# Is minimum when x 4 3 4 24 let r and h be the

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Unformatted text preview: CALCULUS HORIZON MODULE CHAPTER 5 1. The sum of the squares for the residuals for line I is approximately 12 +12 +12 +02 +22 +12 +12 +12 = 10, and the same for line II is approximately 02 + (0.4)2 + (1.2)2 + 02 + (2.2)2 + (0.6)2 + (0.2)2 + 02 = 6.84; line II is the regression line. 2. (a) y = 5.035714286x − 4.232142857 (b) y 14 10 6 2 1 4. (a) 3 4 x r = 0.9907002406 5. 2 S = 2.155239850t + 190.3600714; r = 0.9569426456 (b) yes, because r is close to 1 (c) (d) 6. 244.241068 mi/h It is assumed that the line still gives a good estimate in the year 2000. Y = ln y = bx + ln a has slope b and Y -intercept ln a. y = a + bX has slope b and y -intercept a. (c) Y = ln y = b ln x + ln a = bX + ln a has slope b and Y -intercept ln a. (d) 7. (a) (b) The same algebraic rules hold. (a) y = 3.923208367 + e0.2934589528x (b) y 30 25 20 15 10 5 1 234567 x Calculus Horizon Module 176 It appears that log T = a + b log d, so T = 10a db , an exponential model. log T = 1.719666407 × 10−4 + 1.499661719 log d (c) T = 1.000396046 d1.499661719 (d) 9. (a) (b) 8. “The squares of the periods of revolution of the planets are proportional to the cubes of their mean distances” (a) T = 27 + 57.8 e−0.046t (b) T0 = 84.9◦ C (c) 53.19 min CHAPTER 6 Applications of the Derivative EXERCISE SET 6.1 1. relative maxima at x = 2, 6; absolute maximum at x = 6; relative and absolute minimum at x = 4 2. relative maximum at x = 3; absolute maximum at x = 7; relative minima at x = 1, 5; absolute minima at x = 1, 5 y 3. (a) (b) y (c) y 3 2 10 4. (a) 5 x 7 x 7 x (b) y (c) y y x x -5 5 x 5. f (x) = 8x − 4, f (x) = 0 when x = 1/2; f (0) = 1, f (1/2) = 0, f (1) = 1 so the maximum value is 1 at x = 0, 1 and the minimum value is 0 at x = 1/2. 6. f (x) = 8 − 2x, f (x) = 0 when x = 4; f (0) = 0, f (4) = 16, f (6) = 12 so the maximum value is 16 at x = 4 and the minimum value is 0 at x = 0. 7. f (x) = 3(x − 1)2 , f (x) = 0 when x = 1; f (0) = −1, f (1) = 0, f (4) = 27 so the maximum value is 27 at x = 4 and the minimum value is −1 at x = 0. 8. f (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2), f (x) = 0 when x = −1, 2; f (−2) = −4, f (−1) = 7, f (2) = −20, f (3) = −9 so the maximum value is 7 at x = −1 and the minimum value is −20 at x = 2. 9. √ √ f (x) = 3/(4x2 + 1)3/2 , no critical points; f (−1) = −3/ 5, f (1) = 3/ 5 so the maximum value is √ √ 3/ 5 at x = 1 and the minimum value is −3/ 5 at x = −1. 10. 2(2x + 1) , f (x) = 0 when x = −1/2 and f (x) does not exist when x = −1, 0; 3(x2 + x)1/3 f (−2) = 22/3 , f (−1) = 0, f (−1/2) = 4−2/3 , f (0) = 0, f (3) = 122/3 so the maximum value is 122/3 at x = 3 and the minimum value is 0 at x = −1, 0. f (x) = 11. f (x) = 1 − sec2 x, f (x) = 0 for x in (−π/4, π/4) when x = 0; f (−π/4) = 1 − π/4, f (0) = 0, f (π/4) = π/4 − 1 so the maximum value is 1 − π/4 at x = −π/4 and the minimum value is π/4 − 1 at x = π/4. 177 Exercise Set 6.1 12. 13. 178 f (x) = cos x + sin x, f √x) = 0 for x in (0, π ) when x = 3π/4; f (0) = −1, f (3π/4) = ( the maximum value is 2 at x = 3π/4 and the minimum value is −1 at x = 0. √ 2, f (π ) = 1 so 10 − x2 , |x| ≤ 3 −2x, |x| < 3 , f (x) = thus f (x) = 0 when x = 0, 2x, |x| > 3 −8 + x2 , |x| > 3 f (x) does not exist for x in (−5, 1) when x = −3 because lim − f (x) = lim + f (x) (see Theorem f (x) = 1 + |9 − x2 | = x→−3 x→−3 preceding Exercise 75, Section 3.3); f (−5) = 17, f (−3) = 1, f (0) = 10, f (1) = 9 so the maximum value is 17 at x = −5 and the minimum value is 1 at x = −3. 14. −4, x < 3/2 6 − 4x, x ≤ 3/2 , f (x) does not exist when , f (x) = 4, x > 3/2 −6 + 4x, x > 3/2 x = 3/2 thus 3/2 is the only critical point in (−3, 3); f (−3) = 18, f (3/2) = 0, f (3) = 6 so the maximum value is 18 at x = −3 and the minimum value is 0 at x = 3/2. f (x) = |6 − 4x| = 15. f (x) = 2x − 3; critical point x = 3/2. Minimum value f (3/2) = −13/4, no maximum. 16. f (x) = −4(x + 1); critical point x = −1. Maximum value f (−1) = 5, no minimum. 17. f (x) = 12x2 (1 − x); critical points x = 0, 1. Maximum value f (1) = 1, no minimum because lim f (x) = −∞. x→+∞ 18. f (x) = 4(x3 + 1); critical point x = −1. Minimum value f (−1) = −3, no maximum. 19. No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞. x→+∞ x→−∞ 20. No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞. x→+∞ x→−∞ 21. f (x) = x(x + 2)/(x + 1)2 ; critical point x = −2 in (−5, −1). Maximum value f (−2) = −4, no minimum. 22. f (x) = −6/(x − 3)2 ; no critical points in [−5, 5] (x = 3 is not in the domain of f ). No maximum or minimum because lim+ f (x) = +∞ and lim− f (x) = −∞. x→3 23. x→3 (x2 − 1)2 can never be less than zero because it is the square of x2 − 1; the minimum value is 0 for x = ±1, no maximum because lim f (x) = +∞. 10 x→+∞ -2 2 0 24. (x − 1)2 (x + 2)2 can never...
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