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 1B: Solutions
c
Mon, 04/Oct
2004,
Art Belmonte
9. (c) Note that y =
• As x → 2+ , we have
1. (b) We have
3a − 2b = 3 [2, 3] − 2 [1, −1] = [6, 9] − [2, −2] = [4, 11].
2. (b) From the figure below, we have b + v = a, from which
we conclude v = a − b.
y=
B
1
1
(x + 2)
→ + = +∞.
=
x −2
0
(x − 2) (x + 2)
Thus x = 2 is a vertical asymptote.
v
• As x → −2, we see that
b
C
A
a
y=
3. (b) With a = [2, 3] and b = [1, 2], the scalar projection of b
onto a is
compa b =
1
1
(x + 2)
→ − 6= ±∞.
=
x −2
4
(x − 2) (x + 2)
Hence x = −2 is not a vertical asymptote.
• Here is a plot which corroborates these assertions.
2+6
8
a·b
=√
= √ .
kak
4+9
13
y = (x+2) / (x2−4)
5
4. (e) Since u and v are unit vectors, we have
=
3 (u · u) − 2 (u · v)
3 kuk kuk cos 0◦ − 2 kuk kvk cos 60◦
3 (1) (1) (1) − 2 (1) (1) 12
=
3 − 1 = 2.
=
5. (b) For 0 ≤ t ≤
π,
2
−5
−4
we have x = cos t and y = sin2 t. Thus
sin2 t + cos2 t
=
1
y + x2
=
1
y
=
1 − x 2.
0
y
u · (3u − 2v) =
−2
2
4
10. (b) Resolve the absolute value.
f (x) = |3x − 2| =
This is part of a parabola.
6. (e) 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
3x − 2,
− (3x − 2) ,
3x − 2,
2 − 3x,
3x − 2 ≥ 0;
3x − 2 < 0
x ≥ 2/3;
x < 2/3
Now draw a rough sketch to see that f 0 (x) does not exist for
x = 23 since the graph is sharp or kinked thereat.
7. (b) 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 = | 3x − 2 |
6
4
y
8. (d) Let’s divide numerator
and denominator of the limiting
√
expression by x 2 = x 4 . Then
2
4
2
3+ + 2
3
3x 2 + 2x + 4
x
x
= lim r
=√ .
lim p
x→∞ 2x 4 + 3x 2 + 1
x→∞
3
1
2
2+ 2 + 4
x
x
0
1
−2
0
x
2
11. (b) Use the quotient rule to differentiate f (x) =
0
f (x)
2x + 1
.
x2 + 1
A vector perpendicular to w is v = w⊥ = [2, 3], a direction
vector for our line. A vector equation of the line through
P(1, 3) in this direction is
x 2 + 1 (2) − (2x + 1) (2x)
2
x2 + 1
=
L(t)
2x 2 + 2 − 4x 2 − 2x
2
x2 + 1
2 1 − x − x2
2
x2 + 1
=
=
=
=
[1, 3] + t [2, 3]
[2t + 1, 3t + 3] .
x→1
right-hand limits are equal: lim f (x) = lim f (x).
x→1−
v(t) = s 0 (t) = 2t + 1
x→1−
2
lim f (x) = 2c + 2c for the right-hand limit. Set
x→1+
f (x)
= −1, we have
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
these one-sided limits equal: c + 1 = 2c2 + 2c.
13. (a) With f (0) = 0 and lim
x→0
• Equivalently, 2c2 + c − 1 = 0. Now,
(2c − 1) (c + 1) = 0
whence c = −1, 12 .
(2) f (x) + (2x − 1) f 0 (x)
(2) f (0) + (2(0) − 1) f 0 (0)
(2) (0) + (−1) (−1) = 1.
17.
(i) The derivative of f at a is defined by
f 0 (a) = lim
x→a
14. Let the positive x-axis point east and the positive y-axis point
north. Then the velocity of the man relative to the water is
v = 4i − 20j, with components in mi/h. The man’s speed is
the magnitude of the velocity.
q
kvk =
(4)2 + (−20)2 =
f 0 (a) = lim
h→0
√
416 ≈ 20.40 mi/h
(ii) With f (x) =
=
man
=
θ
−5
=
ship
v
=
−15
−20
−25
=
0
5
10
f (a + h) − f (a)
.
h
1
, we have
3x + 2
f 0 (1) =
5
−10
f (x) − f (a)
x −a
or, equivalently,
His direction θ points south of east into Quadrant 4 and
◦
◦
satisfies tan θ = 20
4 , whence θ ≈ 79 or E79 S (from the
◦
◦
east, 79 toward the south) or S11 E (from the south, 11◦
toward the east).
0
x→1+
• From the left we have lim f (x) = c + 1, whereas
When t = 3 s, the velocity is v(3) = 7 m/s.
=
−
→
P + tv

x < 1;
 cx + 1,
1,
x = 1; In order for
16. Let f (x) =
 2 2
2c x + cx + c, x > 1.
lim f (x) to exist, we must ensure that the left-hand and
12. (d) Velocity is the derivative of position.
g 0 (x) =
g 0 (0) =
=
f (x) − f (1)
x −1
1
1
−
lim 3x + 2 5
x −1
x→1
5 − (3x + 2) 1
lim
5 (3x + 2) x − 1
x→1
3 − 3x
lim
x→1 5 (3x + 2) (x − 1)
−3 (x − 1)
lim
x→1 5 (3x + 2) (x − 1)
3
−3
=− .
lim
25
x→1 5 (3x + 2)
lim
x→1
1
at
(iii) A point on the tangent line to the curve y =
3x
+2
x = 1 is (1, f (1)) = 1, 15 . The slope of the tangent
15. The vector from A(−2, 1) to B(1, −1) is
−→ −
→ −
→
w = A B = B − A = [3, −2] .
2
3 . Hence an equation of
line from part (a) is f 0 (1) = − 25
3
the tangent line is y − 15 = − 25
(x − 1) or
3
8
y = − 25 x + 25 .
(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
1.5
y
18.
slope = 1/2
1
slope = 0
0.5
slope = 2/3
0
0
0.5
1
1.5
x
2
(ii) Clearly g is not differentiable at x =
kink in its graph thereat.
2.5
3
2
3
due to the sharp
(iii) Similarly, f is not differentiable at x = 1 for the same
reason.
(iv) At x = 34 , we have
( f g)0 = f 0 g + f g 0 = (0) g( 34 ) + (1) − 43 = − 43 .
(v) At x = 2, we have
0
g
=
f +g
=
=
( f + g) g 0 − g f 0 + g 0
3
2
( f + g)2
1 + 2
+ 13 23 − 13
2
3
2
3
1
2 + 3
30
≈ 0.2479.
121
3