# 52 by theorem 512 f is increasing on any interval 2n

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Unformatted text preview: x − 2 cos x lim = lim = lim = − = 4 if = ±2 2. x→0 x→0 x→0 x2 2x 2 2 59. If k = −1 then lim (k + cos x) = k + 1 = 0, so lim 60. (a) x→0 (b) 61. x→0 − cos(1/x) + 2x sin(1/x) which does not exist (nor is it ±∞). cos x x x 1 [x sin(1/x)], but lim = lim = 1 and lim x sin(1/x) = 0, thus Rewrite as lim x→0 sin x x→0 sin x x→0 cos x x→0 x [x sin(1/x)] = (1)(0) = 0 lim x→0 sin x Apply the rule to get lim x→0 sin(1/x) sin x , lim = 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between −1 x→0 (sin x)/x x→0+ x x sin(1/x) does not exist. and 1 as x → +∞, so lim+ x→0 sin x lim x→0+ CHAPTER 4 SUPPLEMENTARY EXERCISES 1. (a) f (g (x)) = x for all x in the domain of g , and g (f (x)) = x for all x in the domain of f . (b) They are reﬂections of each other through the line y = x. (c) The domain of one is the range of the other and vice versa. (d) The equation y = f (x) can always be solved for x as a function of y . Functions with no inverses include y = x2 , y = sin x. (e) (f ) 2. Yes, g is continuous; this is evident from the statement about the graphs in part (b) above. Yes, g must be diﬀerentiable (where f = 0); this can be inferred from the graphs. Note that if f = 0 at a point then g cannot exist (inﬁnite slope). (a) For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π ; for tan x, −π/2 < x < π/2; for sec x, 0 ≤ x < π/2 or π/2 < x ≤ π . 135 Chapter 4 (b) y y y = s in -1 x 1 y = s in x π/2 y = c os -1 x x π -1 x y = c os x -1 y y y = t an x 2 y = s ec -1 x y = t an -1 x y = s ec x 2 y = s ec -1 x π/2 - π/2 x π/2 x -1 -2 y = s ec x (a) when the limit takes the form 0/0 or ∞/∞ (b) 3. Not necessarily; only if lim f (x) = 0. Consider g (x) = x; lim g (x) = 0. For f (x) choose x→0 cos x x2 cos x, x2 , and |x|1/2 . There are three possibilities; lim does not exist; lim = 0, and x→0 x x→0 x 1/2 |x| lim = +∞. x→0 x2 4. In the case +∞− (−∞) the limit is +∞; in the case −∞− (+∞) the limit is −∞, because large positive (negative) quantities are being added to large positive (negative) quantities. The cases +∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are being subtracted, and more information about the sizes is needed. 5. (a) x = f (y ) = 8y 3 − 1; y = f −1 (x) = (b) (c) f (x) = (x − 1)2 ; f does not have an inverse because f is not one-to-one, for example f (0) = f (2) = 1. √ x = f (y ) = (ey )2 + 1; y = f −1 (x) = ln x − 1 = 1 ln(x − 1) 2 (d) x = f (y ) = x+1 8 1/3 = 1 (x + 1)1/3 2 y+2 x+2 ; y = f −1 (x) = y−1 x−1 ad − bc ; if ad − bc = 0 then the function represents a horizontal line, no inverse. If ad − bc = 0 (cx + d)2 ay + b b − xd then y = f −1 (x) = . then f (x) > 0 or f (x) < 0 so f is invertible. If x = f (y ) = cy + d xc − a 6. f (x) = 7. (a) (b) 2 −1/3 2 −1/3 x −y y − y = 0. At x = 1 and y = −1, y = 2. The tangent line is 3 3 y + 1 = 2(x − 1). Diﬀerentiating, (xy + y ) cos xy = y . With x = π/2 and y = 1 this becomes y = 0, so the equation of the tangent line is y − 1 = 0(x − π/2) or y = 1. 8. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1 (4/5)] = 3/5, sin[cos−1 (5/13)] = 12/13, cos[sin−1 (4/5)] = 3/5, cos[sin−1 (5/13)] = 12/13 (a) cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13)) − sin(cos−1 (4/5)) sin(sin−1 (5/13)) 33 4 12 3 5 − = . = 5 13 5 13 65 Supplementary Exercises 4 (b) 136 sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13)) 3 12 56 45 + = . = 5 13 5 13 65 9. 3 ln e2x (ex )3 + 2 exp(ln 1) = 3 ln e2x + 3 ln(ex )3 + 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2 10. Y = ln(Cekt ) = ln C + ln ekt = ln C + kt, a line with slope k and Y -intercept ln C 11. (a) ex ex ex = lim = +∞ = lim x→+∞ x→+∞ x→+∞ x2 x→+∞ 2x x→+∞ 2 x 2 2x 2 so lim (e /x − 1) = +∞ and thus lim x (e /x − 1) = +∞ lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim x→+∞ (b) (c) 12. 13. x→+∞ 1 ln x 1/x = lim 3 = ; lim 4−1 x→1 x x→1 4x 4 x→1 lim ln x = x4 − 1 1 ln x = 4−1 x→1 x 2 lim lim ax ln a = ln a x→0 y = aeax sin bx + beax cos bx and y = (a2 − b2 )eax sin bx + 2abeax cos bx, so y − 2ay + (a2 + b2 )y = (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0. √ √ sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y = y + 2 sin y cos3 y = 1 −2x x + 2√ = 0. 2 (1 + x2 )3/2 (1 + x2 )2 1+x 14. ln y = 2x ln 3 + 7x ln 5; 15. Find dy dy /y = 2 ln 3 + 7 ln 5, or = (2 ln 3 + 7 ln 5)y dx dx dy dz = a and = −b. From the ﬁgure dt dt √ sin θ = y/z ; when x = y = 1, z = 2. So θ = sin−1 (y/z ) and 1 y dz a 1 dy dθ = −2 = −b − √ when x = y = 1. 2 /z 2 dt z dt z dt 2 1−y dθ dt 1 −2x ,y = , hence 2 1+x (1 + x2 )2 given x=1 y =1 z y θ x 16. (a) f (x) = −3/(x +...
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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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