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Unformatted text preview: f inﬂection at x = 2 1 x 2 horizontal asymptote y = 0 as x → +∞, lim f (x) = −∞
x→−∞ 0.8 23. f (x) = x(2 − x)e1−x , f (x) = (x2 − 4x + 2)e1−x y
1.8 critical points at x = 0, 2;
relative min at x = 0,
relative max at x = 2
points of inﬂection at x = 2 ± 1 √
2 0.6 horizontal asymptote y = 0 as x → +∞, x
1 lim f (x) = +∞ 2 3 4 x→−∞ 24. f (x) = x2 (3 + x)ex−1 , f (x) = x(x2 + 6x + 6)ex−1 y critical points at x = −3, 0; 1 relative min at x = −3
points of inﬂection at x = 0, −3 ± √
3 0.4 horizontal asymptote y = 0 as x → −∞ 4 lim f (x) = +∞ (a) 1
0.4 x→+∞ 25. 2 (b) 40 5 5 40 1
, f (x) = 2x
400
1
critical points at x = ± ;
20
1
relative max at x = − ,
20
1
relative min at x =
20
f (x) = x2 − x Supplementary Exercises 172 (c) The ﬁner details can be seen when graphing over
a much smaller xwindow. 0.0001 0.1 0.1 0.0001 26. (a) (b) 200 5 5 √
critical points at x = ± 2, 3 , 2;
2
√
relative max at x = − 2,
√
relative min at x = 2,
relative max at x = 3 ,
2
relative min at x = 2 200 (c) 10 2.909 2.2 3.5
1.3
2.912 4 27. (a) (b) 6 5 1.6 Divide y = x2 + 1 into y = x3 − 8 to get the
asymptote ax + b = x 5 6 29. (a) p(x) = x3 − x (b) p(x) = x4 − x2 (c) 28. p(x) = x5 − x4 − x3 + x2 (d) p(x) = x5 − x3 f (x) = 4x3 − 18x2 + 24x − 8, f (x) = 12(x − 1)(x − 2)
f (1) = 0, f (1) = 2, f (1) = 2; f (2) = 0, f (2) = 0, f (2) = 3,
so the tangent lines at the inﬂection points are y = 2x and y = 3. 30. dy
dy dy
dy
cos x
=2 ;
= 0 when cos x = 0. Use the ﬁrst derivative test:
=
and
dx
dx dx
dx
2 + sin y
2 + sin y > 0, so critical points when cos x = 0, relative maxima when x = 2nπ + π/2, relative minima
when x = 2nπ − π/2, n = 0, ±1, ±2, . . .
cos x − (sin y ) 173 31. Chapter 5 (2x − 1)(x2 + x − 7)
x2 + x − 7
=2
,
2 + x − 1)
(2x − 1)(3x
3x + x − 1
horizontal asymptote: y = 1/3, √
vertical asymptotes: x = (−1 ± 13)/6 f (x) = y x = 1/2
10 5 4 2 2 4 x 5 32. (a) (x − 2)(x2 + x + 1)(x2 − 2)
(x − 2)(x2 − 2)2 (x2 + 1)
x2 + x + 1
=2
(x − 2)(x2 + 1) f (x) = y (b)
4 x
2 1 3 2 33. (a) sin x = −1 yields the smallest values, and
sin x = +1 yields the largest 3 O 0 o (b) f (x) = esin x cos x; relative maxima at x = 2nπ + π/2, y = e; relative minima at x = 2nπ − π/2,
y = 1/e; n = 0, ±1, ±2, . . . (ﬁrst derivative test) (c) f (x) = (1 − sin x − sin2 x)esin x ; f (x) = 0 when sin x = t, a root of t2 + t − 1 = 0,
√
√
−1 − 5
−1 ± 5
; sin x =
is impossible. So the points of inﬂection on 0 < x < 2π occur
t=
2
2
√
−1 + 5
, or x = 0.66624, 2.47535
when sin x =
2 34. f (x) = 3ax2 + 2bx + c; f (x) > 0 or f (x) < 0 on (−∞, +∞) if f (x) = 0 has no real solutions so
from the quadratic formula (2b)2 − 4(3a)c < 0, 4b2 − 12ac < 0, b2 − 3ac < 0. If b2 − 3ac = 0, then
f (x) = 0 has only one real solution at, say, x = c so f is always increasing or always decreasing on
both (−∞, c] and [c, +∞), and hence on (−∞, +∞) because f is continuous everywhere. Thus f is
always increasing or decreasing if b2 − 3ac ≤ 0. 35. (a) relative minimum −0.232466 at x = 0.450184 2 1 1.5
0.5 Supplementary Exercises 174 0.2 (b) relative maximum 0 at x = 0;
relative minimum −0.107587 at x = ±0.674841
1.2 1.2 0.15
1 (c) relative maximum 0.876839; at x = 0.886352;
relative minimum −0.355977 at x = −1.244155 1.5 1.5 0.4 36. (a) f (x) = 2+3x2 − 4x3 has one real root at x = 1.14, a relative max; so f is onetoone for x ≤ 1.14 (b) f (1.14) = 3.07 so the domain of f −1 is (−∞, 3.07) and the range is (−∞, 1.14); f −1 (−1) = −0.70
37. (a) (b) 0.5 0 y = 0 at x = 0; lim y = 0
x→+∞ 5
0 (c)
(d) 38. relative max at x = 1/a, inﬂection point at x = 2/a
As a increases, the xcoordinate of the maximum and the inﬂection point move towards the
origin. (a) 4 4 2 2 2 0 2
0 (b) y = 1 at x = a; lim y = +∞, lim y = 0
x→−∞ x→+∞ (c) Since y is always negative and y is always positive, there are
no relative extrema and no inﬂection points.
(d) An increase in b makes the graph ﬂatter.
(e) An increase in a shifts the graph to the right. 175 39. Chapter 5 1
1
and f (x) = 2
; so f > 0 if x > 1 and therefore f is increasing on
x
x (x + 1)
o
[1, +∞). Next, f (1) = ln 2 − 1 < 0. Then by L’Hˆpital’s Rule,
f (x) = ln(1 + 1/x) − ln(1 + 1/x)
−1/x2
= lim
=1
x→+∞
x→+∞ (1 + 1/x)(−1/x2 )
1/x lim x ln(1 + 1/x) = lim x→+∞ and thus lim f (x) = lim
x→+∞ x→+∞ x ln(1 + 1/x) − 1
is indeterminate.
x 1
1
−
= 0.
x
x+1
Thus on [1, +∞) the function f starts negative and increases towards zero, so it is negative
on the whole interval. So f (x) is decreasing, and f (x) > f (x + 1). Set x = n and obtain
ln(1 + 1/n)n+1 > ln(1 + 1/(n + 1))n+2 . Since ln x and its inverse function ex are both increasing,
it follows that (1 + 1/n)n+1 > (1 + 1/(n + 1))n+2 .
By L’Hˆpital’s Rule lim f (x) = lim
o
x→+∞ x→+∞ ln 1 +...
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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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