Solutions to Excercises

# Solutions to Excercises - MIT OpenCourseWare...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . SOLUTIONS TO 18.01 EXERCISES Unit 1. Differentiation 1A. Graphing 1A-1,2 a) y = ( - 1) 2- 2 b) y = 3(2 2 + 2x) + 2 = 3( + 1) 2 -1 2 -1 lb -2 2a '1 2b (-x) 3 - 3x -x 3 - 3x 1A-3 a) f(-x) = 1(-X) 4 1- 4 -f (), so it is odd. b) (sin(-X)) 2 = (sin ) 2 , so it is even. odd c) even ' , so it is odd d) (1 - x) 4 :(1 + X) 4 : neither. e) Jo((-X) 2 ) = Jo(X), so it is even. 1A-4 a) p(x) = pe(x) + po(x), where pe(x) is the sum of the even powers and po(z) is the sum of the odd powers b) f(x) = f() + f(-x) f () - 2 f(-x) 2 f C() + f (-) f_() - f_(-_ ) F(x) = + 2 is even and G(x) = f() - 2 (-x) is odd because F(-x) = f(-x) + f (-(-z)) f() - f(-x) G(-x) = f(- = -G(-). 2 2 c) Use part b: 1 1 2a 2a ---- + --- even- x+a -z + a (x +'a)(-x + a) a2 - 1 1 -2: -2x odd x + a -x + a (x + a)(-x + a) a2 - x: 1 a x z+&amp;quot;a a 2 - 2 a 2 - X 2 @David Jerison and MIT 1996, 2003 S. 18.01 SOLUTIONS TO EXERCISES z-1 1A-5 a) y = 2'+3' Crossmultiply and solve:for x, getting z - 3 y+1 2y so the inverse 3z +1 function is . x 1- 2z b)y = 2 +2z:= (z + 1) 2 -1 (Restrict domain to z &amp;lt; -1, so when it's flipped about the diagonal y = x, you'll still get the graph of a function.) Solving for z, we get z = v - 1,so the inverse function is y = Vr - 1 . I I I I I I Ax) I I I I I Sa 5b 1A-6 a) A = i 3 = 2, tanc= , c= J. So sin + V cos =.2 sin(z'+ 3). b) sin( -1) 1A-7 a) 3sin(2z - r) = 3sin2(z - 2), amplitude 3, period r, phase angle r/2. b) -4 cos(z + 2) = 4 sin z amplitude 4, period 27r, phase angle 0. 3 4 7a 7b. 1A-8 f(z) odd f f(0) = -f(0) = f(0) =0. So f(c) = f(2c) = . = 0, also (by periodicity, where c is the period). 1A-9 9ab period = 4 9c - 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: x/2 - 1 = -1 x = 0 and g(O) = 3f(-1) - 3= -6 z/2-1=1== z=4andg(4)=3f(1)-3=3 z/2 - 1 = 3 = z = 8 and g(8) = 3f(3)- 3 = -6 1. DIFFERENTIATION IB. Velocity and rates of change 1B-1 a) h = height of tube = 400 - 16t 2 . h(2) - h(0) (400 - 16.22) -400 average speed 2 2 2 2 -32ft/sec (The minus sign means the test tube is going down. You can also do this whole problem using the function s(t) = 16t 2 , 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 h(5) - h(3) _0- 2 (400 - 16 32) =-128ft/sec 2 2 c) h(t) - h(5) _ 400 - 16t 2- 0 16(5 - t)(5 + t) t-5 t-5 t-5 = -16(5 + t) -4 -160ft/sec as t -+ 5 1B-2 A tennis ball bounces so that its initial speed straight upwards is b feet per second....
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## This note was uploaded on 10/10/2011 for the course MATH 31 taught by Professor Blake during the Spring '11 term at MIT.

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Solutions to Excercises - MIT OpenCourseWare...

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