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SOLUTIONS TO 18.01 EXERCISES Unit 1. Differentiation 1A. Graphing 2 1A-1,2 a) y = (x − 1) − 2 b) y = 3(x2 +2x)+2 = 3(x + 1)2 − 1 2 2 1 1 -2 -1 -2 1 1a 1b 2a 2b 3 3 1A-3 a) f(−x) = (−x) − 3x = −x − 3x = −f(x), so it is odd. 4 4 1 − (−x) 1 −x 2 2 b) (sin(−x)) = (sin x) , so it is even. c) odd , so it is odd even − x)4 =� ±(1 + x)4: neither. d) (1 2 2 e) J ((−x) ) = J (x ), so it is even. 0 0 1A-4 a) p(x) = p (x)+ p (x), where p (x) is the sum of the even powers and p (x) e o e o is the sum of the odd powers b) f(x) = f(x)+ f(−x) + f(x) − f(−x) 2 2 F (x) = f(x)+ f(−x) is even and G(x) = f(x) − f(−x) is odd because 2 2 COPYRIGHT DAVID JERISON AND MIT 1996, 2003 1 E. Solutions to 18.01 Exercises 1. Differentiation F (−x) = f(−x)+ f(−(−x)) = F (x); G(−x)= f(x) − f(−x) = −G(−x). 2 2 c) Use part b: 1 + 1 = 2a = 2a even 2 2 x + a −x + a (x + a)(−x + a) a − x 1 − 1 = −2x = −2x odd 2 2 x +a −x + a (x + a)(−x + a) a − x =⇒ 1 = a − x 2 2 2 2 x + a a − x a − x x − 1 3y + 1 1A-5 a) y = . Crossmultiply and solve for x, getting x = , so the 2x +3 1 − 2y 3x +1 inverse function is . 1 − 2x 2 2 b) y = x +2x = (x + 1) − 1 (Restrict domain to x ≤ −1, so when it’s flipped about the diagonal y = x, √ you’ll still get the graph of a function.) Solving for x, we get x = y +1 − 1, so √ − 1 . the inverse function is y = x +1 g(x) g(x) f(x) f(x) 5a 5b √ √ √ 1A-6 a) A = 1+3 = 2, tan c = 3, c = π. So sin x + 3cos x = 2 sin(x + π) . 1 3 3 √ π b) 2sin(x − ) 4 π 1A-7 a) 3 sin(2x − π) = 3sin2(x − ), amplitude 3, period π, phase angle π/2. 2 π b) −4cos(x + ) = 4sin x amplitude 4, period 2π, phase angle 0. 2 2 1. Differentiation E. Solutions to 18.01 Exercises 3 4 π 2π π 2π -3 -4 7a 7b 1A-8 f(x) odd =⇒ f(0) = −f(0) =⇒ f(0) = 0. So f(c) = f(2c) = ··· = 0, also (by periodicity, where c is the period). 1A-9 2 3 -8 -4 4 8 12 -7 -5 -3 -1 1 3 5 -1 9ab period = 4 9c -6 c) The graph is made up of segments joining (0, −6) to (4, 3) to (8, −6). It repeats in a zigzag with period 8. * This can be derived using: (1) x/2 − 1 = −1 =⇒ x = 0 and g(0) = 3f(−1) − 3 = −6 (2) x/2 − 1 = 1 =⇒ x = 4 and g(4) = 3f(1) − 3 = 3 (3) x/2 − 1 = 3 =⇒ x = 8 and g(8) = 3f(3) − 3 = −6 (4) 1B. Velocity and rates of change 2 1B-1 a) h = height of tube = 400 − 16t . 2 average speed h(2) − h(0) = (400 − 16 · 2 ) − 400 = −32ft/sec 2 2 (The minus sign means the test tube is going down. You can also do this whole problem using the function s(t) = 16t2, representing the distance down measured from the top. Then all the speeds are positive instead of negative.) b) Solve h(t) = 0 (or s(t) = 400) to find landing time t = 5. Hence the average speed for the last two seconds is 2 h(5) − h(3) = 0 − (400 − 16 · 3 ) = −128ft/sec 2 2 3 E. Solutions to 18.01 Exercises 1. Differentiation c) h(t) − h(5) 400 − 16t2 − 0 16(5 − t)(5 + t) (5) t − 5 = t − 5 = t − 5 (6) = −16(5 + t) → −160ft/sec as t → 5 1B-2 A tennis ball bounces so that its initial speed straight upwards is b feet per second. Its height s in feet at time t seconds is s = bt − 16t2 a) s(t + h) − s(t) b(t + h) − 16(t + h)2 − (bt − 16t2) (7) h = h bt + bh − 16t2 − 32th − 16h2 − bt + 16t2 (8) = h 2 (9) = bh − 32th − 16h h (10) = b − 32t − 16h → b − 32t as h → 0 Therefore, v = b − 32t. b) The ball reaches its maximum height exactly when the ball has finished going up. This is time at which v(t) = 0, namely, t = b/32. c) The maximum height is s(b/32) = b2/64. d) The graph of v is a straight line with slope −32. The graph of s is a parabola with maximum at place where v = 0 at t = b/32 and landing time at t = b/16. v b s b/32 t b/32 b/16 t graph of velocity graph of position e) If the initial velocity on the first bounce was b = b, and the velocity of the 1 √ second bounce is b 2 2 , then b /64 = (1/2)b /64. Therefore, b = b / 2. The second 2 2 1 2 1 bounce is at b /16 + b /16. (continued →) 1 2 4
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