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Math 140 MT1 Sample “C” Solutions
Tyrone Crisp
1 (B): First try direct substitution: you get 00 . So try to cancel common factors. We have
x2 + 2x − 3
(x + 3)(x − 1)
=
,
x−1
x−1
and so the limit as x → 1 is equal to
lim(x + 3) = 4.
x→1
2 (A): As x → 4− , the denominator becomes zero, while the numerator stays away from zero.
So the limit will be infinite; we need to decide whether it’s ∞ or −∞. When x is close to, but a
(−)
little bit less than, 4, we have x − 5 < 0 and x − 4 < 0. So the function is of the form (−)
, so is
positive. Therefore the limit is ∞.
3 (E): Direct substitution gives 00 , so we need to factor. To factor the top, multiply top and bottom
by the conjugate:
√
√
( x2 + 12 − 4) ( x2 + 12 + 4)
x2 − 4
(x − 2)(x + 2)
√
√
· √
=
=
.
(x − 2)
( x2 + 12 + 4)
(x − 2)( x2 + 12 + 4)
(x − 2)( x2 + 12 + 4)
Cancel the (x − 2)’s, and then compute the limit by direct substitution:
1
x+2
4
= .
lim √
=√
2
x→2
2
x + 12 + 4
16 + 4
4 (C): Velocity is the derivative of position. We have s(t) = (t2 + 1)1/2 , and we can differentiate
using the chain rule:
1
1
1
d
1
t
s 0 (t) = (t2 + 1)− 2 · (t2 + 1) = (t2 + 1)− 2 (2t) = √
.
2
2
dt
2
t +1
When t = 1, we therefore have
1
1
s 0 (1) = √
=√ .
1+1
2
5 (D): In questions like this, it’s always a good idea to read all of the options before doing
anything else. From the list of possible answers, we see that the question revolves around the
continuity of this function at 1 and 2. So we ask ourselves:
(i) Is f continuous at x = 1? i.e., is limx→1 f(x) = f(1)? Compute the limit one side at a time:
lim f(x) = lim− (2x + 1) = 3,
x→1−
x→1
and
lim f(x) = lim+ 3x = 3.
x→1+
x→1
So limx→1 f(x) exists and equals 3. But this is not the same as f(1), which we’re told is equal to 4.
We conclude that f has a removable discontinuity at x = 1. (Removable because the limit exists,
but is not equal to f(1)).
(ii) Is f continuous at x = 2? We have
lim f(x) = lim− 3x = 6,
x→2−
x→2
2
while
lim f(x) = lim+ (5 − x) = 3.
x→2+
x→2
So the left- and right-limits don’t match up. So there’s a jump discontinuity at x = 2.
6 (C): Use the chain rule twice:
p
p
d
d p
(sin( 2x2 + 1)) = cos( 2x2 + 1) ( 2x2 + 1)
dx
dx
p
1 d
1
= cos( 2x2 + 1) (2x2 + 1)− 2 (2x2 + 1)
2
dx
p
1
1
= cos( 2x2 + 1) (2x2 + 1)− 2 (4x)
2
√
2x cos( 2x2 + 1)
√
=
.
2x2 + 1
7 (B): Since we’re only looking at x → 2+ , we only care about what happens when x ≥ 2. For
these x we have 2 − x ≤ 0, and so |2 − x| = −(2 − x). So
x2 (2 − x)
x2 (2 − x)
= lim+
= lim+ −x2 = −4.
lim
x→2 −(2 − x)
x→2
x→2+ |2 − x|
8 (B): f has a horizontal tangent at x = −1, so f 0 will cut the x-axis there. But this is of no help,
since all the given options have this property.
For x < −1, the slope of f is positive, so the graph of f 0 will be above the x-axis. This rules out
options (d) and (e), leaving only (a), (b) and (c).
The difference between these three graphs is what happens at x = 0. The graph of f has a corner
at zero, so the derivative will have a jump discontinuity (because the slope suddenly changes). This
rules out (a), leaving only (b) and (c).
For x ≥ 0, the slope of f is positive, so the graph of f 0 will lie above the x-axis. This eliminates
(c), so the correct answer must be (b).
9 (E): First note that regardless of a, the function f will certainly be continuous everywhere
except possibly at x = 1. (This is because away from x = 1, the function looks like a polynomial,
so is continuous everywhere.) So we just need to decide for which value of a does limx→1 f(x) =
f(1). In particular, we need the limit to exist. We have
lim f(x) = lim− ax2 + 3 = a + 3,
x→1−
x→1
while
lim f(x) = lim+ x − 3 = −2.
x→1+
x→1
So in order for limx→1 f(x) to exist, we must have a + 3 = −2, or in other words a = −5. For this
value of a, the limit is equal to −2, which is equal to f(1), and so f is continuous for a = −5.
10 (C): Direct substitution yields 00 , so we want to try to factor. We notice that there’s a sin in
the top and an x in the bottom, so we try to use the formula limθ→0 sinθ θ = 1. This requires some
3
devious manoeuvering:
sin(2x) 2
1
· ·
2x
1 cos x + sec x
sin(2x)
2
1
= lim
· · lim
x→0
2x
1 x→0 cos x + sec x
1
= 1 · 2 · = 1.
2
sin(2x)
lim
= lim
x→0 x(cos x + sec x)
x→0
(Note: the limit of
x = 0.)
1
cos x+sec x
can be found by direct substitution, since this function is continuous at
11 (A): The tangent is parallel to y = 12 x − 1 when it has the same slope as that line, namely
1/2. So we want to solve the equation dy
= 21 . Using the quotient rule, we find
dx
dy
d x−1
(x + 1)(1) − [(x − 1)(1)]
2
=
=
=
.
dx
dx x + 1
(x + 1)2
(x + 1)2
So we need to solve
1
2
=
(x + 1)2
2
⇒
(x + 1)2 = 4
⇒
x + 1 = ±2
⇒
x = 1 or − 3.
12 (B): We need to find the second derivative. Begin with the first derivative, using the product
rule:
dy
d
=
(sin x cos x) = sin x(− sin x) + cos x(cos x) = − sin2 x + cos2 x.
dx
dx
(You might notice that this is equal to cos 2x; but since none of the given answers involves any
2x’s, it’s probably best to leave it as-is.) Then differentiate again, using the chain rule:
d
d2 y
=
(− sin2 x + cos2 x) = −(2 sin x(cos x)) + 2 cos x(− sin x) = −4 sin x cos x.
2
dx
dx
(Note that this is equal to −2 sin 2x, which is the answer you would have got had you used cos 2x
above.)
13 (C): The slope of the tangent is given by the derivative, y 0 . Use implicit differentiation to
find this: First differentiate both sides of the equation with respect to x:
d 2
d
(x + xy − y2 ) =
(1) ⇒ 2x + (xy 0 + y) − 2yy 0 = 0.
dx
dx
(We used the product rule to differentiate xy, and the chain rule to differentiate y2 .) Then solve
this equation for y 0 :
xy 0 − 2yy 0 = −2x − y
⇒
y 0 (x − 2y) = −2x − y
The slope of the tangent at the point (2, 3) is therefore equal to
y0 =
−2(2) − 3
−7
7
=
= .
2 − 2(3)
−4
4
⇒
y0 =
−2x − y
.
x − 2y
4
14 FALSE: We’ve seen that continuity need not imply differentiability. For example, the function f(x) = |x| is continuous at 0 but not differentiable at 0.
15 FALSE: In order for f to be continuous at 2, we require also that f(2) exists, and is equal to
the limit.
16 (a) To differentiate x2 f(x), use the product rule:
y 0 = x2 f 0 (x) + f(x)2x.
So when x = 2, we have
y 0 (2) = 4f 0 (2) + 4f(2) = 4(−1) + 4(−3) = −16.
16 (b) Use the chain rule (keeping in mind that the positions of f and g are the opposite of what
they are in our formula for the chain rule):
d
d
(3g(f(x))) = 3 (g(f(x))) = 3g 0 (f(x))f 0 (x).
y0 =
dx
dx
So when x = 4, we have
y 0 (4) = 3g 0 (f(4))f 0 (4) = 3g 0 (4)(−2) = 3(5)(−2) = −30.
16 (c) Use the quotient rule (being careful to keep in mind that f and g are not in their usual
places!):
d g(x)
f(x)g 0 (x) − [g(x)f 0 (x)]
0
y =
.
=
dx f(x)
f(x)2
So when x = 0, we have
f(0)g 0 (0) − [g(0)f 0 (0)]
1(−6) − [3(2)]
y 0 (0) =
=
= −12.
2
f(0)
12
17 (a) Either
f(a + h) − f(a)
,
h→0
h
(Both are acceptable answers.)
f 0 (a) = lim
or
f(x) − f(a)
.
x→a
x−a
f 0 (a) = lim
17 (b) Using the first definition, the strategy is to find a common factor of h in the numerator, to
cancel the denominator:
2(a + h)2 − (a + h) − [2a2 − a]
f 0 (a) = lim
h→0
h
2
2a + 4ah + 2h2 − a − h − 2a2 + a
= lim
h→0
h
4ah + 2h2 − h
= lim
h→0
h
h(4a + 2h − 1)
= lim
h→0
h
= lim (4a + 2h − 1) = 4a − 1.
h→0
5
Alternatively, using the second definition (if you prefer), the strategy is to find a common factor of
(x − a) in the numerator, to cancel the denominator:
2x2 − x − (2a2 − a)
x→a
x−a
2
2(x − a2 ) − x + a
= lim
x→a
x−a
2(x − a)(x + a) − (x − a)
= lim
x→a
x−a
(x − a)(2(x + a) − 1)
= lim
x→a
x−a
= lim (2(x + a) − 1) = 4a − 1.
f 0 (a) = lim
x→a
(Note that this illustrates the advantage of the “limh→0 ” definition: the factoring is often easier.)
18 (a) Step 1: Draw a picture and introduce variables. Let x be the distance from the bottom of
the ladder to the bottom of the wall, and let y be the distance up the wall from the ground to the
top of the ladder.
dx
dy
Step 2: Rate that we’re told:
= 5 ft/s. Rate that we’re asked to find:
when x = 12.
dt
dt
Step 3: Find an equation relating the quantities x and y: Since the ladder, the wall and the
ground form a right triangle, Pythagoras tells us that x2 + y2 = 169.
Step 4: Differentiate both sides with respect to t, using the chain rule:
2x
(?)
dx
dy
+ 2y
= 0.
dt
dt
Step 5: Plug in the known information, and solve for the unknown: We know dx
= 5 and
dt
dy
2
2
x = 12. In order to solve (?) for dt we also need to know y. Since x + y = 169, we have
√
√
y = 169 − x2 = 169 − 144 = 5. Put these three numbers into (?) and solve for dy
:
dt
dy
dy
dy
= 0 ⇒ 120 + 10
=0 ⇒
= −12.
dt
dt
dt
Step 6: Interpret your answer: The height is decreasing at a rate of 12 feet per second.
2 · 12 · 5 + 2 · 5 ·
18(b) Step 1: Same picture, but we now let θ denote the angle in question.
dx
dθ
Step 2: Rate we’re told:
. Rate we’re asked to find:
when x = 12.
dt
dt
Step 3: Equation relating θ and x:
x
cos θ = .
13
Step 4: Differentiate both sides with respect to t:
(??)
− sin θ ·
dθ
1 dx
=
.
dt
13 dt
Step 5: Plug in the known information and solve for the unknown: We know dx
= 5 and x = 12.
dt
dθ
To solve (??) for dt , we’ll also have to know sin θ. When x = 12, we already know that y = 5,
6
and so sin θ =
5
.
13
Now plug all the knowns into equation (??) and solve:
1
dθ
5 dθ
·
=
·5 ⇒
= −1.
13 dt
13
dt
Step 6: Interpret: The rate of change of θ is −1 radian per second.
−