# CalcReview b.pdf

## Calculus 1571 with Haught at Trumbull Career and Technical Center *

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Solutions to the Mathematics 1571 Final Exam Review Problems 1. Consider the function f defined by f (x) = 2x 2 − 5x . Determine the following: (a) f (a + b) (b) f (2x) − 2 f (x) Solution: (a) Substituting in (a + b) for x , gives f (a + b) =2(a + b)2 − 5(a + b) =2(a2 + 2ab + b2) − 5(a + b) =2a2 + 4ab + 2b2 − 5a − 5b (b) Again, substituting in appropriate quantities for x , gives f (2x) − 2 f (x) =2(2x)2 − 5(2x) − 2(2x2 − 5x) =2 · 4x2 − 10x − [ 4x2 − 10x ] =8x2 − 4x2 − 10x + 10x =4x2 2. Find the domain of g if (a) g(x) = √ x2 − 3x − 4 (b) g(x) = x + 2 x3 − x Solution: (a) The domain will be the set of real numbers such that x 2 − 3x − 4 ≥ 0. Factoring, this gives (x − 4)(x + 1) ≥ 0 One way to solve this inequality is to use sign analysis. Page 2 of 68 x − 4 − − − − − − − − + + + + x + 1 − − − − + + + + + + + + Product + + + + − − − − + + + + −1 4 So the domain is given by x ≥ 4 or x ≤ −1. In interval notation this set is (−∞, 1] ∪ [4,∞). (b) First factor the denominator g(x) = x + 2 x3 − x = x + 2 x(x2 − 1) = x + 2 x(x + 1)(x − 1). The domain is all real numbers except those for which the denominator is zero. The denominator will be zero when : x = 0 or x + 1 = 0 or x − 1 = 0 that is, when x = 0 or x = −1 or x = +1 In interval notation, the domain is (−∞,−1) ∪ (−1, 0) ∪ (0, 1). 3. The graph of f (x) = √x + 3 − 1 is the same as the graph of f (x) = √x except that it is translated in what way? Solution: Since the one is negative, the graph is moved downward 1 unit, and since the 3 is positive, it is moved 3 units to the left. 4. For the functions f and g, find f ◦ g, g ◦ f , and the domains of each. Page 3 of 68 (a) f (x) = 1 x2 − 1 and g(x) = √ x (b) f (x) = x2 + 1 and g(x) = x x − 1 Solution: (a) f (x) = 1 x2 − 1 , g(x) = √ x ( f ◦ g)(x) = f (g(x)) = 1 (g(x))2 − 1 = 1 ( √ x)2 − 1 = 1 x − 1 To find the domain of f ◦ g: • f ◦ g is not defined where g is not defined. • f ◦ g is not defined where f is not defined at the value g takes at x . So: g(x) is not defined for x < 0 [eliminate x < 0 from domain] and the denominator of f is not defined at x = ±1, that is f ◦ g is not defined at g(x) = ±1, that is, at x = 1 (since √x is never negative). The range is given by x ≥ 0, x = 1, that is, [0, 1) ∪ (1,∞) g(x) = √x , f (x) = 1 x2 − 1 (g ◦ f )(x) =g( f (x)) =g ( 1 x2 − 1 ) = √ 1 x2 − 1 For the domain, we must have 1 x2 − 1 ≥ 0, so we must have x 2 − 1 ≥ 0. x2 − 1 = (x + 1)(x − 1) Sign analysis: Page 4 of 68 x − 1 − − − − − − − − + + + + x + 1 − − − − + + + + + + + + x2 − 1 + + + + − − − − + + + + −1 1 x2 − 1 ≥ 0 exactly when x ≤ −1 or x ≥ +1, that is, |x | ≥ 1. Also, x2 − 1 must not be zero, so x = ±1. So, |x | > 1. Domain: (−∞,−1) ∪ (1,∞). (b) f (x) = x2 + 1 , g(x) = x x − 1. ( f ◦ g)(x) = f (g(x)) = f ( x x − 1 ) = ( x x − 1 )2 + 1 = x 2 (x − 1)2 + 1 = x 2 (x − 1)2 + (x − 1)2 (x − 1)2 = x 2 + (x − 1)2 (x − 1)2 = x 2 + x2 − 2x + 1 (x − 1)2 =2x 2 − 2x + 1 (x − 1)2 The domain will be all real numbers x such that x = 1, that is (−∞, 1) ∪ (1,∞). {x ∈ R|x = 1} (g ◦ f )(x) =g ( f (x)) =g(x2 + 1) = x 2 + 1 x2 + 1 − 1 = x2 + 1 x2 Page 5 of 68 Domain: All real numbers such that x = 0, that is (−∞, 0) ∪ (0,∞). {x ∈ R|x = 0} 5. Find the intercepts of the following equations. Also determine whether the equations are symmetric with respect to the y-axis or the origin. (a) y = x4 + x3 + x2 (b) y = 1 x3 − 3x (c) y = 2 − |x | Solution: (a) y-intercepts: Let x = 0. Then y = 04 + 03 + 02 = 0. y-intercept is 0. x-intercepts: y =0 x4 + x3 + x2 =0 x2(x2 + x + 1) =0 x2 = 0 or x2 + x + 1 = 0 x = 0 or by quadratic formula x = −1 ± √ 1 − 4 · 1 · 1 2 x = 0 is only x-intercept. Symmetry: Substitute −x for x : (−x)4 + (−x)3 + (−x)2 =x4 − x3 + x2 =x4 + x3 + x2 So not symmetric about y-axis. Origin: y = x4 + x3 + x2 Page 6 of 68 Substitute −x for x and −y for y. −y =(−x)4 + (−x)3 + (−x)2 y = − x4 + x3 − x2 Not symmetric about origin. (b) y = 1 x3 − 3x y-intercepts: y does not exist when x = 0, so there are no y-intercepts. x-intercepts: 0 = 1 x3−3x has no solutions. No intercepts. Exchange −x for x : 1 (−x)3 − 3(−x) = 1 −x3 + 3x = − 1 x3 − 3x Not symmetric with respect to y-axis. (c) y = 2 − |x | y-intercepts: Let x = 0. Then y = 2 − |0| = 2 (0, 2) x-intercepts: Let y = 0. Then 0 =2 − |x | |x | =2 x = ± 2 (2, 0), (−2, 0) Intercepts: (0, 2), (2, 0), (−2, 0) Symmetry: Page 7 of 68 Check for symmetry with respect to y-axis: 2 − | − x | = 2 − |x | Symmetric about y-axis. Check for symmetry about origin: Exchange x and y: x = 2 − |y| This doesn’t even give a function, so there is not symmetry about the origin. 6. Determine (a) lim x→1 f (x), if f (x) = ⎧⎪⎨ ⎪⎩ 4x − 2, if x < 1 7, if x = 1 2x − x2, if x > 1 (b) lim x→−2 g(x), if g(x) = { x + 5, if x ≥ −2 2x2 − x − 7, if x < −2 Solution: (a) The limit does not depend on the value at x = 1. It exists iff lim x→1+ f (x) and lim x→1− f (x) both exist, and are equal. lim x→1+ f (x) = lim x→1+ 2x − x2 = 2 − 1 = 1 lim x→1− f (x) = lim x→1− 4x − 2 = 4 − 2 = 2 = 1 lim x→1 f (x) does not exist. (b) This limit exists iff both directed limits exist and are equal: lim x→−2− g(x) = lim x→−2− 2x2 − x − 7 = 2(−2)2 − (−2) − 7 = 2 · 4 + 2 − 7 = 3 lim x→−2+ g(x) = lim x→−2+ x + 5 = −2 + 5 = 3 So lim x→−2 g(x) =3 7. Determine (a) lim x→2 (√ x − 1 − √3x − 2 ) (b) lim x→−1 x3 + 1 x2 − 4x − 5 Page 8 of 68 (c) lim x→∞ 4x3 + x − 5 x3 − 2 (d) limx→7− x2 + 49 x − 7 (e) lim x→7+ x2 + 49 x − 7 (f) limx→7 x2 + 49 x − 7 (g) lim x→ 12 + |2x − 1| 2x − 1 (h) limx→ 12 − |2x − 1| 2x − 1 (i) lim x→ 12 |2x − 1| 2x − 1 (j) limx→−∞ x2 + 4x − 1 x4 (k) lim x→−∞ 3√ x3 + 2x2 + 1 3x − 5 (l) lim x→−3+ √ x2 − 9 (m) lim x→∞ sin ( 1 x ) x (n) lim x→π2 − sec x − 1 tan x (o) lim x→1 √ 8 + x2 − 3 x − 1 (p) limh→0 tan ( π 3 + h )− tan π3 h (q) lim h→0 sin ( π 6 + h )− sin (π6 ) h (r) lim h→0 cos (x + h) − cos x h Solution: (a) lim x→2 (√ x − 1 − √3x − 2 ) =√2 − 1 − √6 − 2 = √ 1 − √ 4 =1 − 2 = − 1 (b) lim x→−1 x3 + 1 x2 − 4x − 5 Simply substituting x = −1 doesn’t work. It yields 00 lim x→−1 x3 + 1 x2 − 4x − 5 = limx→−1 (x + 1)(x2 − x + 1) (x + 1)(x − 5) = lim x→−1 x2 − x + 1 x − 5 Page 9 of 68 =(−1) 2 − (−1) + 1 −1 − 5 = 3 −6 = − 1 2 (c) lim x→∞ 4x3 + x − 5 x3 − 2 = 4 (d) lim x→7− x2 + 49 x − 7 = −∞ Substituting x = 7 yields 49 + 49 0 , but x −7 approaches 0 from below as x approaches 7 from below, so answer is negative. (e) lim x→7+ x2 + 49 x − 7 = +∞ since substituting x = 7 yields form 49 + 49 0 and denominator approaches 0 from above. (f) lim x→7 x2 + 49 x − 7 does not exist, by combining results of (d) and (e). (g) lim x→ 12 + |2x − 1| 2x − 1 = +1 Explanation: For x > 12 , 2x − 1 is positive, so |2x − 1| = 2x − 1. (h) lim x→ 12 − |2x − 1| 2x − 1 = −1 Explanation: For x < 12 , 2x − 1 is negative, so |2x − 1| = −(2x − 1). (i) lim x→ 12 |2x − 1| 2x − 1 does not exist, by answers to (g) and (h). (j) lim x→−∞ x2 + 4x − 1 x4 = 0 Explanation: Both leading terms are positive, x 4 dominates. (k) lim x→−∞ 3√ x3 + 2x2 + 1 3x − 5 = limx→−∞ ( 3√ x3 + 2x2 + 1 ) · 1 x 3 − 5 x = lim x→−∞ ( 3 √ 1 + 2 x + 1 x3 ) 3 − 5 x = 3√1 3 = 1 3 (l) lim x→−3+ √ x2 − 9 At first glance, this may appear to be 0. However, note that for values slightly above Page 10 of 68 −3, x2 − 9 is negative. Since the expression does not exist as x approaches −3 from above, neither does the limit. (m) lim x→∞ sin ( 1 x ) x = 0 (Remark: As x → ∞, 1 x → 0 from above, so sin ( 1 x ) → 0 also. So this has form 0∞ , so answer is 0.) (n) lim x→π2 − sec x − 1 tan x sec π2 − = ∞ tan π2 − =undefined (∞) Substituting x = π2 gives ∞∞ . lim x→π2 − sec x − 1 tan x = lim x→π2 − 1 cos x − 1 sin x cos x = lim x→π2 − 1 − cos x sin x =1 − 0 1 =1 (o) lim x→1 √ 8 + x2 − 3 x − 1 Substituting x = 1 yields 00: √ 8 + 1 − 3 1 − 1 = √ 9 − 3 0 = 0 0 Multiplying by the conjugate of the numerator gives lim x→1 √ 8 + x2 − 3 x − 1 = limx→1 (8 + x2)1/2 − 3 x − 1 · (8 + x2)1/2 + 3 (8 + x2)1/2 + 3 = lim x→1 (8 + x2) − 9 (x − 1)((8 + x2)1/2 + 3) = lim x→1 x2 − 1 (x − 1)((8 + x2)1/2 + 3) = lim x→1 (x − 1)(x + 1) (x − 1)((8 + x2)1/2 + 3) Page 11 of 68 = lim x→1 x + 1 (8 + x2)1/2 + 3 = 2 6 = 1 3 (p) lim h→0 tan ( π 3 + h )− tan π3 h The limit is the derivative of the tangent function at π3 . Hence, lim h→0 tan ( π 3 + h )− tan π3 h = ( sec (π 3 ))2 =4 (q) lim h→0 sin ( π 6 + h )− sin (π6 ) h The limit is the derivative of the sine function at π6 , so lim h→0 sin ( π 6 + h )− sin (π6 ) h = cos (π 6 ) = √ 3 2 (r) lim h→0 cos(x + h) − cos x h The limit is the derivative of the cosine function , so lim h→0 cos(x + h) − cos x h = − sin(x) 8. Determine P such that the following functions are continuous at x = 3. (a) f (x) = ⎧⎨ ⎩ 3(x4 − 81) x2 − 9 , if x = 3 Px + 9, if x = 3 (b) g(x) = ⎧⎨ ⎩ x3 − 27 x2 − 9 , if x = 3 P, if x = 3 Solution: (a) f (x) = ⎧⎨ ⎩ 3(x4 − 81) x2 − 9 , if x = 3 Px + 9, if x = 3 Page 12 of 68 Factoring the top expression: lim x→3 f (x) = lim x→3 3(x2 − 9)(x2 + 9) x2 − 9 = limx→3 3(x 2 + 9) = 3(9 + 9) = 54 Since 3(x2 + 9) is continuous and equal to the first expression except at x = 3, the function may be made continuous by setting Px + 9 equal to 54. So, Px + 9 = 54 and P = 15 (b) g(x) = ⎧⎨ ⎩ x3 − 27 x2 − 9 , if x = 3 P, if x = 3 Factoring: lim x→3 x3 − 27 x2 − 9 = limx→3 (x2 + 3x + 9)(x − 3) (x + 3)(x − 3) = limx→3 x2 + 3x + 9 x + 3 = 9 + 9 + 9 3 + 3 = 27 6 = 9 2 Since x2 + 3x + 9 x + 3 is continuous except at x = −3 and equal to x3 − 27 x2 − 9 except at x = 3, the function may be made continuous by letting P equal to 92 . 9. Determine the largest intervals on which the functions defined are continuous. (a) f (x) = { 8 − 7x, if x ≤ 4 −x − 16, if x > 4 (b) f (x) = { x2 + 1, if x ≤ −3 5 − x, if x > −3 Solution: (a) Since linear functions are always continuous, the only possibility for discontinuity is at x = 4: lim x→4− f (x) = lim x→4− 8 − 7x = 8 − 28 = −20 lim x→4+ f (x) = lim x→4+ −x − 16 = −20 f (4) =8 − 28 = −20 Thus, the function is everywhere continuous, i.e. on (−∞,+∞). (b) Since linear and quadratic functions are continuous, the only possibility for disconti- Page 13 of 68 nuity is at x = −3. lim x→3− f (x) = lim x→3− (x2 + 1) = 32 + 1 = 10 f (3) =32 + 1 = 10 lim x→3+ f (x) = lim x→3+ (5 − x) = 5 − 3 = 2 So intervals lim x→3 f (x) does not exist. The intervals are (−∞,−3) and (−3,∞) 10. Identify all asymptotes of the graphs following. (a) y = x − 2 x − 1 (b) y = x2 − 3x + 2 x2 − 4 (c) y = √ x2 + 4 x (d) y = x 2 − 9 x2 − 5x + 6 Solution: (a) y = x − 2 x − 1 Horizontal: lim x→∞ x − 2 x − 1 =1 lim x→−∞ x − 2 x − 1 =1 Horizontal asymptotes in both directions: y = 1 Vertical asymptotes: where denominator is 0: x − 1 =0 x =1 (b) y = x 2 − 3x + 2 x2 − 4 Page 14 of 68 Horizontal asymptotes: lim x→∞ x2 − 3x + 2 x2 − 4 =1 lim x→−∞ x2 − 3x + 2 x2 − 4 =1 y =1 Vertical asymptotes: where denominator is 0 x + 2 =0 x = − 2 (x − 2) does not yield an asymptote, since it results only in a point discontinuity. (c) y = √ x2 + 4 x Vertical: Denominator is 0 at x = 0. Vertical asymptote: x = 0 Horizontal asymptotes: For x > 0: y = √ x2 + 4 x = √ x2 + 4 · 1 x x · 1 x = √ x2+4 x2 1 = √ 1 + 4 x2 lim x→∞ y = limx→∞ √ 1 + 4 x2 = 1 For x < 0: y = √ x2 + 4 x = √ x2 + 4 · 1 x x · 1 x = − √ x2+4 x2 1 = − √ 1 + 4 x2 lim x→−∞ y = limx→−∞ − √ 1 + 4 x2 = −1 Asymptotes: y = 1, y = −1 (d) y = x 2 − 9 x2 − 5x + 6 = (x + 3)(x − 3) (x − 2)(x − 3) = x + 3 x − 2 (x = 3) Page 15 of 68 Vertical asymptotes: at x = 2 Horizontal asymptotes: For x = 3, y =x + 3 x − 2 = x − 2 x − 2 + 5 x − 2 = 1 + 5 x − 2 lim x→∞ y = limx→∞ 1 + 5 x − 2 = 1 lim x→−∞ y = limx→−∞ 1 + 5 x − 2 = 1 x = 2, y = 1 11. Give a specific example to show that it is possible for lim x→a f (x) to exist when f (a) is unde- fined. Solution: f (x) = x 2 − 1 x − 1 , a = 1. f (1) does not exist, but limx→1 f (x) = 2. Other Examples: f (x) = x 2 + 2x + 1 x + 1 , a = −1. f (−1) does not exist, but limx→−1 f (x) = 0. f (x) =sin x x , a = 0. f (0) does not exist, but lim x→0 f (x) = 1. 12. Let y = 7√ x4 − 15. Determine y ′ when x = 2. Solution: y = 7√ x4 − 15 = 7 (x4 − 15)1/2 =7(x4 − 15)−1/2 y′ =7 · −1 2 · (x4 − 15)−3/2 · 4x3 Page 16 of 68 = − 7 2 · 4 · (x4 − 15)−3/2 · x3 = − 14x3(x4 − 15)−3/2 So y′(2) = − 14 · 23(24 − 15)−3/2 = − 14 · 8 · (16 − 15)−3/2 = − 14 · 8 · 1 = − 112 13. Let f (x) = 3 √ (2x + 17)2. Determine f ′(5). Solution: f (x) = [ (2x + 17)2 ]1/3 =(2x + 17)2·1/3 =(2x + 17)2/3 f ′(x) =2 3 (2x + 17)−1/3 · 2 f ′(5) =2 3 (10 + 17)−1/3 · 2 =2 3 (27)−1/3 · 2 =2 3 · 1 3 · 2 =4 9 14. Determine y ′ when x = 1 if y = 8x 3 − 3 8x . Solution: y =8 3 x − 3 8 x−1 y′ =8 3 + 3 8 x−2 Page 17 of 68 y′(1) =8 3 + 3 8 (1)−2 =8 3 + 3 8 =8 · 8 + 3 · 3 3 · 8 =64 + 9 24 = 73 24 15. Determine y ′ when x = 1 if y = (2 − x)√x2 + 8. Solution: y =(2 − x)(x2 + 8)1/2 y′ =(2 − x) · 1 2 (x2 + 8)−1/2 · 2x + (−1)(x2 + 8)1/2 y′(1) =(2 − 1) · 1 2 (12 + 8)−1/2 · 2 · 1 + (−1)(12 + 8)1/2 =1 · 1 2 (9)−1/2 · 2 − (9)1/2 =9−1/2 − 91/2 =1 3 − 3 = 1 3 − 9 3 = −8 3 16. Find an equation of the tangent line to the curve defined by y = 2x 2 − 5x + 8 when x = 2. Solution: Find the first derivative: y′ = 4x − 5 The slope of the tangent curve is given by y ′(2): m = y′(2) = 4 · 2 − 5 = 8 − 5 = 3. We also need a point on the line: (Use the point where the tangent line intersects the curve.) y =2 · 22 − 5 · 2 + 8 =8 − 10 + 8 = 16 − 10 = 6 Page 18 of 68 So (2, 6) is on the line. The equation has the form (y − 6) =3(x − 2) y − 6 =3(x − 2) y =3(x − 2) + 6 y =3x − 6 + 6 y =3x 17. At what point (x, y) is the tangent line to the curve y = 2x 2 − 5x + 8 parallel to the line y = 3x − 7 Solution: • Slope of the given line is 3. • Need to find x such that y ′ = 3. dy dx = 4x − 5 Let 4x − 5 =3 4x =8 x =2 Find y(2): y(2) =2 · 22 − 5 · 2 + 8 =8 − 10 + 8 = 6 At (x, y) = (2, 6), the tangent line is parallel to y = 3x − 7. 18. Find the slope of the line tangent to the graph of x 2 + 2xy2 + 3y = 31 at the point (2,−3). Solution: A. Verify that (2,−3) is on the graph: (2)2 + 2(2)(−3)2 + 3(−3) =31 4 + 4 · 9 + (−9) =31 4 + 36 − 9 =31 40 − 9 =31 Page 19 of 68 B. Differentiate Implicitly d dx (x2 + 2xy2 + 3y) = d dx (31) 2x + 2x2yy′ + 2y2 + 3y′ = 0 C. Let x = 2 and y = −3. 2(2) + 2(2)(2)(−3)y ′ + 2(−3)2 + 3y′ =0 4 + −24y ′ + 2 · 9 + 3y ′ =0 4 + 18 − 24y ′ + 3y′ =0 22 − 21y ′ =0 −21y ′ = − 22 y′ =22 21 ← slope of the desired tangent line To find the tangent line itself: (y − (−3)) =22 21 (x − 2) y + 3 =22 21 (x − 2) 19. Let y = √x2 − 1. Find y ′′ when x = 2. Solution: y =(x2 − 1)1/2 y′ =1 2 (x2 − 1)−1/2 · 2x y′ =(x2 − 1)−1/2x y′′ = − 1 2 (x2 − 1)−3/22x · x + (x2 − 1)−1/2 · 1 = − (x2 − 1)−3/2x2 + (x2 − 1)−1/2 = − (x2 − 1)−1 · (x2 − 1)−1/2 · x2 + (x2 − 1)−1/2 = [ −(x2 − 1)−1 · x2 + 1 ] (x2 − 1)−1/2 = [ − x 2 x2 − 1 + 1 ] (x2 − 1)−1/2 Page 20 of 68 = [ −x2 x2 − 1 + x2 − 1 x2 − 1 ] (x2 − 1)−1/2 = −1 x2 − 1(x 2 − 1)−1/2 y′′(2) = −1 4 − 1(2 2 − 1)−1/2 = −1 3 (3)−1/2 = − 1 3 √ 3 = − √ 3 9 20. Let f (x) = √ x + √x . Find f ′(1). Solution: f (x) = ( x + x1/2 )1/2 f ′(x) =1 2 ( x + x1/2 )−1/2 [ 1 + 1 2 x−1/2 ] f ′(1) =1 2 ( 1 + 11/2 )−1/2 [ 1 + 1 2 · 1−1/2 ] f ′(1) =1 2 · 2−1/2 [ 3 2 ] f ′(1) =1 2 · 1√ 2 · 3 2 = 1 2 · √ 2 2 · 3 2 = 3 √ 2 8 21. Determine f ′(1) if f (x) = 3 √ x x3 + 1. Solution: Recall ( f g )′ = f ′g − g′ f g2 . f (x) = ( x x3 + 1 )1/3 f ′(x) =1 3 ( x x3 + 1 )−2/3 · 1 · (x 3 + 1) − (x3 + 1)′ · x (x3 + 1)2 =1 3 ( x x3 + 1 )−2/3 · (x 3 + 1) − 3x2 · x (x3 + 1)2 Page 21 of 68 f ′(1) =1 3 ( 1 13 + 1 )−2/3 · (1 3 + 1) − 3 · 12 · 1 (13 + 1)2 =1 3 ( 1 2 )−2/3 · 2 − 3 22 =1 3 ( 1 2 )−2/3 · −1 4 =−1 12 (4)1/3 = − 3√4 12 22. Determine f ′(3) if f (x) = 1 x − √x2 − 5. Solution: f (x) = [ x − (x2 − 5)1/2 ]−1 f ′(x) = − 1 · [ x − (x2 − 5)1/2 ]−2 [ 1 − 1 2 (x2 − 5)−1/2 · 2x ] f ′(3) = − 1 · [ 3 − (9 − 5)1/2 ]−2 [ 1 − 1 2 (9 − 5)−1/2 · 2 · 3 ] f ′(3) = − 1 · [ 3 − 41/2 ]−2 [ 1 − 1 2 (4)−1/2 · 2 · 3 ] f ′(3) = − 1 · [3 − 2]−2 [ 1 − 1 2 · 1 2 · 2 · 3 ] f ′(3) = − 1 · [1]−2 [ 1 − 3 2 ] = − 1 · 1 · −1 2 = +1 2 23. Determine y ′′ at x = 1 if y = 3 3 √ x4 − 1 3x3 . Solution: y =3x4/3 − 1 3 x−3 y′ =3 · 4 3 x1/3 − 1 3 (−3)x−4 y′ =4x1/3 + x−4 Page 22 of 68 y′′ =4 3 x−1/3 − 4x−5 y′′(1) =4 3 (1)−1/3 − 4(1)−5 =4 3 − 4 = 4 3 − 12 3 = −8 3 24. Let y = [cos(2x − π)]3. Find y ′ at x = π6 . Solution: y = [cos(2x − π)]3 y′ =3 [cos(2x − π)]2 · [− sin(2x − π)] · 2 y′ (π 6 ) =3 [ cos ( 2 · π 6 − π )]2 · [− sin(2 · π 6 − π )] · 2 =3 [ cos (π 3 − π )]2 · [− sin(π 3 − π )] · 2 =3 [ cos (−2π 3 )]2 · [ − sin (−2π 3 )] · 2 =6 · [ −1 2 ]2 · [ − ( − √ 3 2 )] =6 · 1 4 · √ 3 2 =6 8 √ 3 = 3 4 √ 3 25. Determine f ′(π4 ) if f (x) = tan x 1 + cos x . Solution: Use Quotient Rule: f ′(x) =(sec x) 2 [1 + cos x] − tan x [− sin x] [1 + cos x]2 f ′(x) = ( sec π4 )2 [1 + cos π4 ]− tan π4 [− sin π4 ][ 1 + cos π4 ]2 Page 23 of 68 Recall: cos π 4 = sin π 4 = √ 2 2 tan π 4 =1 sec π 4 = √ 2( sec π 4 )2 =2 f ′ (π 4 ) = 2 · [ 1 + √ 2 2 ] − 1 · [ − √ 2 2 ] [ 1 + √ 2 2 ]2 = 2 + √ 2 + √ 2 2 1 + 2 (√ 2 2 ) + [√ 2 2 ]2 =2 + √ 2 + √ 2 2 1 + √2 + 12 Noticing similarity between numerator and denominator: f ′ (π 4 ) = ( 2 + √2 + √ 2 2 )√ 2( 1 + √2 + 12 )√ 2 = ( 2 + √2 + √ 2 2 )√ 2(√ 2 + 2 + √ 2 2 ) = √2 Alternatively, to do the simplification we could write: f ′(x) =2 + √ 2 + √ 2 2 1 + √2 + 12 =2 + 3 2 √ 2 3 2 + √ 2 · 3 2 − √ 2 3 2 − √ 2 Page 24 of 68 =3 + (9 4 − 2 )√ 2 − 32 · 2 9 4 − 2 = 1 4 √ 2 1 4 = √ 2 26. Let f (x) = sin 3x cos 2x . Find f ′ (π6 ). Solution: Use Product Rule f ′(x) = [sin 3x cos 2x]′ = [sin 3x]′ (cos 2x) + (sin 3x) [cos 2x]′ =3 cos 3x · cos 2x + sin 3x · (−2 sin 2x) f ′ (π 6 ) =3 cos ( 3 · π 6 ) · cos ( 2 · π 6 ) + sin ( 3 · π 6 ) · ( −2 sin 2 · π 6 ) =3 cos (π 2 ) · cos (π 3 ) + sin (π 2 ) · ( −2 sin π 3 ) =3 · 0 · cos (π 3 ) + 1 · ( −2 · √ 3 2 ) =0 + ( − √ 3 ) = − √ 3 cos π 2 =0 1 = 0 sin π 2 =1 1 = 1 sin π 3 = √ 3 2 27. Let y4 + x4 − 2x2y + 9x = 9. Find y′ at (1, 1). Solution: Use implicit differentiation, but group terms involving y together first. (This isn’t all that important, but it makes it less confusing to do it before the equation gets messier.) y4 − 2x2y + x4 + 9x =9 Page 25 of 68 d dx (y4 − 2x2y + x4 + 9x) = d dx (9) 4y3y′ − 2x2y′ − 4xy + 4x3 + 9 =0 Plug in y = 1 and x = 1: 4 · 13 · y′ − 2 · 12 · y′ − 4 · 1 · 1 + 4 · 13 + 9 =0 4y′ − 2y′ − 4 + 4 + 9 =0 2y′ + 9 =0 2y′ = − 9 y′ = − 9 2 28. Suppose that f ′(x) = x2(4x − 3) and f ′′(x) = 6x(2x − 1). Determine the intervals on which f is increasing. Solution: Use sign analysis (or another method if you prefer): Need: f ′(x) >0 x2(4x − 3) >0 x2 · ( x − 3 4 ) · 4 >0 x2 + + + + + + + + + + + + x − 34 − − − − − − − − + + + + x2(4x − 3) − − − − − − − − + + + + 0 34 x2(4x − 3) > 0 exactly when x > 34 . So f ′(x) > 0 on (34 ,+∞). So f is increasing on [34,+∞). 29. Suppose that f ′(x) = x − 1 2x + 3 and f ′′(x) = 5 (2x + 3)2 . Determine the intervals on which f is decreasing. Solution: f is decreasing whenever f ′(x) < 0: Top, x − 1 is > 0 when Page 26 of 68 x − 1 − − − − − − − − + + + + 2x + 3 − − − − + + + + + + + + f ′(x) + + + + − − − − + + + + −32 1 f is decreasing on (−32, 1] Note that −32 is excluded because f ′(−32) is undefined and f ′ → −∞ as x → −32 from above. 30. Determine the intervals on which f is concave upward if f (x) = 1 10 x5 + 1 6 x4 − 4x3 + 87x + 69. Solution: Differentiating: f ′(x) = 1 10 · 5x4 + 1 6 · 4x3 − 4 · 3x2 + 87 =1 2 x4 + 2 3 x3 − 12x2 + 87 f ′′(x) =1 2 · 4x3 + 2 3 · 3x2 − 12 · 2x =2x3 + 2x2 − 24x Factor f ′′(x), if possible: f ′′(x) =2(x3 + x2 − 12x) =2x(x2 + x − 12) =2x(x + 4)(x − 3) Do sign analysis or otherwise determine where f ′′(x) is positive: Page 27 of 68 x − − − − − − − − + + + + + + + + x + 4 − − − − + + + + + + + + + + + + x − 3 − − − − − − − − − − − − + + + + 2x(x + 4)(x − 3) − − − − + + + + − − − − + + + + −4 0 3 From sign analysis, f ′′(x) > 0 for −4 < x < 0 and x > 3 This gives as intervals for concave up: (−4, 0) and (3,∞) 31. Let f ′(x) = −3 2(x2 − 4)2 and f ′′(x) = 6x (x2 − 4)3 . Determine the intervals on which f is concave downward. Solution: f ′(x) = −3 2(x2 − 2)2 = − 3 2 (x2 − 2)−2 f ′′(x) = − 3 2 (−2)(x2 − 2)−3(2x) = 6x(x2 − 4)3 = 6x (x2 − 4)3 f is concave downward when f ′′(x) < 0. 6x (x2 − 4)3 = 6x (x + 2)3(x − 2)3 Zeros at: 0, −2, +2 6x − − − − − − − − + + + + + + + + x + 2 − − − − + + + + + + + + + + + + x − 2 − − − − − − − − − − − − + + + + f ′′(x) − − − − + + + + − − − − + + + + −2 0 2 Page 28 of 68 f is concave downward on (−∞,−2) and (0, 2). 32. Determine all points of inflection for f (x) = 1 12 x4 + 1 6 x3 − 6x2. Solution: f ′(x) =1 3 x3 + 1 2 x2 − 12x f ′′(x) =x2 + x − 12 f ′′(x) =0 x2 + x − 12 =0 (x − 4)(x + 3) =0 x =4,−3 Since x2 + x − 12 is a quadratic having two zeros, it must change sign at both zeros. hence points of inflection occur when x = 4 and x = −3. 33. Determine all points of inflection for f (x) = 1 12 x4 − 7 3 x3 + 49 2 x2 + 88. Solution: f ′(x) =1 3 x3 − 7x2 + 49x f ′′(x) =x2 − 14x + 49 f ′′(x) =0 x2 − 14x + 49 =0 (x − 7)(x − 7) =0 x =7 is the only root. It is necessary to see if f ′′(x) actually changes sign at x = 7, but it doesn’t since f ′′(x) = x2 − 14x + 49 = (x − 7)2 is a perfect square and cannot be negative. No inflection points 34. Let f (x) = 1 3 x3 + 3x2 + x − 2. Determine all local minima for f . Solution: Page 29 of 68 f ′(x) = x2 + 6x + 1 f ′′(x) = 2x + 6 x2 + 6x + 1 = 0 x =−6 ± √ 36 − 4 · 1 · 1 2 = − 3 ± √ 32 2 = − 3 ± 4 √ 2 2 = − 3 ± 2 √ 2 f ′′(−3 + 2 √ 2) =2(−3 + 2 √ 2) + 6 = − 6 + 4 √ 2 + 6 =4 √ 2 > 0 minimum f ′′(−3 − 2 √ 2) =2(−3 − 2 √ 2) + 6 = − 6 − 4 √ 2 + 6 = − 4 √ 2 < 0 maximum Minimum at −3 + 2√2 = 2√2 − 3. 35. Let f (x) = 1 4 x4 − 5 2 x2 + 3. Determine all local maxima for f . Solution: f ′(x) =x3 − 5x f ′(x) =0 x3 − 5x =0 x(x2 − 5) =0 x =0,± √ 5 f ′′(x) =3x2 − 5 f ′′(0) =3 · 02 − 5 = −5 maximum Page 30 of 68 f ′′(+ √ 5) =3( √ 5)2 − 5 = 15 − 5 = 10 > 0 minimum f ′′(− √ 5) =3(− √ 5)2 − 5 = 15 − 5 = 10 > 0 minimum Local maximum at 0 36. Find the maximum and minimum values of the functions on the given intervals. (a) f (x) = x3 − 12x on the interval [0, 3]. (b) f (x) = x3 − 12x on the interval [−3, 0]. Solution: (a) f ′(x) =3x2 − 12 f ′′(x) =6x Interval [0, 3]: Let f ′(x) = 0 3x2 − 12 =0 x2 − 4 =0 x2 =4 x = + 2 or − 2 2 lies in the interval, −2 doesn’t. So check 0, 3, and 2: f (0) =03 − 12 · 0 = 0 f (3) =33 − 12 · 3 = 27 − 36 = −9 f (2) =23 − 12 · 2 = 8 − 24 = −16 Minimum is −16, maximum is 0. (b) Interval [−3, 0]: Let f ′(x) = 0 3x2 − 12 =0 x2 − 4 =0 x2 =4 x = + 2 or − 2 Page 31 of 68 −2 lies in the interval, 2 doesn’t. So check −3, 0, and −2: f (−3) =(−3)3 − 12(−3) = −27 + 36 = 9 f (0) =03 − 12 · 0 = 0 f (−2) =(−2)3 − 12(−2) = −8 + 24 = 16 Minimum is 0, maximum is 16. 37. A rock thrown from the top of a cliff is s(t) = 192 + 64t − 16t 2 feet above the ground t seconds after being thrown. Determine (a) the height of the cliff, (b) the time it takes the rock to reach the ground, and (c) the velocity of the rock when it strikes the ground. Solution: (a) The height of the cliff is the same as the height of the rock immediately after being thrown. This is given by s(0) = 192 + 64 · 0 − 16 · 02 = 192. The cliff is 192 feet high. (b) t to reach the ground is given by solving for t in s(t) = 0, i.e. solving for the rock being zero feet above the ground. s(t) =0 0 =192 + 64t − 16t 2 Note: 192 = 12 · 16 and 64 = 4 · 16, so 0 =16(12 + 4t − t2) t2 − 4t − 12 =0 (t − 6)(t + 2) =0 t = 6 or t = −2 t = −2 is not appropriate to question. So, it takes 6 seconds for the rock to reach the ground. Page 32 of 68 (c) To find the velocity of the rock when it strikes the ground, find s ′(6): s(t) =192 + 64t − 16t 2 s ′(t) =0 + 64 − 2 · 16t = 64 − 32t s ′(6) =64 − 32 · 6 = 64 − 192 = −128 The rock is traveling at 128 feet per second when it strikes the ground. [The minus sign indicates that the direction of the motion is downward, because the expression for s(t) gives altitude above the ground.] 38. A rock is thrown vertically upward from the roof of a house 32 feet high with an initial velocity of 128 ft/sec. (a) What is the speed of the rock at the end of 2 seconds? (b) What is the maximum height the rock will reach? Solution: (a) Leaving off units: a = − 32 v0 = + 128 s0 = + 32 a = dv dt , v = ds dt s = 32 So, v = ∫ a(t) dt + c1 v = ∫ −32 dt + c1 = − 32t + c1 v(0) = + 128 Page 33 of 68 v(0) = − 32 · 0 + c1 c1 = + 128 v(t) =128 − 32t s(t) = ∫ v(t) dt + c2 = ∫ (128 − 32t) dt + c2 =128t − 32 2 t2 + c2 32 = s0 =128 · 0 − 322 · 0 2 + c2 32 =c2 So, s(t) =128t − 32 2 t2 + 32 s(t) =128t − 16t 2 + 32 v(t) =128 − 32t Recall from previous work: a(t) = −32 To find speed after 2 seconds, first check s to see that rock has not hit the ground: s(2) =128 · 2 − 16 · 22 + 32 =256 − 64 + 32 =256 − 32 = 224 > 0 v(2) =128 − 32 · 2 = 128 − 64 = 64 Velocity after 2 seconds is 64 feet/sec. upward. Remark: Because s(t) = −16t2 + 128t + 32 has form of parabola, opening down- wards, and the initial velocity is directed upward, we know that once s(t) reaches 0, it will continue decreasing. Thus, s(2) > 0 guarantees that the rock does not hit the ground before or at 2 seconds. Note that once the rock hits the ground, none of the equations apply. Alternatively: Since v(2) > 0, we know that the maximum height has not been reached, so the rock cannot have hit the ground. (b) To find maximum height: Maximize s(t): s ′(t) = v(t) =128 − 32t Page 34 of 68 Let v(t) =0 128 − 32t =0 128 =32t 32t =128 t =128 32 = 4 Maximum height is at 4 seconds. s(t) = − 16t2 + 128t + 32 max height = s(4) = − 16 · 42 + 128 · 4 + 32 = − 16 · 16 + 128 · 4 + 32 = − 256 + 512 + 32 =256 + 32 = 288 Maximum height is 288 feet. 39. What is the maximum area which can be enclosed by 200 ft. of fencing if the enclosure is in the shape of a rectangle and one side of the rectangle requires no fencing? Solution: Let w be the length of one of the sides whose opposite requires fencing and l the length of the side whose opposite does not. w l Then we want to maximize A = l · w where l + 2w = 200. l =200 − 2w A = l · w =(200 − 2w)w = 200w − 2w2 To maximize A, find the derivative and set it equal to 0: d A dw = 200 − 4w =0 Page 35 of 68 200 =4w 4w =200 w =50 Plug in to find maximum area: Amax = [ 200w − 2w2 ] w=50 = [ 200 · 50 − 2 · 502 ] =10000 − 5000 = 5000 The maximum area that can be enclosed is 5000 sq. feet. 40. A woman throws a ball vertically upward from the ground. The equation of its motion is given by s(t) = −16t 2 + ct , where c is the initial velocity of the ball. She wants the ball to reach a maximum height of 100 ft. Find c. Solution: To find maximum height given c, differentiate s(t) and set the derivative equal to 0. s ′(t) = −32t + c =0 −32t = − c t = −c−32 = c 32 Find s ( c 32 ) and set it equal to 100. s ( c 32 ) =100 −16 ( c 32 )2 + c ( c 32 ) =100 −16 c 2 322 + c 2 32 =100 −c2 64 + c 2 32 =100 −c2 64 + 2c 2 64 =100 c2 64 =100 Page 36 of 68 c2 =64 · 100 = 6400 c = √ 6400 = 80 c =80 41. A rectangular open tank is to have a square base, and its volume is to be 125 yd3. The cost per square yard for the base is $8 and for the sides is $4. Find the dimensions of the tank in order to minimize the cost of the material. Solution: Let B be the length of the side of the base and H be the height (in yards). Volume is given by V = B · B · H = B2H Requirement: V = 125, B2 · H = 125 B H B Also, must minimize total cost: C =8 · B2 + 4 · (4 · B H) C =8B2 + 16B H H =125 B2 C =8B2 + 16B 125 B2 C =8B2 + 16 · 125 B = 8B2 + 16 · 125B−1 Minimize total cost: dC d B = 16B + (−1)16 · 125B−2 dC d B = 16B − 16 · 125B−2 =0 B − 125B−2 =0 B =125B−2 B3 =125 Page 37 of 68 B =5 Total cost is minimized when B = 5. So, H = 125 B2 = 125 25 = 5 So, make a tank 5yd × 5yd × 5yd to minimize total cost. 42. A power station is on one side of a river which is 1/2 mile wide, and a factory is 1 mile downstream on the other side of the river. It costs $300 per foot to run power lines overland and $500 per foot to run them under water. Find the most economical way to run the power lines from the power station to the factory. Solution: V L Power station Factory 1/2 mi 1 mi Let V be the length under water, and L the length above water, in miles. Thus: V 2 = (1 − L)2 + ( 1 2 )2 Also, let C be total cost, and α be the number of feet in a mile. Thus: C = 300αL + 500αV Our goal is to maximize C with respect to L (and V ). We know C =300αL + 500αV and also that: V 2 =(1 − L)2 + ( 1 2 )2 V = √ (1 − L)2 + ( 1 2 )2 Page 38 of 68 so that: C =300αL + 500α √ (1 − L)2 + ( 1 2 )2 and differentiating: dC d L =300α + 500α 1 2 1√ (1 − L)2 + (12)2 2(1 − L)(−1) =300α + 500α (−1)(1 − L)√ (1 − L)2 + (12)2 =300α − 500α (1 − L)√ (1 − L)2 + (12)2 dC d L =0 300α − 500α (1 − L)√ (1 − L)2 + (12)2 =0 500α (1 − L)√ (1 − L)2 + (12)2 =300α 5 (1 − L)√ (1 − L)2 + (12)2 =3 5(1 − L) =3 √ (1 − L)2 + ( 1 2 )2 25(1 − L)2 =9 [ (1 − L)2 + ( 1 2 )2] 25(1 − L)2 =9(1 − L)2 + 9 4 16(1 − L)2 =9 4 (1 − L)2 = 9 64 1 − L = √ 9 64 = 3 8 1 − L =3 8 Page 39 of 68 L =5 8 The power line should run 5 8 mile overland. 43. A cardboard box manufacturer needs to make open boxes from pieces of cardboard 12 in. square by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out in order to obtain a box of the largest possible volume. What is the largest possible volume? Solution: 12 - 2x x Volume is V =x · (12 − 2x)2 =x · [2(6 − x)]2 =4x(6 − x)2 To maximize volume, differentiate: V = 4x(6 − x)2 dV dx =4(6 − x)2 + 4x(2)(6 − x)(−1) =4(6 − x)2 + (−8x)(6 − x) =(6 − x) · 4 · ((6 − x) − 2x) =4(6 − x)(6 − 3x) Set dV dx = 0 4(6 − x)(6 − 3x) =0 6 − x = 0 or 6 − 3x = 0 x = 6 or x = 2 Check these: V |x=6 =6 · (12 − 2 · 6)2 = 6 · (0)2 = 0 V |x=2 =2 · (12 − 2 · 2)2 = 2 · (12 − 4)2 = 2 · 82 = 128 Maximum volume of 128in3 is achieved when the squares have sides of length 2. Page 40 of 68 44. A train leaves a station traveling north at the rate of 60 mph. One hour later, a second train leaves the same station traveling east at the rate of 45 mph. Find the rate at which the trains are separating 2 hours after the second train leaves the station. Solution: Let t be time in hours, after second train leaves station. At t = −1, 1 hour before second train leaves station, first train leaves station going north At t = 0, First train is at 60 miles N and second train leaves station going east At t > 0, First train is 60 + 60t north, second train is 45t east Let x(t) and y(t) be the distance of the eastbound and northbound trains from the station at time t . Then, respectively, x(t) = 45t and y(t) = 60 + 60t. The separation distance s(t) at time t is s(t) = √ (x(t))2 + (y(t))2 So ds dt = 1 2 √ (x(t))2 + (y(t))2 [ 2x(t) dx dt + 2y(t)dy dt ] Because x(2) = 90, y(2) = 180, dxdt = 45, and dydt = 60, we have ds dt (t = 2) = 1 2 √ (90)2 + (180)2 [2(90)(45) + 2(180)(60)] = 1 180 √ 5 [180(45 + 120)] =165√ 5 = 33 √ 5 Answer: 33 √ 5 mph is the rate of separation at the time two hours after the second train leaves the station. 45. A street light hangs on a pole 24 ft. above the sidewalk. A man 6 ft. tall walks away from the light at the rate of 3 ft/sec. (a) At what rate is the length of his shadow increasing? (b) At what rate is the tip of the shadow moving away from the pole? Page 41 of 68 6 distance from lamp = 3t 6 foot man shadow 24 lamp Solution: (a) • Let s be the length of the man’s shadow (in feet). • Note that the two triangles are similar. • Let t denote times in seconds. Then the distance from the point below the street lamp is given by 3t . • Since triangles are similar, 3t + s 24 = s 6 Solve for s: 3t 24 + s 24 = s 6 t 8 = s 6 − s 24 = 4s 24 − s 24 = 3s 24 = s 8 t 8 = s 8 , t = s The shadow is increasing at the rate ds dt = 1, i.e. 1 foot/second. (b) From above we have: • The distance from the street lamp is given by 3t , where t denotes the time in seconds. • The length of the shadow is given by t . • The distance from the pole to the tip of the shadow is 3t + t = 4t . Page 42 of 68 6 distance from lamp = 3t 6 foot man length of shadow = t 24 lamp Let the distance from the pole to the tip of the shadow be given by x = 4t . The distance is increasing at the rate dxdt = 4, i.e. 4 feet/second. 46. A barge is pulled toward a dock by means of a taut cable. If the barge is 20 ft. below the level of the dock, and the cable is pulled in at the rate of 36 ft/min, find the speed of the barge when the cable is 52 ft. long. Solution: Let l be the length of the cable in feet. l distance 20 ft We know dl dt = 36 (in ft/min) Also, it is known that the distance of the barge from the dock is given by (using the quadratic formula): s = √ l2 − (20)2 = √ l2 − 400 = (l2 − 400)1/2 The speed of the barge is given by ds dt = 1 2 (l2 − 400)−1/2(2l)dl dt Page 43 of 68 When the cable is 52 ft. long (l = 52), ds dt =1 2 (522 − 400)−1/2(2 · 52) · 36 =1 2 (2704 − 400)−1/2(104) · 36 =1 2 (2304)−1/2 · 104 · 36 = 1872√ 2304 = 1872 48 =39 So, the speed of the barge when the length of the rope is 52 ft. is 39 ft/min. 47. Let y = 2x2 − x , x = 2, and dx = x = 0.01. Find the values of (a) y (b) dy Solution: (a) dy dx =4x − 1 dy =(4x − 1) dx dy =(4 · 2 − 1) · 0.01 = (8 − 1) · 0.01 = 0.07 (b) y =y(2 + 0.01) − y(2) =y(2.01) − y(2) = [ 2(2.01)2 − (2.01) ] − [ 2(2)2 − 2 ] = [2 · 4.0401 − 2.01] − [8 − 2] = [8.0802 − 2.01] − 6 =6.0702 − 6 = 0.0702 48. Use differentials to approximate the maximum possible error that can be produced when calculating the volume of a cube if the length of an edge is known to be 2 ± 0.005 ft. Solution: x =2, dx = x = ±0.005 Page 44 of 68 y =x3 y′ =3x2 Maximum error | y| ≈ |dy| =|y ′ · dx | =|3x2 · dx | =|3(2)2 · (0.005)| = |12 · 0.005| = 0.060 Maximum possible error is approximately ±0.06 ft. 49. Use differentials to approximate √ 50. Solution: Let f (x) =x1/2 Then d f dx =1 2 x−1/2 d f =1 2 x−1/2 dx To approximate √ 50, let x = 49 and dx = 1. Then f (x) = f (49) = √49 = 7 and d f =1 2 49−1/2 · 1 =1 2 · 1 7 · 1 = 1 14 √ 50 ≈ f (49) + d f =7 + 1 14 ≈7.07 50. The moment of inertia of an annular cylinder is I = .5M(R22 − R21), where M is the mass of the cylinder, R2 is its outer radius, and R1 is its inner radius. Suppose that R1 = 2, and R2 changes from 4 to 4.01. Use differentials to estimate the resulting change in the moment of inertia. Solution: d I d R2 =.5M(2R2) Page 45 of 68 d I =.5M(2R2) d R2 and d R2 =0.01, R2 = 4 So: d I =.5M(2 · 4)(.01) =.5M(8)(.01) =0.04M 51. The range of a shell shot from a certain ship is R = 300 sin(2θ) meters, where θ is the angle above horizontal of the gun when it is shot. The gun is intended to be fired at an angle of π/6 radians to hit its target, but due to waves it actually shot .05 radians too low. Use differentials to estimate how far short of its target the shell will fall. Solution: θ =π 6 dθ = − .05 d R dθ =300 cos(2θ) · 2 = 600 cos(2θ) d R =600 cos(2θ)dθ d R =600 cos ( 2 · π 6 ) · (−.05) =600 cos π 3 · (−.05) =600 cos 60◦ · (−.05) =600 · 1 2 · −.05 = − 300 · .05 = −15 It will fall about 15 m short of the target. 52. In each of the following, determine whether the Intermediate Value Theorem guarantees the equation has a solution in the specified interval. (a) x5 + 2x2 − 10x + 5 = 0, [1, 2] (b) x − √x = 5, [4, 9] (c) x = cos x, [0, π ] (d) x = tan x, [π/4, 3π/4] Solution: Intermediate Value Theorem: Suppose that f is continuous on the closed interval [a, b] and let N be any number strictly between f (a) and f (b). Then there exists a number c in (a, b) such that f (c) = N . Page 46 of 68 (a) x5 + 2x2 − 10x + 5 =0, [1, 2] Let f (x) =x5 + 2x2 − 10x + 5 Note f is continuous f (1) =15 + 2 · 12 − 10 · 1 + 5 =1 + 2 − 10 + 5 = −2 f (2) =25 + 2 · 22 − 10 · 2 + 5 =32 + 8 − 10 + 5 = 25 Now, f (1) < 0 < f (2), so there is a c such that f (c) = 0, i.e. x 5 +2x2 −10x +5 = 0 is guaranteed a solution by Intermediate Value Theorem. (b) x − √x =5, [4, 9] Let f (x) =x − √x f (4) =4 − √ 4 = 4 − 2 = 2 f (9) =9 − √ 9 = 9 − 3 = 6 Since f (x) = x −√x is continuous and 2 < 5 < 6, Intermediate Value Theorem does guarantee a solution. (c) x = cos x , [0, π ] Let f (x) = x − cos x . Note that f (x) is continuous. Need c such that f (c) = 0. Now, f (0) =0 − cos 0 = −1 f (π) =π − cos π = π − (−1) = π + 1 Now, f (0) < 0 < f (π), so f (x) = 0 has a solution in 0, π , so x = cos x has a solution in [0, π ], by Intermediate Value Theorem. (d) x = tan x , [ π 4 , 3π 4 ] Let f (x) = x − tan x . Recall that tan x has discontinuities at π 2 , 3π 2 , 5π 2 , etc. and − π 2 , −3π 2 , etc. and π 4 < 2π 4 < 3π 4 and 2π 4 = π 2 , Page 47 of 68 so f (x) has a discontinuity in the range indicated, so Intermediate Value Theorem does not apply. Remark: This does not necessarily mean no solution exists, just that we aren’t guar- anteed one by Intermediate Value Theorem. 53. Suppose that f ′′(x) = 4x + 3, f (1) = 2, and f ′(1) = −3. Find f (x). Solution: f ′′(x) =4x + 3 f ′(x) = ∫ f ′′(x) dx + C = ∫ (4x + 3) dx + C =1 2 4x2 + 3x + C = 2x2 + 3x + C Find C: f ′(1) = 1 2 · 4 · 12 + 3 · 1 + C = − 3 1 2 · 4 + 3 + C = − 3 2 + 3 + C = − 3 C = − 8 f ′(x) =2x2 + 3x − 8 f (x) = ∫ f ′(x) dx + D = ∫ (2x2 + 3x − 8) dx + D =2 3 x3 + 3 2 x2 − 8x + D f (1) =2 2 3 · 13 + 3 2 · 12 − 8 · 1 + D =2 2 3 + 3 2 − 8 + D =2 D =2 + 8 − 2 3 − 3 2 Page 48 of 68 D =10 − 4 6 − 9 6 = 10 − 13 6 =60 6 − 13 6 = 47 6 f (x) =2 3 x3 + 3 2 x2 − 8x + 47 6 54. Suppose that f ′(x) = x2 + 3x + 2, and f (−3) = −32 . Find f (x). Solution: f (x) = ∫ f ′(x) dx + C = ∫ (x2 + 3x + 2) dx + C =x 3 3 + 3x 2 2 + 2x + C f (−3) = (−3) 3 3 + 3(−3) 2 2 + 2(−3) + C = − 3 2 −27 3 + 27 2 + (−6) + C = − 3 2 −9 + 27 2 − 6 + C = − 3 2 −15 + 27 2 + C = − 3 2 −30 2 + 27 2 + C = − 3 2 −3 2 + C = − 3 2 C =0 f (x) =x 3 3 + 3x 2 2 + 2x 55. Let f (t) be the function defined by the graph shown. Estimate the following to the nearest integer. (a) The instantaneous rate of change of f at t = 3. (b) The average rate of change of f over the interval [0, 4]. (c) The intervals where f (t) is increasing Page 49 of 68 and where it is decreasing. (d) The inflection point or points of f (t). (e) The intervals where f ′(t) is increasing and where it is decreasing. Solution: (a) 4 (b) 3 (c) f (t) is increasing in [1, 4] and decreasing in [0, 1] and [4, 5] (d) (2, 2) (e) f ′(t), the derivative of the function shown, is increasing on [0, 2] and is decreasing in [2, 5], as can be seen by noting the concavity of the function, which indicates the sign of the derivative of f ′(t). 56. Let v(t), as shown in the graph, be the velocity of a car in meters per second at time t in seconds, where positive velocity means the car is moving forward. Estimate to the nearest integer. (a) When the car stopped? (b) How far the car traveled in the time in- terval 8 to 12 seconds? (c) How far the car traveled in the time in- terval 12 to 14 seconds? (d) How far the car traveled in the time in- terval 8 to 14 seconds? Page 50 of 68 Solution: (a) When time is 12 seconds (b) 16 meters (c) 4 meters (d) 20 meters (Note: The total displacement is approximately 12 meters.) 57. Consider the car whose motion is described in Problem 56. (a) Is the car moving forward or backward at the time 2 seconds? (b) Is the driver’s foot on the gas or the brake at 2 seconds? (c) Is the car moving forward or backward at the time 16 seconds? (d) Is the driver’s foot on the gas or the brake at 16 seconds? Solution (a) Forward (b) Foot on the gas (c) Backward (d) Foot on the brake 58. Integrate the following: (a) ∫ 1 0 x(3x2 + 4)4 dx (b) ∫ (2x4 + 4x)3(2x3 + 1) dx (c) ∫ 2 1 (2x3 + 3x + 5) x5 dx (d) ∫ (3x2 + 6x + 2) x4 dx (e) ∫ 8 3 √ 12 − x dx (f) ∫ (3x − 1) √ 3x2 − 2x + 1 dx (g) ∫ 1 (x + 5)4 dx (h) ∫ (5x + 12) (10x2 + 2x + 40)4 dx (i) ∫ 3x2 √ 6x3 + 1 dx (j) ∫ x √ 2 − x dx (k) ∫ 3√3 + z−1 z2 dz (l) ∫ sec x tan x cos(sec x) dx Page 51 of 68 (m) ∫ sin x 3 ( cos x 3 )3 dx (n) ∫ tan x (sec x)2 dx Solution: (a) ∫ 1 0 x(3x2 + 4)4 dx = ∫ 7 4 1 6 du [ u4 ] =1 6 ∫ 7 4 u4 du =1 6 [ u5 5 ]7 4 =1 6 [ 75 5 − 4 5 5 ] =1 6 [ 16807 − 1024 5 ] =15783 30 = 5261 10 Let u =3x2 + 4 du =6x dx 1 6 du =x dx Page 52 of 68 (b) ∫ (2x4 + 4x)3(2x3 + 1) dx = ∫ 1 4 (u3) du =1 4 ∫ u3 du =1 4 [ u4 4 + C ] =1 4 [ (2x4 + 4x)4 4 + C ] =(2x 4 + 4x)4 16 + C Let u =2x4 + 4x du =(8x3 + 4) dx du =4(2x3 + 1) dx 1 4 du =(2x3 + 1) dx (c) ∫ 2 1 (2x3 + 3x + 5) x5 dx = ∫ 2 1 [ 2x3 x5 + 3x x5 + 5 x5 ] dx =2 ∫ 1 x2 dx + 3 ∫ 1 x4 dx + 5 ∫ 1 x5 dx =2 ∫ x−2 dx + 3 ∫ x−4 dx + 5 ∫ x−5 dx =2 [ x−1 −1 + c1 ] + 3 [ x−3 −3 + c2 ] + 5 [ x−4 −4 + c3 ] Let c1 + c2 + c3 =C =−2 x − 1 x3 − 5 4x4 + C = [−2 x − 1 x3 − 5 4x4 ]2 1 = [−2 2 − 1 (2)3 − 5 4(2)4 ] − [−2 1 − 1 (1)3 − 5 4(1)4 ] = − 1 − 1 8 − 5 64 + 2 + 1 + 5 4 =195 6 (d) ∫ (3x2 + 6x + 2) x4 dx = ∫ [3x2 x4 + 6x x4 + 2 x4 ] dx Page 53 of 68 =3 ∫ 1 x2 dx + 6 ∫ 1 x3 dx + 2 ∫ 1 x4 dx =3 ∫ x−2 dx + 6 ∫ x−3 dx + 2 ∫ x−4 dx =3 [ x−1 −1 + c1 ] + 6 [ x−2 −2 + c2 ] + 2 [ x−3 −3 + c3 ] Let c1 + c2 + c3 =C =−3 x − 3 x2 − 2 3x3 + C (e) ∫ 8 3 √ 12 − x dx = ∫ 4 9 (−du)(u)1/2 = − ∫ 4 9 u1/2 du = − [ 2u3/2 3 ]9 4 = − [ 2(4)3/2 3 − 2(9) 3/2 3 ] = − 2(2) 3 3 + 2(3) 3 3 = − 2(8) 3 + 2(27) 3 =−16 + 54 3 = 38 3 Let u =12 − x du = − dx −du =dx (f) ∫ (3x − 1) √ 3x2 − 2x + 1 dx = ∫ 1 2 du(u)1/2 =1 2 ∫ u1/2 du =1 2 [ 2u3/2 3 + C ] =1 2 [ 2(3x2 − 2x + 1)3/2 3 + C ] =(3x 2 − 2x + 1)3/2 3 + C =1 3 (3x2 − 2x + 1)3/2 + C Let u =3x2 − 2x + 1 du =(6x − 2) dx du =2(3x − 1) dx 1 2 du =(3x − 1) dx Page 54 of 68 (g) ∫ 1 (x + 5)4 dx = ∫ 1 u4 = ∫ u−4 =u −3 −3 + C =(x + 5) −3 −3 + C = − 1 3 ( 1 x + 5 )3 + C = −1 3(x + 5)3 + C Let u =x + 5 du =dx (h) ∫ (5x + 12) (10x2 + 2x + 40)4 dx = ∫ 1 4 du(u)−4 =1 4 ∫ u−4 du =1 4 [ u−3 −3 + C ] =1 4 [ −1 12(10x2 + 2x + 40)3 + C ] Let u =10x2 + 2x + 10 du =(20x + 2) dx du =2(10x + 2) dx du =4(5x + 1 2 ) dx 1 4 du =(5x + 1 2 ) dx (i) ∫ 3x2 √ 6x3 + 1 dx = ∫ 1 6 du(u)1/2 =1 6 ∫ u1/2 du =1 6 [ 2u3/2 3 + C ] =1 9 ( 6x3 + 1 )3/2 + C Let u =6x3 + 1 du =18x2 dx du =6(3x2) dx 1 6 du =3x2 dx Page 55 of 68 (j) ∫ x √ 2 − x dx = ∫ (2 − u)(u)1/2(−du) = ∫ (u − 2)(u)1/2 du = ∫ (u3/2 − 2u1/2)du =2u 5/2 5 − 2 ( 2 3 u3/2 ) + C =2(2 − x) 5/2 5 − 4 3 (2 − x)3/2 + C Let u =2 − x x =2 − u du = − dx −du =dx (k) ∫ 3√3 + z−1 z2 dz = ∫ u1/3(−du) = ∫ −u1/3 du = − ∫ u1/3 du = − [ 3u4/3 4 + C ] = − 3 4 (3 + z−1)4/3 + C Let u =3 + z−1 du = − z−2 dz −du =dz z2 (l) ∫ sec x tan x cos(sec x) dx = ∫ cos(u) du = sin(u) + C = sin(sec x) + C Let u = sec x du = sec x tan x Page 56 of 68 (m) ∫ sin x 3 ( cos x 3 )3 dx = ∫ u3(−3du) = − 3 ∫ u3 du = − 3 [ u4 4 + C ] =−3 [ cos ( x 3 )]4 4 + C =−3 4 [ cos (x 3 )]4 + C Let u = cos ( x 3 ) du = − sin ( x 3 ) [1 3 ] dx −3 du = sin (x 3 ) dx (n) ∫ tan x (sec x)2 dx = ∫ u du = [ u2 2 + C ] =(tan x) 2 2 + C Let u = tan x du =(sec x)2 dx Alternatively, ∫ tan x (sec x)2 dx = ∫ u du = [ u2 2 + C ] =(sec x) 2 2 + C Let u = sec x du = sec x tan x dx 59. Differentiate the following: (a) f (x) = ∫ x 3 (t2 − 8) dt (b) f (x) = ∫ x2 4 (t2 + 6)5/2 dt Page 57 of 68 Solution: (a) Dx ∫ x 3 (t2 − 8)dt = x2 − 8 (b) Dx ∫ x2 4 (t2 + 6)5/2dt = ( (x2)2 + 6 )5/2 2x = 2x(x4 + 6)5/2 60. Find the area of the region bounded by the graphs of y = x 2 and y = x4. Solution: y =x2 y =x4 Points of Intersection: x4 =x2 x2(x2 − 1) =0 x2 =0, (x − 1)(x + 1) = 0 x =0, x = 1,−1 They are: (0, 0), (1, 1), (−1, 1) (-1,1) (1,1) Area = 2 ∫ 1 0 (x2 − x4) dx =2 [ x3 3 − x 5 5 ]1 0 =2 [ 1 3 − 1 5 ] =2 [ 5 − 3 15 ] =2 [ 2 15 ] = 4 15 61. Find the area of the region bounded by the graphs of y = x 2 and y = 2 − x . Page 58 of 68 Solution: y =x2 y =2 − x Points of Intersection: x2 + x − 2 =0 (x + 2)(x − 1) =0 x = −2, x =1 A = ∫ 1 −2 (2 − x) − x2dx = [ 2x − 1 2 x2 − 1 3 x3 ]1 −2 = ( 2 − 1 2 − 1 3 ) − ( 2(−2) − 1 2 (−2)2 − 1 3 (−2)3 ) = ( 12 6 − 3 6 − 2 6 ) − ( −4 − 2 + 8 3 ) =12 6 − 3 6 − 2 6 + 36 6 − 16 6 =27 6 = 9 2 62. Find the area of the region bounded by the graphs ofy = x 4 and y = 8x . Page 59 of 68 Solution: y =8x y =x4 Points of Intersection: x(x3 − 8) =0 x(x − 2)(x2 + 2x + 4) =0 x = 0, x =2 A = ∫ 2 0 (8x − x4)dx = [ 4x2 − 1 5 x5 ]2 0 =16 − 32 5 =48 5 63. Find the volume of the solid generated by revolving about the x-axis the region bounded by the graph of y = x3/2, the x-axis, and the line x = 4. Solution: V =π ∫ b a [ f (x)]2 dx V =π ∫ 4 0 x3dx = [π 4 x4 ]4 0 =64π 64. Find the volume of the solid generated by revolving about the x-axis the region bounded by Page 60 of 68 y2 = 4x and x2 = 4y. Solution: y =1 4 x2 x =1 4 y2 Points of Intersection: x = 1 64 x4 x ( 1 64 x3 − 1 ) =0 x = 0, x =4 V =π ∫ b a [ f (x)]2 − [g(x)]2 dx V =π ∫ 4 0 (( 2 √ x )2 − (1 4 x2 )2) dx V =π ∫ 4 0 ( 4x − 1 16 x4 ) dx =π [ 2x2 − x 5 80 ]4 0 = ( 32 − 64 5 ) =96π 5 65. Find the volume of the solid generated by revolving the region in the first quadrant bounded by the graph of y = x2, y = 4, and the y-axis about a) the y-axis; b) y = 4. Page 61 of 68 Solution: y =4 y =x2 Points of Intersection: x2 − 4 =0 x = ± 2 (-2,4) (2,4) (a) About the y-axis: V =2π ∫ 2 0 x [ 4 − x2 ] dx =2π ∫ 2 0 ( 4x − x3 ) dx =2π [ 2x2 − 1 4 x4 ]2 0 =2π(8 − 4) =8π (b) About the axis where y = 4: 2π ∫ 4 0 (4 − y)√y dy =2π ∫ 4 0 (4y1/2 − y3/2)dy =2π [ 4(2)y3/2 3 − 2y 5/2 5 ]4 0 =2π [ 8(4)3/2 3 − 2(4) 5/2 5 − 0 ] =2π [ 8(8) 3 − 2(2) 5 5 ] =2π [ 26 3 − 2 6 5 ] Page 62 of 68 =2 7π 15 [5 − 3] =2 8π 15 = 256π 15 66. Find the volume of the solid generated by revolving about the y-axis the region bounded by y = x + 4 x , the x-axis, and the lines x = 1 and x = 3. Solution: V =2π ∫ b a x [ f (x)] dx V =2π ∫ 3 1 x ( x + 4 x ) dx =2π ∫ 3 1 ( x2 + 4 ) dx =2π [( x3 3 + 4x )]3 1 =2π [( 27 3 + 12 ) − ( 1 3 + 4 )] =2π [ 26 3 + 8 ] =2π [ 26 + 24 3 ] =2π ( 50 3 ) = 100π 3 67. A solid has for its base the region in the first quadrant bounded by x 2 + y2 = 25. Every plane section of the solid taken perpendicular to the x-axis is a square. Find the volume of the solid. Solution: A = y · y x2 + y2 =25 y = √ 25 − x2 Page 63 of 68 V = ∫ x=5 x=0 y2dx = ∫ 5 0 (25 − x2)dx = [ 25x − x 3 3 ]5 0 =125 − 125 3 =375 − 125 3 =250 3 68. A solid has as its base the region in the xy-plane bounded by the graphs of y = x and y2 = x . Find the volume of the solid if every cross section by a plane perpendicular to the x-axis is a semicircle with diameter in the xy-plane. Solution: y =x y2 =x Points of Intersection: (0, 0), (1, 1) For any x , diameter of the semicircle is: d =√x − x ⇒ radius = √ x − x 2 A =1 2 π (√ x − x 2 )2 V = ∫ b a A(x)dx Page 64 of 68 V =1 8 π ∫ 1 0 (√ x − x)2 dx =π 8 ∫ 1 0 ( x − 2x3/2 + x2 ) dx = [ π 8 ( x2 2 − 4 5 x5/2 + x 3 3 )]1 0 =π 8 ( 1 2 − 4 5 + 1 3 ) = π 240 69. Find the average value of the function f (x) = x2 + x + 1 on the interval [−1, 2]. Solution: 1 2 − (−1) ∫ 2 −1 (x2 + x + 1)dx =1 3 ∫ 2 −1 (x2 + x + 1)dx =1 3 [ x3 3 + x 2 2 + x ]2 −1 =1 3 [( 8 3 + 4 2 + 2 ) − ( −1 3 + 1 2 − 1 )] =1 3 [ 8 3 + 1 3 + 2 + 2 − 1 2 + 1 ] =1 3 [ 8 − 1 2 ] = 1 3 [ 15 2 ] = 15 6 = 5 2 70. Find the average value of the function f (x) = sin x on the interval [0, π ]. Solution: 1 π − 0 ∫ π 0 sin2 x dx = 1 π ∫ π 0 1 − cos 2x 2 dx = 1 2π [ x − 1 2 sin(2x) ]π 0 = 1 2π [( π − 1 2 sin(2π) ) − ( 0 − 1 2 sin(0) )] = 1 2π [π − 0 − 0 + 0] =1 2 Page 65 of 68 71. A cylindrical water tank with a circular base has radius 3 feet and height 10 feet. How much work is required to empty the tank by pumping the water out of the top if a) the tank is full? b) the tank is half full? (Assume that the density of water is 62.5 lb/ft3.) Solution: (a) F = (density)(volume) F = (density)(area thickness) F = (62.5) (π(3)2) y F = 562.5πy W = (F)(distance) W = ∫ 10 0 562.5π(10 − y) dy =562.5π ∫ 10 0 (10 − y) dy =562.5π [ 10y − y 2 2 ]10 0 =562.5π [ 10(10) − 10 2 2 − ( 10(0) − 0 2 2 )] =562.5π [ 100 − 100 2 ] = 562.5π(50) = 28125π (b) W = ∫ 5 0 562.5π(10 − y) dy =562.5π ∫ 5 0 (10 − y) dy =562.5π [ 10y − y 2 2 ]5 0 =562.5π [ 10(5) − 5 2 2 − ( 10(0) − 0 2 2 )] =562.5π [ 50 − 25 2 ] =562.5π [ 100 − 25 2 ] = 562.5π ( 75 2 ) = 562.5π(37.5) = 21093.75π Page 66 of 68 72. A bucket with 24 lb of water is raised 30 feet from the bottom of a well. Find the work done assuming that a) the weight of the empty bucket is 4 lb and the weight of the rope is negligible; b) the bucket weighs 4 lb and the rope weighs 4 oz/ft; c) in addition, the water leaks out of the bucket at a constant rate and that only 18 lb are left at the top. Solution: (a) F =(weight of water + weight of bucket) =24 + 4 =28 W = ∫ 30 0 28 dx = [28x]300 =28(30) = 840 (b) F =(weight of water + weight of bucket + weight of rope) =24 + 4 + 1 4 x =28 + 1 4 x W = ∫ 30 0 ( 28 + 1 4 x ) dx = [ 28x + 1 8 x2 ]30 0 =28(30) + 1 8 (30)2 = 840 + 112.5 = 952.5 (c) F =(weight of water + weight of bucket + weight of rope) − (weight leaking out) =24 + 4 + 1 4 x − 6 30 x =28 + 1 20 x W = ∫ 30 0 ( 28 + 1 20 x ) dx Page 67 of 68 = [ 28x + 1 40 x2 ]30 0 =28(30) + 1 40 (30)2 = 840 + 22.5 = 862.5 73. Match each numbered item with a lettered item. (There are more lettered items than num- bered items. Some lettered items don’t match any numbered item.) 1. Definition of lim x→a f (x) = L . 2. Definition of lim x→a+ f (x) = L . 3. Definition of lim x→∞ f (x) = L . 4. Definition of “ f is continuous at a”. 5. The Intermediate Value Theorem. 6. Definition of the derivative of f at a. 7. Definition of a differentiable function f at a. 8. Theorem relating differentiability and continuity. 9. The power rule for differentiation. 10. Definition of the differential. 11. Definition of a function f having an ab- solute maximum at c. 12. Definition of a function f having a local maximum at c. 13. The Extreme Value Theorem. 14. Definition of a function that is increasing on an interval I . 15. The Mean Value Theorem. 16. Definition of an antiderivative of f on an interval I . A. lim x→a f (x) = f (a), limx→a f (x) and f (a) both exist. B. For every > 0 there is a corresponding number N such that | f (x) − L| < whenever x > N . C. If f is continuous on the closed interval [a, b], and N is a number strictly between f (a) and f (b), then there exists a number c in (a, b) such that f (c) = N . D. The limit lim h→0 f (a + h) − f (a) h exists. E. If f is differentiable at a, then f is continuous at a. F. If f is continuous at a, then f is differentiable at a. G. d dx xn = nxn−1. H. The function g has the property g′(x) = f (x) for all x in I . I. lim h→0 f (a + h) − f (a) h J. f (c) ≥ f (x) for all x in the domain of f . K. For every > 0 there is a corresponding number δ > 0 such that | f (x) − L| < whenever a < x < a + δ. L. For every > 0 there is a corresponding number δ > 0 such that | f (x) − L| < whenever 0 < |x − a| < δ. M. There is an open interval I containing c such that f (c) ≥ f (x) for all x in I . N. f ′(x) > 0 for all x in I . O. If f is continuous on [a, b] then there are numbers c and d in [a, b] such that f (c) is an absolute maximum for f in [a, b] and f (d) is an absolute minimum for f in [a, b]. P. If f is differentiable, dy = f ′(x)dx . Q. f (x1) < f (x2) whenever x1 < x2 in I . R. If f is continuous on [a, b] and differentiable in (a, b), then there is a number c in (a, b) such that f ′(c) = f (b) − f (a) b − a . Page 68 of 68 Solution: 1. L 2. K 3. B 4. A 5. C 6. I 7. D 8. E 9. G 10. P 11. J 12. M 13. O 14. Q 15. R 16. H CalcReview.dvi

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