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x2
dx x2
x2
x4
x2
x4
2
2
If f (x) = 0 then (3x + 1) (3x + 2x) = 0. The tangent line is horizontal at x = −1/3, −2/3 (x = 0 is
ruled out from the deﬁnition of f ). 8. f (x) = 3 10. (a) x = −2, −1, 1, 3 (d) 9. g (x) = f (x) sin x + 2f (x) cos x − f (x) sin x; g (0) = 2f (0) cos 0 = 2(2)(1) = 4 (a) f (1)g (1) + f (1)g (1) = 3(−2) + 1(−1) = −7 (b)
(c)
11. (b) (−∞, −2), (−1, 1), (3, +∞) g (1)f (1) − f (1)g (1)
−2(3) − 1(−1)
5
=
=−
2
2
g (1)
(−2)
4
1
1
3
(d)
f (1) = √ 3 =
2
21
2 f (1) (c) (−2, −1), (1, 3) 0 (because f (1)g (1) is constant) The equations of such a line has the form y = mx. The points (x0 , y0 ) which lie on both the line and
the parabola and for which the slopes of both curves are equal satisfy y0 = mx0 = x3 − 9x2 − 16x0 ,
0
0
so that m = x2 − 9x0 − 16. By diﬀerentiating, the slope is also given by m = 3x2 − 18x0 − 16.
0
0
Equating, we have x2 − 9x0 − 16 = 3x2 − 18x0 − 16, or 2x2 − 9x0 = 0. The root x0 = 0 corresponds
0
0
0
to m = −16, y0 = 0 and the root x0 = 9/2 corresponds to m = −145/4, y0 = −1305/8. So the line
y = −16x is tangent to the curve at the point (0, 0), and the line y = −145x/4 is tangent to the curve
at the point (9/2, −1305/8). 12. The slope of the line x + 4y = 10 is m1 = −1/4, so we set the negative reciprocal √
d
1 ± 1 + 24
(2x3 − x2 ) = 6x2 − 2x and obtain 6x2 − 2x − 4 = 0 with roots x =
= 1, −2/3.
4 = m2 =
dx
6
13. 14. d
(3x − tan x) = 3 − sec2 x = 1, or sec2 x = 2,
The line y − x = 2 has slope m1 = 1 so we set m2 =
dx
√
sec x = ± 2 so x = nπ ± π/4 where n = 0, ±1, ±2, . . ..
f (x) is continuous and diﬀerentiable at any x = 1, so we consider x = 1.
(a)
(b) lim (x2 − 1) = lim+ k (x − 1) = 0 = f (1), so any value of k gives continuity at x = 1. x→1− x→1 lim f (x) = lim− 2x = 2, and lim+ f (x) = lim+ k = k , so only if k = 2 is f (x) diﬀerentiable at x→1− x = 1. x→1 x→1 x→1 Supplementary Exercises 3 15. 96 The slope of the tangent line is the derivative
y = 2x x= 1 (a+b)
2 y = a + b. The slope of the secant is a2 − b2
= a + b, so they are equal.
a−b (b, b)2 (a, a)2
x
a a+b
2 b 16. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles are
completed after two hours. Thus time is up, and the speed for the remaining 20 miles would have to
be inﬁnite. 17. (a) ∆x = 1.5 − 2 = −0.5; dy =
∆y = −1
−1
∆x =
(−0.5) = 0.5; and
(x − 1)2
(2 − 1)2 1
1
−
= 2 − 1 = 1.
(1.5 − 1) (2 − 1) (b) ∆x = 0 − (−π/4) = π/4; dy = sec2 (−π/4) (π/4) = π/2; and ∆y = tan 0 − tan(−π/4) = 1. (c) ∆x = 3 − 0 = 3; dy = √
∆y = 18. (a) 19. (a)
(b) 20. 21. √ 25 − 32 − √ −0
25 − (0)2 (3) = 0; and 25 − 02 = 4 − 5 = −1. 56
43 − 23
=
= 28
4−2
2 (b) (dV /d ) =5 = 3 2
=5 = 3(5)2 = 75 dW
dW
= 200(t − 15); at t = 5,
= −2000; the water is running out at the rate of 2000 gal/min.
dt
dt
W (5) − W (0)
10000 − 22500
=
= −2500; the average rate of ﬂow out is 2500 gal/min.
5−0
5 46π
46π
π
; let x0 = and x =
. Then
180
4
180
π
π
π
x−
=1−2
cot 46◦ = cot x = cot − csc2
4
4
4
with a calculator, cot 46◦ = 0.9657. cot 46◦ = cot (a) (a) 1
21
dT
=√ √ =√
dL
g2 L
gL (d)
(e) = 0.9651; 51
π
π radians and dφ = ±0.5◦ = ±0.5
180
180
radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so the height lies between
139.48 m and 144.55 m.
If dh ≤ 5 then dφ ≤ (c) 46π π
−
180
4 h = 115 tan φ, dh = 115 sec2 φ dφ; with φ = 51◦ = (b)
22. −x
=
25 − x2 5
115 cos2 51
180 π ≈ 0.017 radians, or dφ ≤ 0.98◦ .
(b) s/m dT
> 0 an increase in L gives an increase in T , which is the period. To speed up a clock,
Since
dL
decrease the period; to decrease T , decrease L.
√
dT
L
= − 3/2 < 0; a decrease in g will increase T and the clock runs slower
dg
g
√
√
dT
−1
L
−3/2
=2 L
g
= − 3/2
(f ) s3 /m
dg
2
g 97 23. Chapter 3 (a) f (x) = 2x, f (1.8) = 3.6 (b) f (x) = (x2 − 4x)/(x − 2)2 , f (3.5) ≈ −0.777778 24. (a) f (x) = 3x2 − 2x, f (2.3) = 11.27 25. f (x) = 2x ln 2; f (2) ≈ 2.772589 (b) f (x) = (1 − x2 )/(x2 + 1)2 , f (−0.5) = 0.48 26. f (x) = xsin x (cos x ln x + sin x/x); f (2) = 0.312141
27.
28. 3(h + 1)2.5 + 580h − 3
1 d 2.5
= 58 +
3x
h→0
10h
10 dx vinst = lim
164 ft/s 29. = 58 +
x=1 1
(2.5)(3)(1)1.5 = 58.75 ft/s
10 Solve 3x2 − cos x = 0 to get x = ±0.535428. 2500 1 20
0 30. When x4 − x − 1 > 0, f (x) = x4 − 2x − 1; when x4 − x − 1 < 0,
f (x) = −x4 + 1, and f is diﬀerentiable in both cases. The roots of
x4 − x − 1 = 0 are x1 = −0.724492, x2 = 1.220744. So x4 − x − 1 > 0
on (−∞, x1 ) and (x2 , +∞), and x4 − x − 1 < 0 on (x1 , x2 ). Then
lim− f (x) = lim− (4x3 − 2) = 4x3 − 2 and
1
x→x1 1.5 2 x→x1 x→x+
1 1.5 x→x1 lim f (x) = lim+ −4x3 = −4x3 which is not equal to 4x3 − 2, so
1
1 f is not diﬀerentiable at x = x1 ; similarly f is not diﬀerentiable at
x = x2 .
(a) f (x) = 5x...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14
 The Land

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