Unformatted text preview: t; x < +∞
(f ) ln x + x, x > 0
ex
(h)
,x>0
x x2 = lim x→+∞ 1+ y = 2x, lim (1 + y )2/y = lim (1 + y )1/y (a) y = 3x, lim lim (1 + x)1/3x = lim (1 + x)1/x y →+∞ x→0 1
y y /3 = lim (a) 16. (a) 17. F (x) = 1/3 (a)
18. = e2
1
y y 1/3 = lim y →+∞ (b) eln x = e1/3 − 1
x2 sin 1
x cos x
−(x2 + 3) sin x − 2x cos x
, F (x) =
2+3
x
(x2 + 3)2 0 (b) 1/3
√
3x
3x2 + 1, F (x) = √
3x2 + 1 F (x) =
0 (b) (a) d
dx
x2
1 y 1/3 1
y 1
=1
x (b) √
2x x2 + 1 (b) 1+ 14. g (x) = 1 − cos x 3
1
(3x2 ) =
3
x
x (a)
19. 2 ln 2 = e1/3 g (x) = x2 − x 15. 1+ y →+∞ x→0 13. 2 y →0 1+ √ =e = e2 (b) 12. 2 x2 1
x (b) y →0 √ x2
1 √
t 1 + tdt = x2 (c) (c) √
13 1 + x2 (2x) = 2x3 1 + x2 √
22
22
42
3/2
5/2
t 1 + tdt = − (x + 1) + (x + 1) −
3
5
15
√ 0 √
6/ 13 247 20. Chapter 7 (a)
(b) a d
dx f (t)dt = − x
a d
dx d
dx f (t)dt = − g (x) x f (t)dt = −f (x) a
g (x) d
dx f (t)dt = −f (g (x))g (x) a tan2 x
sec2 x = − tan2 x
1 + tan2 x 21. (a) − sin x2 (b) − 22. (a) −(x2 + 1)40 (b) − cos3 23. −3 1
x − 1
x2 = cos3 (1/x)
x2 x2 − 1
3x − 1
+ 2x 4
9x2 + 1
x +1 24. If f is continuous on an open interval I and g (x), h(x), and a are in I then
g (x) a g (x) f (t)dt = f (t)dt + h(x) so h(x)
g (x) d
dx h(x) f (t)dt = − a g (x) f (t)dt +
a f (t)dt
a f (t)dt = −f (h(x))h (x) + f (g (x))g (x) h(x) 25. sin2 (x3 )(3x2 ) − sin2 (x2 )(2x) = 3x2 sin2 (x3 ) − 2x sin2 (x2 )
1
2
1
(1) −
(−1) =
(b)
1+x
1−x
1 − x2 26. F (x) = 27. from geometry, (a) 1
1
(3) − (1) = 0 so F (x) is constant on (0, +∞). F (1) = ln 3 so F (x) = ln 3 for all x > 0.
3x
x
3 5 f (t)dt = 0,
0 (a) 7 f (t)dt = 6,
3 10 f (t)dt = 0; and
5 10 f (t)dt =
7 (4t − 37)/3dt = −3 7 F (0) = 0, F (3) = 0, F (5) = 6, F (7) = 6, F (10) = 3 (b) F is increasing where F = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2] and
[6, 37/4]
(c)
(d) critical points when F (x) = f (x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum
−9/4 at x = 3/2
F(x)
6
4
2
2 4 6 8 10 x 2 28. fave = 1
10 − 0 10 f (t)dt =
0 1
F (10) = 0.3
10 x 29. x < 0 : F (x) =
x ≥ 0 : F (x) = 1
(−t)dt = − t2
2
−1
0
−1 x =
−1 x (−t)dt + t dt =
0 1
(1 − x2 ),
2 1 12
+ x ; F (x) =
22 (1 − x2 )/2, x < 0
(1 + x2 )/2, x ≥ 0 Exercise Set 7.9 30. 248
x 0 ≤ x ≤ 2 : F (x) = t dt =
0 2 x > 2 : F (x) = x t dt +
0 x 31. y (x) = 2 +
1
x 32. y (x) = 3
t1/3 dt = 2 + t4/3
4 (t1/2 + t−1/2 )dt =
x x = 2x − 2, x > 2 1 5 3 4/3
+x
44 2 3/2 2
x − + 2x1/2 − 2
3
3 (sec2 t − sin t)dt = tan x + cos x − y (x) = 1 + x2 /2, 0 ≤ x ≤ 2 2 dt = 2 + 2(x − 2) = 2x − 2; F (x) = 2 1 33. 12
x,
2 √
2/2 π/4
x 34. 2 tet dt = y (x) =
0 x 1 −x2 1
e
−
2
2 35. P (x) = P0 + r(t)dt individuals
0 T 36. s(T ) = s1 + v (t)dt
1 37. II has a minimum at x = 1, and I has a zero there, so I could be the derivative of II; on the other
hand I has a minimum near x = 1/3, but II is not zero there, so II could not be the derivative of I
38. (b)
39. 1
o
lim (xk − 1) = ln x lim xk = ln x by L’Hˆpital’s rule (with respect to k )
k →0
k k →0 (a) where f (t) = 0; by the ﬁrst derivative test, at t = 3 (b) where f (t) = 0; by the ﬁrst derivative test, at t = 1 (c) at t = 0, 1 or 5; from the graph it is evident that it is at t = 5 (d) at t = 0, 3 or 5; from the graph it is evident that it is at t = 3 (e) F is concave up when F = f is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4);
it is concave down on (1/2, 2) and (4, 5) (f ) F(x)
1
0.5
1 2 3 4 2 4 5 x 0.5
1 40. (a) erf (x)
1 4 2 x 1 (c)
(e)
(g) erf (x) > 0 for all x, so there are no relative extrema
2√
erf (x) = −4xe−x / π changes sign only at x = 0 so that is the only point of inﬂection
lim erf(x) = +1, lim erf(x) = −1 x→+∞ x→−∞ 249 41. Chapter 7 C (x) = cos(πx2 /2), C (x) = −πx sin(πx2 /2)
cos t goes from negative to positive at 2kπ − π/2, and from positive to negative at t = 2kπ + π/2,
√
so C (x) has relative minima when πx2 /2 = 2kπ − π/2, x = ± 4k − 1, k = 1, 2, . . ., and C (x)
√
has relative maxima when πx2 /2 = (4k + 1)π/2, x = ± 4k + 1, k = 0, 1, . . ..
√
(b) sin t changes sign at t = kπ , so C (x) has inﬂection points at πx2 /2 = kπ , x = ± 2k , k = 1, 2, . . .;
the case k = 0 is distinct due to the factor of x in C (x), but x changes sign at x = 0 and
sin(πx2 /2) does not, so there is also a point of inﬂection at x = 0
(a) F (x + h) − F (x)
1
= lim
h→0
h→0 h
h x 42. Let F (x) =
1 1
h→0 h ln tdt = ln x
x
x Diﬀerentiate: f (x) = 3e3x , so 2 + (a) (c) = 2 + e3x − e3a = e3x provided 3e3t dt = 2 + e3t
a a The area under 1/t for x ≤ t ≤ x + 1 is less than the area of the rectangle with altitude 1/x and
base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1.
x+1 (b) x x f (t)dt = 2 +
a e3a = 2, a = (ln 2)/3.
44. ln tdt; but F (x) = ln x so
x x+h lim 43. x+h ln tdt, F (x) = lim x+1 1
dt = ln t
= ln(x + 1) − ln x = ln(1 + 1/x), so
t
x
x
1/(x + 1) < ln(1 + 1/x) < 1/x for...
View
Full
Document
 Spring '14
 The Land

Click to edit the document details