x −2
(x − 2)
. Therefore,
=
2
(x − 2) (x + 2)
x −4
candidates for vertical asymptotes are x = 2 and x = −2.
Now comes the election!
Fall 2004 Math 151
Exam 1A: Solutions
c
Mon, 04/Oct
2004,
Art Belmonte
9. (b) Note that y =
1. (a) We have
2a − 3b = 2 [2, 3] − 3 [1, −1] = [4, 6] − [3, −3] = [1, 9].
• As x → 2, we have
2. (c) From the figure below, we have a + v = b, from which
we conclude v = b − a.
y=
B
1
1
(x − 2)
→ 6= ±∞.
=
x +2
4
(x − 2) (x + 2)
Thus x = 2 is not a vertical asymptote.
v
• As x → −2+ , we see that
a
C
A
b
y=
3. (d) With a = [−2, 3] and b = [1, 2], the scalar projection of
b onto a is
compa b =
1
1
(x − 2)
→ + = +∞.
=
x +2
0
(x − 2) (x + 2)
Hence x = −2 is a vertical asymptote.
• Here is a plot which corroborates these assertions.
−2 + 6
4
a·b
=√
= √ .
kak
4+9
13
y = (x−2) / (x2−4)
5
4. (a) Since u and v are unit vectors, we have
=
2 (v · u) − 3 (v · v)
2 kvk kuk cos 60◦ − 3 kvk kvk cos 0◦
2 (1) (1) 12 − 3 (1) (1) (1)
=
1 − 3 = −2.
=
5. (a) For 0 ≤ t ≤
π,
2
−5
−4
we have x = sin t and y = cos2 t. Thus
sin2 t + cos2 t
=
1
x2 + y
=
1
y
=
1 − x 2.
0
y
v · (2u − 3v) =
−2
2
4
10. (c) Resolve the absolute value.
f (x) = |2x − 3| =
This is part of a parabola.
6. (d) With f defined on an open interval containing 2 and
f (2) = 3, it is always true that if lim f (x) = 3 = f (2),
then f is continuous at x = 2.
0
x
=
x→2
2x − 3,
− (2x − 3) ,
2x − 3,
3 − 2x,
2x − 3 ≥ 0;
2x − 3 < 0
x ≥ 3/2;
x < 3/2
Now draw a rough sketch to see that f 0 (x) does not exist for
x = 32 since the graph is sharp or kinked thereat.
7. (c) Let f (c) = c3 + c − 1 − π 2 . Then f (2) = 9 − π 2 < 0
and f (3) = 29 − π 2 > 0. Now f is a polynomial and thus
continuous everywhere. Therefore, by the Intermediate Value
Theorem f (c) = 0 for some c ∈ (2, 3). So for some number
c ∈ (2, 3), we have c3 + c − 1 = π 2 .
y = | 2x − 3 |
3
2
y
8. (c) Let’s divide numerator
and denominator of the limiting
√
expression by x 2 = x 4 . Then
r
3
1
p
√
2+ 2 + 4
2x 4 + 3x 2 + 1
2
x
x
.
= lim
=
lim
4
2
x→∞ 3x 2 + 2x + 4
x→∞
3
3+ + 2
x
x
1
0
0
1
2
x
1
3
11. (b) Use the quotient rule to differentiate f (x) =
0
f (x)

 2c2 x 2 + cx + c, x < 1;
16. Let f (x) =
1,
x = 1; In order for

cx + 1,
x > 1.
lim f (x) to exist, we must ensure that the left-hand and
2x − 1
.
x2 + 1
x 2 + 1 (2) − (2x − 1) (2x)
2
x2 + 1
=
x→1
right-hand limits are equal: lim f (x) = lim f (x).
x→1−
2x 2 + 2 − 4x 2 + 2x
2
x2 + 1
2 1 + x − x2
2
x2 + 1
=
=
• From the left we have lim f (x) = 2c2 + 2c, whereas
x→1−
lim f (x) = c + 1 for the right-hand limit. Set these
x→1+
one-sided limits equal: 2c2 + 2c = c + 1.
• Equivalently, 2c2 + c − 1 = 0. Now,
12. (b) Velocity is the derivative of position.
0
v(t) = s (t) = 2t + 1
(2c − 1) (c + 1) = 0
When t = 2 s, the velocity is v(2) = 5 m/s.
whence c = −1, 12 .
f (x)
= −1, we have
x→0 x
f (x) − f (0)
f (x)
= lim
= −1; that is,
f 0 (0) = lim
x −0
x→0
x→0 x
f 0 (0) = −1. Since g(x) = (2x − 1) f (x), we have
13. (a) With f (0) = 0 and lim
g 0 (x)
0
=
g (0) =
=
17.
(i) The derivative of f at a is defined by
f 0 (a) = lim
x→a
(2) f (x) + (2x − 1) f 0 (x)
(2) f (0) + (2(0) − 1) f 0 (0)
(2) (0) + (−1) (−1) = 1.
f 0 (a) = lim
h→0
(ii) With f (x) =
=
25
=
woman
20
=
15
ship
θ
=
5
=
0
woman
−5
−5
0
f (a + h) − f (a)
.
h
1
, we have
2x + 3
f 0 (1) =
Her direction θ points west of north into Quadrant 2 and
3
, whence θ ≈ 8◦ or N8◦ W (from the
satisfies tan θ = 22
◦
north, 8 toward the west).
10
f (x) − f (a)
x −a
or, equivalently,
14. Let the positive x-axis point east and the positive y-axis point
north. Then the velocity of the woman relative to the water is
v = −3i + 22j, with components in mi/h. The woman’s
speed is the magnitude of the velocity.
q
√
kvk = (−3)2 + 222 = 493 ≈ 22.20 mi/h
v
x→1+
f (x) − f (1)
x −1
x→1
1
1
−
2x
+
3
5
lim
x −1
x→1
5 − (2x + 3) 1
lim
5 (2x + 3) x − 1
x→1
2 − 2x
lim
x→1 5 (2x + 3) (x − 1)
−2 (x − 1)
lim
x→1 5 (2x + 3) (x − 1)
2
−2
=− .
lim
25
x→1 5 (2x + 3)
lim
5
1
at
(iii) A point on the tangent line to the curve y =
2x
+3
x = 1 is (1, f (1)) = 1, 15 . The slope of the tangent
15. The vector from A(1, 1) to B(2, −1) is
−→ −
→ −
→
w = A B = B − A = [1, −2] .
A vector perpendicular to w is v = w⊥ = [2, 1], a direction
vector for our line. A vector equation of the line through
P(3, 2) in this direction is
→
−
L(t) = P + tv
=
=
2
line from part (a) is f 0 (1) = − 25
. Hence an equation of
1
2
the tangent line is y − 5 = − 25 (x − 1) or
2
7
y = − 25
x + 25
.
18.
[3, 2] + t [2, 1]
[2t + 3, t + 2] .
2
(i) Here is a sketch of the graphs of f and g on the same
figure. We label the slopes of the piecewise-linear
components.
Problem 18
f
g
2
slope = −4/3
y
1.5
slope = 1/2
1
slope = 0
0.5
slope = 2/3
0
0
0.5
1
1.5
x
2
2.5
3
(ii) Clearly f is not differentiable at x = 1 due to the sharp
kink in its graph thereat.
(iii) Similarly, g is not differentiable at x =
reason.
3
2
for the same
(iv) At x = 12 , we have
( f g)0 = f 0 g + f g 0 = (0) g( 12 ) + (1) − 43 = − 43 .
(v) At x = 2, we have
0
f
=
f +g
( f + g) f 0 − f f 0 + g 0
=
=
−
3
2
( f + g)2
1
2
+ 13 12 − 32
+
2
3
2
3
1
2 + 3
30
≈ −0.2479.
121
3