 −5.8h − 4.9h2
= lim
(b) (2,7.389) and (3,20.085) 12.696 h→0 h
(c) (1.5,4.481) and (2,7.389) 5.814 h (−5.8 − 4.9h)
= lim = −5.8
(d) (2,7.389) and (2.5,12.182) 9.586 h→0 h
(e) (1.9,6.686) and (2,7.389) 7.03
17. (a) Velocity at time t = 0 is,
(f) (2,7.389) and (2.1,8.166) 7.77
s(0 + h) − s(0)
lim
(g) Slope seems to be approximately 7.4 h→0 √ h √
h + 16 − 4 h + 16 + 4
13. C, B, A, D. At the point labeled C, the slope = lim ·√
h→0 h h + 16 + 4
is very steep and negative. At the point B, (h + 16) − 16
the slope is zero and at the point A, the slope = lim √
h→0 h( h + 16 + 4)
is just more than zero. The slope of the line 1 1
tangent to the point D is large and positive. = lim √ =
h→0 h + 16 + 4 8
14. In order of increasing slope: D (large nega- (b) Velocity at time t = 2 is,
tive), C (small negative), B (small positive), s(2 + h) − s(2)
and A (large positive). lim
h→0 √ h √
15. (a) Velocity at time t = 1 is, 18 + h − 18
= lim
s(1 + h) − s(1) h→0 h√ √
lim h + 18 + 18
h→0 h Multiplying by √ √ gives
2
−4.9(1 + h) + 5 − (0.1) h + 18 + 18
= lim (h + 18) − 18
h→0 h = lim √ √
−4.9(1 + 2h + h2 ) + 5 − (0.1) h→0 h( h + 18 + 18)
= lim 1 1
h→0 h = lim √ √ = √
−9.8h − 4.9h2 h→0 h + 18 + 18 2 18
= lim
h→0 h
h (−9.8 − 4.9h) 18. (a) Velocity at time t = 2 is,
= lim = −9.8. s(2 + h) − s(2)
h→0 h lim
(b) Velocity at time t = 2 is, h→0 h
4 4−4−2h
s(2 + h) − s(2) (2+h) − 2 (2+h)
lim = lim = lim
h→0 h h→0 h h→0 h

4. 2.1. TANGENT LINE AND VELOCITY 81
√
−2h −2 (c) Second point: (1.9, 18.81)
= lim = lim = −1.
h→0 h(2 + h) h→0 2 + h Average velocity:
√ √
20 − 18.81
(b) Velocity at time t = 4 is, = 1.3508627
4
−1 2 − 1.9
s(4 + h) − s(4) (4+h)
lim = lim √
h→0 h h→0 h (d) Second point: (1.99, 19.8801)
4−1(4+h) 4−4−h
(4+h) (4+h) Average velocity:
√ √
= lim = lim 20 − 19.88
h→0 h h→0 h = 1.3425375
−h −1 1 2 − 1.99
= lim = lim =−
h→0 h(4 + h) h→0 4 + h 4
(e) One might conjecture that these num-
19. (a) Points: (0, 10) and (2, 74) bers are approaching 1.34. The exact
74 − 10 6
Average velocity: = 32 limit is √ ≈ 1.341641.
2 20
(b) Second point: (1, 26) 22. (a) Points: (0, −2.7279) and (2, 0)
74 − 26 Average velocity:
Average velocity: = 48
1 0 − (−2.7279)
= 1.3639
2−0
(c) Second point: (1.9, 67.76)
74 − 67.76 (b) Second point: (1, −2.5244)
Average velocity: = 62.4
0.1 Average velocity:
0 − (−2.5244)
(d) Second point: (1.99, 73.3616) = 2.5244
74 − 73.3616 2−1
Average velocity: = 63.84
0.01 (c) Second point: (1.9, −0.2995)
(e) The instantaneous velocity seems to be Average velocity:
64. 0 − (−0.2995)
= 2.995
2 − 1.9
20. (a) Points: (0, 0) and (2, 26) (d) Second point: (1.99, −0.03)
26 − 0 0 − (−0.03)
Average velocity: = 13 Average velocity: =3
2−0 2 − 1.99
(e) The instantaneous velocity seems to be
(b) Second point: (1, 4) 3.
26 − 4
Average velocity: = 22
2−1
23. A graph makes it apparent that this function
(c) Second point: (1.9, 22.477)
26 − 22.477 has a corner at x = 1.
Average velocity: = 35.23 5
2 − 1.9 4
3
(d) Second point: (1.99, 25.6318)
Average velocity: 2
26 − 25.6318 1
= 36.8203
2 − 1.99 0
−5 −4 −3 −2 −1 0 1 2 3 4 5
x −1
(e) The instantaneous velocity seems to be
−2
approaching 37. y
−3
√ −4
21. (a) Points: (0, 0) and √
(2, 20) −5
20 − 0
Average velocity: = 2.236068 Numerical evidence suggests that,
2−0 f (1 + h) − f (1)
lim+ =1
(b) Second point: (1, 3) h→0 h
√ f (1 + h) − f (1)
20 − 3 while lim− = −1.
Average velocity: = 1.472136 h→0 h
2−1
Since these are not equal, there is no tangent

5. 82 CHAPTER 2. DIFFERENTIATION
5
line.
4
3
24. Tangent line does not exist at x = 1 because 2
the function is not defined there. −5 −4 −3 −2 −1
1
0 1 2 3 4 5
10
0
8
−1
6
−2
4
−3
2
−4
0 −5
−10 −8 −6 −4 −2 0 2 4 6 8 10
−2
x
Also,
−4
y f (0 + h) − f (0) −2h
−6 lim− = lim− = −2
h→0 h h→0 h
−8
f (0 + h) − f (0)
lim+ = lim+ (h − 4) = −4.
−10
h→0 h h→0
25. From the graph it is clear that, curve is not Numerical evidence suggest that,
continuous at x = 0 therefore tangent line f (0 + h) − f (0)
lim−
at y = f (x) at x = 0 does not exist. h→0 h
10.0 f (0 + h) − f (0)
= lim+ .
h→0 h
Therefore tangent line does not exist at x =
7.5
0.
5.0 27. Tangent line at x = π to y = sin x as below:
3
2.5
2
0.0 1
x
−10 −8 −6 −4 −2 0 2 4 6 8 10
0 1 2 3 4 5 6
0
−2.5
Also, −1
f (0 + h) − f (0)
lim −2
h→0− h
h2 − 1 − (−1) −3
= lim−
h→0 h
h2
= lim− = lim− h = 0
h→0 h h→0 28. Tangent line at x = 0 to y = tan−1 x as be-
Similarly, low:
f (0 + h) − f (0) 5.0
lim+
h→0 h
h + 1 − (1) h
= lim+ = lim+ = 1. 2.5
h→0 h h→0 h
Numerical evidence suggest that,
f (0 + h) − f (0)
lim− 0.0
h→0 h −10 −5 0 5 10
f (0 + h) − f (0)
= lim .
h→0+ h −2.5
Therefore tangent line does not exist at
x = 0.
−5.0
26. From the graph it is clear that, the curve of
y = f (x) is not smooth at x = 0 therefore 29. Since the graph has a corner at x = 0, tan-
tangent line at x = 0 does not exist. gent line does not exist.

6. 2.1. TANGENT LINE AND VELOCITY 83
30. The tangent line overlays the line: f (s) − f (r)
33. vavg =
s−r
f (s) − f (r)
vavg =
2
s−r
as + bs + c − (ar2 + br + c)
2
=
1.5
s−r
a(s2 − r2 ) + b(s − r)
=
1
s−r
a(s + r)(s − r) + b(s − r)
=
0.5
s−r
= a(s + r) + b
0
0 0.5 1 1.5 2
Let v(r) be the velocity at t = r. We have,
x v(r) =
f (4) − f (2) f (r + h) − f (r)
31. (a) = 21,034 lim
2 h→0 h
f (b)−f (a)
Since b−a is the average rate of 2
a(r + h) + b(r + h) + c − (ar2 + bh + c)
change of function f between a and b. = lim
h→0 h
The expression tells us that the average
a(r + 2rh + h ) + bh − ar2
2 2
rate of change of f between a = 2 to = lim
b = 4 is 21,034. That is the average h→0 h
rate of change in the bank balance be- h(2ar + ah + b)
= lim
tween Jan. 1, 2002 and Jan. 1, 2004 was h→0 h
21,034 ($ per year). = lim (2ar + ah + b) = 2ar + b
h→0
(b) 2 [f (4) − f (3.5)] = 25,036 So, v(r) = 2ar + b.
Note that 2[f (4) − f (3.5)] = f (4) − The same argument shows that v(s) =
f (3.5)/2. The expression says that the 2as + b.
average rate of change in balance be- Finally
tween July 1, 2003 and Jan. 1, 2004 v(r) + v(s) (2ar + b) + (2as + b)
=
was 25,036 ($ per year). 2 2
f (4 + h) − f (4) 2a(s + r) + 2b
(c) lim = 30,000 =
h→0 h 2
The expression says that the instanta- = a(s + r) + b = vavg
neous rate of change in the balance on
Jan. 1, 2004 was 30,000 ($ per year).
34. f (t) = t3 −t works with r = 0, s = 2. The av-
f (40) − f (38) 6−0
32. (a) = −2103 erage velocity between r and s is, = 3.
2 2−0
Since f (b)−f (a)
is the average rate of The instantaneous velocity at r is
b−a 3
change of function between a and b. The (0 + h) − (0 + h) − 0
lim =0
expression tells us that the average rate h→0 h
of change of f between a = 38 to b = 40 and the instantaneous velocity at s is,
3
is −2103. That is the average rate of de- (2 + h) − (2 + h) − 6
lim
preciation between 38 and 40 thousand h→0 h
miles. 8 + 12h + 6h2 + h3 − 2 − h − 6
= lim
(b) f (40) − f (39) = −2040 h→0 h
= lim 11 + 6h + h2 = 11
The expression says that the average h→0
rate of depreciation between 39 and 40 so, the average between the instantaneous
thousand miles is −2040. velocities is 5.5.
f (40 + h) − f (40)
(c) lim = −2000 35. (a) y = x3 + 3x + 1
h→0 h
The expression says that the instanta- y = 3x2 + 3
neous rate of depreciation in the value Since the slope needed to be 5, y = 5.
of the car when it has 40 thousand miles 3x2 + 3 = 5
is −2000. ⇒ 3x2 = 5 − 3

7. 84 CHAPTER 2. DIFFERENTIATION
2 gent line is y = 6x − 1.
⇒ x2 =
3 Given that y = x3 + 3x + 1.
2 Therefore, we write
⇒x=±
3 x3 + 3x + 1 = 6x − 1
Therefore, slope of tangent line at x = x3 − 3x + 2 = 0
2 2 (x − 1) x2 + x − 2 = 0
and x = − to y = x3 + 3x + 1
3 3 (x − 1)(x − 1) (x + 2) = 0
equals 5. 2
(x − 1) (x + 2) = 0.
(b) Since the slope needed to be 1, y = 1. Therefore, tangent line intersects y =
3x2 + 3 = 1 which has no real roots. x3 + 3x + 1 at more then one point that
Therefore slope of tangent line to y = is at x = 1 and x = −2.
x3 + 3x + 1 cannot equals 1.
36. (a) From the graph it is clear that y = x2 +1 (c) y = x2 + 1
and y = x do not intersect. f (c + h) − f (c)
10 lim
h→0 h
(c + h)2 + 1 − c2 + 1
= lim
5 h→0 h
(c2 + 2ch + h2 ) + 1 − c2 + 1
= lim
h→0 h
0 c2 + 2ch + h2 + 1 − c2 − 1
= lim
−10 −5 0 5 10 h→0 h
2ch + h2
−5 = lim
h→0 h
h (2c + h)
= lim = 2c
−10
h→0 h
The point correponding to x = c is
c, c2 + 1 . So, line with slope 2c
(b) y = x2 + 1 and y = x through point c, c2 + 1 has equation
y = x2 + 1 ⇒ y = 2x y = 2c (x − c)+c2 +1 or y = 2cx−c2 +1.
y=x⇒y =1 Given that y = x2 + 1
For, y = x2 + 1 Therefore,
y = 2x = 1. x2 + 1 = 2cx − c2 + 1
2x = 1 x2 − 2cx + c2 = 0
1 2
⇒x= (x − c) = 0.
2
1 Therefore, tangent line intersects y =
Therefore, tangent line at x = to x2 + 1 only at one point that is at x = c.
2
2
y = x +1 is parallel to the tangent lines
to y = x. 38. Let x = h + a. Then h = x − a and clearly
f (a + h) − f (a) f (x) − f (a)
37. (a) y = x3 + 3x + 1 = .
h x−a
f (1 + h) − f (x) It is also clear that, x → a if and only if
lim
h→0 h h → 0. Therefore, if one of the two limits
(1 + h)3 + 3(1 + h) + 1 − 5 exists, then so does the other and
= lim
h→0 h f (a + h) − f (a) f (x) − f (a)
2 3 lim = lim .
= lim (1+3h+3h +h h)+(3+3h)+1−5 h→0 h x→a x−a
h→0
6h + 3h2 + h3 39. The slope of the tangent line at p = 1 is ap-
= lim
h→0 h proximately
h 6 + 3h + h2 −20 − 0
= lim =6 = −10
h→0 h 2−0
The point correponding to x = 1 is which means that at p = 1 the freezing tem-
(1, 5). So, line with slope 6 through perature of water decreases by 10 degrees
point (1,5) has equation y = 6 (x − 1)+5 Celsius per 1 atm increase in pressure. The
or y = 6x − 1. slope of the tangent line at p = 3 is approx-
(b) From part (a) we have, equation of tan- imately

8. 2.2. THE DERIVATIVE 85
−11 − (−20)
= 4.5 x
4−2 0
0 5 10 15 20
which means that at p = 3 the freezing tem-
perature of water increases by 4.5 degrees
Celsius per 1 atm increase in pressure. -4
y
-8
40. The slope of the tangent line at v = 50 is
47 − 28 -12
approximately = .95.
60 − 40
This means that at an initial speed of 50mph
the range of the soccer kick increases by .95
yards per 1 mph increase in initial speed.
44. Possible graph of bungee-jumpers height:
350
41. The hiker reached the top at the highest
300
point on the graph (about1.75 hours). The
hiker was going the fastest on the way up at 250
about 1.5 hours. The hiker was going the
200
fastest on the way down at the point where
the tangent line has the least (i.e most neg- 150
ative) slope, at about 4 hours at the end of 100
the hike. Where the graph is level the hiker
was either resting or walking on flat ground. 50
0 5 10 15 20
x
A graph of the bungee-jumper s velocity:
42. The tank is the fullest at the first spike (at
around 8 A.M.). The tank is the emptiest
just before this at the lowest dip (around 50
x
7 A.M.). The tank is filling up the fastest 0 5 10 15 20
0
where the graph has the steepest positive
slope (in between 7 and 8 A.M.). The tank
-50
is emptying the fastest just after 8 A.M.
where the graph has the steepest negative -100
slope. The level portions most likely repre-
sent night when waterusage is at a minimum. -150
43. A possible graph of the temperature with
respect to time:
100
2.2 The Derivative
80
1. Using (2.1):
f (1 + h) − f (1)
f (1) = lim
60 h→0 h
y 3(1 + h) + 1 − (4)
= lim
40 h→0 h
3h
20 = lim = lim 3 = 3
h→0 h h→0
Using (2.2):
0
0 5 10 15 20 f (b) − f (1)
x lim
b→1 b−1
Graph of the rate of change of the tem- 3b + 1 − (3 + 1)
perature: = lim
b→1 b−1

9. 86 CHAPTER 2. DIFFERENTIATION
3b − 3 f (b) − f (1)
= lim f (1) = lim
b→1 b − 1 b→1 b−1
3(b − 1) 3
= lim = lim √
b→1 b − 1 b→1 3b + 1 + 2
= lim 3 = 3
b→1 3 3
=√ =
2. Using (2.1): 4+2 4
f (1 + h) − f (1) 4. Using (2.1):
f (1) = lim
h→0 h f (2 + h) − f (2)
2 f (2) = lim
3(1 + h) + 1 − 4 h→0 h
= lim 3
(2+h)+1 − 1
h→0 h
6h + 3h 2 = lim
= lim h→0 h
h→0 h 3 3+h
− 3+h
3+h
= lim 6 + 3h = 6 = lim
h→0 h→0 h
Using (2.2): −h
3+h
f (x) − f (1) = lim
f (1) = lim h→0 h
x→1 x−1 −1 1
(3x2 + 1) − 4 = lim =−
= lim h→0 3 + h 3
x→1 x−1 Using (2.2):
3(x − 1)(x + 1) f (x) − f (2)
= lim f (2) = lim
x→1 x−1 x→2 x−2
3
= lim 3(x + 1) = 6 −1
x→1 = lim x+1
x→2 x − 2
3. Using (2.1): Since 3
− x+1
f (1 + h) − f (1) = lim x+1 x+1
h
x→2 x−2
−(x−2)
3(1 + h) + 1 − 2 x+1
= = lim
√ h √ x→2 x−2
4 + 3h − 2 4 + 3h + 2 −1 1
= ·√ = lim =−
h 4 + 3h + 2 x→2 x + 1 3
4 + 3h − 4 3h
= √ = √ f (x + h) − f (x)
h( 4 + 3h + 2) h( 4 + 3h + 2) 5. lim
3 3 h→0 h
=√ =√ , 2 2
3(x + h) + 1 − (3(x) + 1)
4 + 3h + 2 4 + 3h + 2 = lim
we have h→0 h
f (1 + h) − f (1) 3x2 + 6xh + 3h2 + 1 − (3x2 + 1)
f (1) = lim = lim
h→0 h h→0 h
3 6xh + 3h2
= lim √ = lim
h→0 4 + 3h + 2 h→0 h
= lim (6x + 3h) = 6x
3 3 h→0
= =
4 + 3(0) + 2 4 f (x + h) − f (x)
Using (2.2): Since 6. f (x) = lim
h→0 h
f (b) − f (1) 2
(x + h) − 2(x + h) + 1 − f (x)
√− 1
b = lim
3b + 1 − 2 h→0 h
= 2xh + h2 − 2h
√ b−1 √ = lim
( 3b + 1 − 2)( 3b + 1 + 2) h→0 h
= √
(b − 1)( 3b + 1 + 2) h(2x + h − 2)
= lim = 2x − 2
(3b + 1) − 4 h→0 h
= √
(b − 1) 3b + 1 + 2 f (b) − f (x)
3(b − 1) 3 7. lim
= √ =√ , b→x b−x
(b − 1) 3b + 1 + 2 3b + 1 + 2 b3 + 2b − 1 − x3 + 2x − 1
we have = lim
b→x b−x

10. 2.2. THE DERIVATIVE 87
(b − x) b2 + bx + x2 + 2 (3b + 1) − (3t + 1)
= lim f (t) = lim √ √
b→x b−x b→t (b − t) 3b + 1 + 3t + 1
= lim b2 + bx + x2 + 2 3(b − t)
b→x
= lim √ √
= 3x2 + 2 b→t (b − t)
3b + 1 + 3t + 1
3
= lim √ √
b→t 3b + 1 + 3t + 1
8. f (x) = 3
f (x + h) − f (x) = √
lim 2 3t + 1
h→0 h √
(x+h)4 −2(x+h)2 +1−f (x) 12. f (t) = 2t + 4
= lim h
h→0 f (b) − f (t)
= lim 4x3 + 6x2 h + 4xh2 + h3 − 4x − 2h f (t) = lim
h→0 b→t (b − t)
√ √
= 4x3 − 4x 2b + 4 − 2t + 4
= lim
b→t
√ (b − t) √
2b + 4 + 2t + 4
f (b) − f (x) Multiplying by √ √ gives
9. lim 2b + 4 + 2t + 4
b→x b−x (2b + 4) − (2t + 4)
3 3 f (t) = lim √ √
b+1 − x+1 b→t (b − t) 2b + 4 + 2t + 4
= lim
b→x b−x
3(x+1)−3(b+1) 2(b − t)
(b+1)(x+1) = lim √ √
= lim b→t (b − t)
2b + 4 + 2t + 4
b→x b−x
−3(b − x) 2
= lim = lim √ √
b→x (b + 1)(x + 1)(b − x) b→t 2b + 4 + 2t + 4
−3 2 1
= lim = √ =√
b→x (b + 1)(x + 1) 2 2t + 4 2t + 4
−3
= 13. (a) The derivative should look like:
(x + 1)2 10
8
6
f (x + h) − f (x) 4
10. f (x) = lim
h→0 h 2
2 2
2(x+h)−1 − 2x−1 0
= lim −10 −8 −6 −4 −2 0 2 4 6 8 10
h→0 h −2
2(2x−1)−2(2x+2h−1) −4
(2x+2h−1)(2x−1)
= lim −6
h→0 h −8
−4h
(2x+2h−1)(2x−1) −10
= lim
h→0 h
−4
= lim
h→0 (2x + 2h − 1)(2x − 1) (b) The derivative should look like:
−4
= 5.0
(2x − 1)2
2.5
√
11. f (t) = 3t + 1
f (b) − f (t) 0.0
f (t) = lim −5 −4 −3 −2 −1 0 1 2 3 4 5
b→t (b − t)
√ √
3b + 1 − 3t + 1 −2.5
= lim
b→t
√ (b − t) √
3b + 1 + 3t + 1 −5.0
Multiplying by √ √ gives
3b + 1 + 3t + 1

11. 88 CHAPTER 2. DIFFERENTIATION
5
4
3
2
1
14. (a) The derivative should look like:
0
5
−5 −4 −3 −2 −1 0 1 2 3 4 5
x −1
4
−2
3
y
−3
2
−4
1
−5
0
−5 −4 −3 −2 −1 0 1 2 3 4 5
−1
x
−2
y
−3
−4
−5
16. (a) The derivative should look like:
10
8
(b) The derivative should look like:
6
5
4
4
2
3
0
2
−5 −4 −3 −2 −1 0 1 2 3 4 5
x −2
1
−4
0 y
−5 −4 −3 −2 −1 0 1 2 3 4 5 −6
−1
x
−8
−2
y
−10
−3
−4
−5
(b) The derivative should look like:
4.0
3.2
15. (a) The derivative should look like: 2.4
3 1.6
0.8
2
0.0
−4 −2 0 2 4
x −0.8
1
−1.6
y
0 −2.4
−3 −2 −1 0 1 2 3
−3.2
x
−1
−4.0
y
−2
−3
(b) The derivative should look like: 17. (a) The function should look like:

12. 2.2. THE DERIVATIVE 89
10 25
8
20
6
15
4
y
2 10
0
−10 −8 −6 −4 −2 0 2 4 6 8 10 5
x −2
−4 0
y −5.0 −2.5 0.0 2.5 5.0 7.5 10.0
−6 x
−5
−8
−10
−10
19. The left-hand derivative is
f (h) − f (0)
(b) The function should look like: D− f (0) = lim
10
h→0− h
2h + 1 − 1
8
= lim− =2
6
h→0 h
4 The right-hand derivative is
2
0 f (h) − f (0)
−10 −8 −6 −4 −2 0 2 4 6 8 10 D+ f (0) = lim+
x −2 h→0 h
−4 3h + 1 − 1
y = lim+ =3
−6 h→0 h
−8
Since the one-sided limits do not agree (2 =
−10
3), f (0) does not exist.
20. The left-hand derivative is
f (h) − f (0)
D− f (0) = lim
h→0+ h
0−0
= lim− =0
h→0 h
18. (a) The function should look like: The right-hand derivative is
10
8 f (h) − f (0)
D+ f (0) = lim−
6 h→0 h
4 2h
= lim+ =2
2 h→0 h
−10 −8 −6 −4 −2
0
0 2 4 6 8 10
Since the one-sided limits do not agree (0 =
x −2
2), f (0) does not exist.
−4
y
−6 21. The left-hand derivative is
−8
f (h) − f (0)
−10
D− f (0) = lim
h→0− h
h2 − 0
= lim− =0
h→0 h
The right-hand derivative is
(b) The function should look like:

13. 90 CHAPTER 2. DIFFERENTIATION
f (h) − f (0) f (x) − f (2)
D+ f (0) = lim+ x f (x)
h→0 h x−2
h3 − 0 1.1 172.7658734 635.6957329
= lim =0 1.01 114.2323717 503.6071639
h→0+ h
1.001 109.6888867 492.5866054
Since the one-sided limits are same (0 = 0), 1.0001 109.2454504 491.5034872
f (0) exist. 1.00001 109.201214 491.3953621
1.000001 109.1967915 491.3845515
1.0000001 109.1963492 491.3834702
22. The left-hand derivative is 1.00000001 109.1963050 491.3833622
The evidence of this table strongly suggests
that the difference quotients (essentially) in-
f (h) − f (0) distinguishable from the values (themselves)
D− f (0) = lim+
h→0 h 491.383. If true, this would mean that f (2)
2h ≈ 491.383.
= lim− =2
h→0 h
25. f (x) = cos 3x
f (x) − f (0)
The right-hand derivative is x f (x)
x−0
0.1 0.9553365 −0.4466351
f (h) − f (0) 0.01 0.9995500 −0.0449966
D+ f (0) = lim−
h→0 h 0.001 0.9999955 −0.0045000
h2 + 2h 0.0001 1.0000000 −0.0004500
= lim 0.00001 1.0000000 −0.0000450
h→0+ h
h(h + 2) The evidence of this table strongly suggests
= lim that the difference quotients (essentially) in-
h→0+ h
distinguishable from the values (themselves)
= lim+ h + 2 = 2 0. If true, this would mean that f (0) ≈ 0.
h→0
26. f (x) = ln 3x
Since the one-sided limits are same (2 = 2), f (x) − f (2)
f (0) exist. x f (x)
x−2
2.1 1.8405496 0.4879016
2.01 1.7967470 0.4987542
x
23. f (x) = √ 2.001 1.7922593 0.4998757
x2 +1 2.0001 1.7918095 0.4999875
f (x) − f (1) 2.00001 1.7917645 0.4999988
x f (x)
x−1 2.000001 1.7917600 0.4999999
1.1 0.7399401 0.3283329 2.0000001 1.7917595 0.5000000
1.01 0.7106159 0.3509150 The evidence of this table strongly suggests
1.001 0.7074601 0.3532884 that the difference quotients (essentially) in-
1.0001 0.7071421 0.3535268 distinguishable from the values (themselves)
1.00001 0.7071103 0.3535507 0.5. If true, this would mean that f (2)
≈ 0.5.
27. Compute average velocities:
The evidence of this table strongly suggests
Time Interval Average Velocity
that the difference quotients (essentially) in-
(1.7, 2.0) 9.0
distinguishable from the values (themselves)
0.353. If true, this would mean that f (1) (1.8, 2.0) 9.5
≈ 0.353. (1.8, 2.0) 10.0
(2.0, 2.1) 10.0
(2.0, 2.2) 9.5
(2.0, 2.3) 9.0
2 Our best estimate of velocity at t = 2 is 10.
24. f (x) = xex
28. Compute average velocities:

14. 2.2. THE DERIVATIVE 91
3
(b) g(x) = e−2/(x −x)
Time Interval Average Velocity 5
(1.7, 2.0) 8 4
(1.8, 2.0) 8.5 3
(1.8, 2.0) 9.0 2
(2.0, 2.1) 8.0 1
(2.0, 2.2) 8.0 0
−5 −4 −3 −2 −1 0 1 2 3 4 5
(2.0, 2.3) 7.67 x −1
A velocity of between 8 and 9 seems to be a −2
y
good guess. −3
29. (a) f (x) = |x| + |x − 2| −4
10 −5
8
g(x) is not differentiable at x = 0 and
6 x = ±1.
p
4
(0 + h) − 0p hp
2 31. lim = lim = lim hp−1
h→0 h h→0 h h→0
0 The last limit does not exist when p < 1,
−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6
−2 equals 1 when p = 1 and is 0 when p > 1.
−4 Thus f (0) exists when p ≥ 1.
−6
x2 + 2x x < 0
32. f (x) =
−8
ax + b x ≥ 0
−10
For h < 0, f (h) = h2 + 2h, f (0) = b
f (x) is not differentiable at x = 0 and f (h) − f (0)
D− f (0) = lim
x = 2. h→0− h
(b) f (x) = |x2 − 4x| 2
h + 2h − b
= lim−
5.0 h→0 h
For f to be differentiable D− f (0) must ex-
ist.
2.5
D− f (0) exists if and only if b = 0.
Substituting b = 0, we get
0.0 h2 + 2h
D− f (0) = lim− = lim− (h + 2) = 2
−6 −5 −4 −3 −2 −1 0 1 2 3
x
4 5 6
h→0 h h→0
For h > 0, f (h) = ah + b, f (0) = b
y
−2.5
f (h) − f (0)
D+ f (0) = lim+
h→0 h
−5.0 ah + b − b
= lim+
h→0 h
f (x) is not differentiable at x = 0 and ah
x = 4. = lim+ =a
h→0 h
30. (a) g(x) = e−2/x D+ f (0) = 2 if and only if a = 2.
33. Let f (x) = −1 − x2 then for all, we have
5
4
f (x) ≤ x. But at x = −1, we find f (−1) =
3
−2 and
2 f (−1 + h) − f (−1)
f (−1) = lim
1 h→0 h
2
0
−1 − (−1 + h) − (−2)
−5 −4 −3 −2 −1
−1
0 1 2 3 4 5 = lim
x h→0 h
y
−2
1 − (1 − 2h + h2 )
−3 = lim
h→0 h
−4
2h − h2
−5 = lim = lim (2 − h) = 2.
h→0 h h→0
g(x) is not differentiable at x = 0. So, f (x) is not always less than 1.