Ch 5 solutions

# 2 14 23 1 2 1 2 t t t c s0 04 03 0 c

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Unformatted text preview: x2 on does not, so there it is concave down /2)(1/2, 2) and (4, 5) is also a point of inﬂection at x = 0 F(x￿ ) (f ) 1) = 46. Let F (x 1 lim h→0 h 0.5 ￿ x F (x + h) − F (x) 1 ln tdt, F (x) = lim = lim h→0 h→0 h h ￿ 1 x+h –0.5x 1ln tdt = 23 ln x 5 x –1 47. Diﬀerentiate: f (x) = 2e2x , so 4 + (f ) provided e2a = 4, a = (ln 4)/2. ￿ x f (t)dt = 4 + a ￿ ￿ Chapter 5 x+h ln tdt; but F ￿ (x) = ln x so x x 2e2t dt = 4 + e2t a ￿x a = 4 + e2x − e2a = e2x 2 2 45. (a) erf( x 44. C (x) = cos(πx /2), C (x)) = −πx sin(πx /2). 1 48. (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 at kπ − rectangle with altitude 1 negative at base kπ (a) cos t goes from negative to positivearea2of theπ/2, and from positive to/(x + 1) and t = 21. + π/2, so C (x) √ ￿x+1 ￿ x+1 has relative minima when πx2 /2 = 2kπ − π/2, x = ± 4k − 1, k = 1, 2, . . ., and C (x) has relative maxima when x √ 1 πx2 /2 –4 (4k + dt π/2, 4t = ±= 4k + + 1) = 0, 1, . . .ln(1 + 1/x), so (b) = –2 1) = ln x ln(x 1, k − ln x = . 2 t x x √ 1 t changes ln(1 + t = < /x for x > 0. (b) sin/(x + 1) <sign at 1/x)kπ ,1so C (x) has inﬂection points at πx2 /2 = kπ , x = ± 2k , k = 1, 2, . . .; the case –1 k = 0 is distinct due to /(x+1) the factorln(1+1/x)C (x/x but/(x+1) of x in ), x changes sign at x 1= 0 and sin(πx2 /2) does not, so there is (c) from Part (b), e1 <e < e1 , e1 < 1 + 1/x < e /x , also a point of inﬂection at x = 0. ex/(x+1) < (1 + 1/x)x < e; by the Squeezing Theorem, lim (1 + 1/x)x = e. x x x→+∞ x 47. Diﬀerentiate: f (x) = 2e2x , x/(x4 + so +1) f (t)dt = 4 + 2e2t dt = 4 + e2t = 4 + e2x − e2a = e2x provided e2a = 4, (d) Use the inequality e < (1 + 1/x)x to get e < (1 + 1/x)x+1 a so a a a = (ln(1 + 1= ln 2. e < (1 + 1/x)x+1 . 4)/2 /x)x < ￿ ￿ ￿0 ￿ ￿ 5 50 ￿ 11 ￿ ￿ 49. From Exercise 48(d) ￿ − ￿ < (50), and from the graph y (50) < 0.06. < y y (50), and from the graph y (50) < 0.06 + 49. From Exercise 48(d) e e − 1 1 + ￿ ￿ 50 50 0.2 0 100 0 50. F ￿ (x) = f (x), thus F ￿ (x) has a value at each x in I because f is continuous on I so F is continuous Chapter because a function that is diﬀerentiable at a point is also continuous at that point on I 5 Review Exercises 1. − 1 8 + x3/2 + C . 2 4x 3 3. −4 cos x + 2 sin x + C . 5. 3x1/3 − 5ex + C . 7. tan−1 x + 2 sin−1 x + C . √ √ 2 4 2 4 9. (a) y (x) = 2 x − x3/2 + C ; y (1) = 0, so C = − , y (x) = 2 x − x3/2 − . 3 3 3 3 (b) y (x) = sin x − 5ex + C, y (0) = 0 = −5 + C, C = 5, y (x) = sin x − 5ex + 5. x (c) y (x) = 2 + 1 x 2 3 t1/3 dt = 2 + t4/3 4 tet dt = (d) y (x) = 0 1 x2 1 e−. 2 2 x = 1 5 3 4/3 +x . 44 Chapter 5 Review Exercises 155 11. (a) If u = sec x, du = sec x tan xdx, sec2 x tan xdx = du = sec2 xdx, sec2 x tan xdx = udu = u2 /2 + C1 = (sec2 x)/2 + C1 ; if u = tan x, udu = u2 /2 + C2 = (tan2 x)/2 + C2 . (b) They are equal only if sec2 x and tan2 x diﬀer by a constant, which is true. 13. u = x2 − 1, du = 2x dx, du 1 1 √ = sec−1 |u| + C = sec−1 |x2 − 1| + C . 2 2 u u2 − 1 1 2 1 1 1√ √ du = u1/2 + C = 5 + 2 sin 3x + C . 3 3 6u 15. u = 5 + 2 sin 3x, du = 6 cos 3x...
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