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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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