# If x 0 4 3 4 dadx 32 6x2 then a 0 2563 3 0

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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 x-window. 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 one-to-one 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 x-coordinate 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.

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