Download Exam 1 Sol

APPM 1345
Exam 1 Solutions
Spring 2012
4. y = 2x − 3x2/3
(a) The domain of the function is all real numbers: (−∞, ∞) .
(b) If x = 0, y = 0.
Now solve y = 0.
1. There is an infinite number of solutions.
(a) S = 3 − 5 + 7 − 9 + 11 − 13 + 15 =
7
X
(−1)i+1 (2i + 1)
2x − 3x2/3 = 0
x2/3 (2x1/3 − 3) = 0
i=1
5
X
xi+1
x2
x3
x4
x5
x6
(b) S =
+
+
+
+
=
2
4
8
16 32
2i
i=1
2.
(a)
x = 0 is one solution. The other is
2x1/3 − 3 = 0
3
x1/3 =
2
27
.
x=
8
27
There are two intercepts: (0, 0) and
,0 .
8
y á f HxL
y
-3
-2
-1 x2
1
2 x3 3
x
2
(c) y 0 = 2 − √
is undefined at x = 0. Solve y 0 = 0.
3
x
2
=0
2− √
3
x
(b) No, x10 would be a worse approximation because each successive
approximation is farther from the root.
3.
(a)
2
2= √
3
x
√
3
x=1
f (x) = 2x3 − 4x + 1
x=1
f 0 (x) = 6x2 − 4
There are two critical numbers: x = 0, 1.
f (x1 )
x2 = x1 − 0
f (x1 )
=0−
Intervals
x<0
0<x<1
x>1
f (0)
1
1
=−
=
f 0 (0)
−4
4
1
(b) f ( 14 ) = 2( 41 )3 − 4( 14 ) + 1 = 32
. Since f ( 14 ) is close to 0, x2 =
is a good approximation of the root.
1
4
y0
+
–
+
y
increasing on (−∞, 0)
decreasing on (0, 1)
increasing on (1, ∞)
(d) There is a local maximum at x = 0, y = 0 . There is a local minimum at x = 1, y = −1 .
2
is undefined at x = 0. There are no solutions for
3x4/3
y 00 = 0. Note that y 00 is positive for all non-zero x.
(e) y 00 =
Intervals
x<0
x>0
y 00
+
+
7. f (x) = −x2 + 9 on [−3, 3]
(a) Each rectangle will have a width of ∆x = 3−(−3)
= 32 .
4
3
The partition points are x0 = −3, x1 = − 2 , x2 = 0, x3 = 32 .
y
concave up on (−∞, 0)
concave up on (0, ∞)
R4 = 32 f (x0 ) + 32 f (x1 ) + 23 f (x2 ) + 23 f (x3 )
2 = 32 9 − (−3)2 + 9 − − 23
+ 9 − 02 + 9 −
(f) There are no inflection points .
(g)
(b)
y á 2 x - 3 x2ê3
y
1
-3
-1
-1
1
y á 9 - x2
27
8
y
10
x
-3
5.
√
7x2
3
1
4
−5 x−8+ 2 − √ + √
3
3
x
2 x
x2
7
3
= x2 − 5x1/2 − 8 + 4x−2 − x−1/2 + x−2/3
3
2
10
7
y = x3 − x3/2 − 8x − 4x−1 − 3x1/2 + 3x1/3 + C
9
3
y0 =
6. Find the antiderivative twice.
-3 - 32
8.
(a)
Rn =
=
a(t) = − sin t + 2 cos t
v(t) = cos t + 2 sin t + C
s(t) = sin t − 2 cos t + Ct + D
Now use the two initial values to solve for constants C and D.
s(0) = −2 = 0 − 2 + C(0) + D ⇒ D = 0
s(π) = 2 − π = 0 + 2 + Cπ + 0 ⇒ C = −1
The position function is s(t) = sin t − 2 cos t − t .
3
2
3
x
n 3 X
i
1
i=1
n
X
i=1
n
n
i3
n4
n
1 X 3
i
n4 i=1
2
1 n(n + 1)
= 4
n
2
=
=
n2 (n + 1)2
(n + 1)2
=
4n4
4n2
3 2
2
(b)
Now check the second derivative.
A = lim Rn
n→∞
(n + 1)2
n→∞
4n2
2
n + 2n + 1
1
= lim
=
n→∞
4n2
4
= lim
By dominance of powers, since the polynomial in the numerator
has the same degree as the polynomial in the denominator, the limit
equals the ratio of the leading coefficients.
Pn
(c) Since Rn = i=1 f (xi )∆x for ∆x = (b−a)/n and xi = a+i∆x,
then f (x) = x3 on [0, 1] , where ∆x = 1/n and xi = i/n.
9. Let x represent the side length of the square top and bottom, and y represent the height of the cage. The volume is V = x2 y = 32 ⇒ y =
32/x2 . The combined area of the top and bottom of the cage is 2x2 . The
combined area of the four sides of the cage is 4xy. We wish to minimize
the cost C.
C = 10(2x2 ) + 20(4xy)
= 20x2 + 80xy
= 20x2 + 80x ·
2560
x
2560
0
C = 40x − 2
x
= 20x2 +
Solve C 0 = 0.
0 = 40x −
40x3 = 2560
x3 = 64
x=4
2560
x2
32
x2
C 00 = 40 +
5120
x3
Since C 00 > 0 for all positive values of x, then the graph of C is concave
up and there is a minimum point at x = 4, y = 32/42 = 2. The optimal
dimensions are 4 × 4 × 2 meters .