Review for Exam 1
Fall 2013 MTH 132-036/ MTH 152H
Professor: Mandi Schaeffer Fry
This review will provide a good indication of whether you are ready for the exam, but it is by no
means a complete list of the types of questions that will be found on the actual exam. There are
questions here that will not be on the exam, and will be questions on the exam that are not quite
found here. You should also go back through all homework, WeBWorK, quizzes, and examples
from class.
1. What is the definition of continuity at an interior point? an endpoint? What is the definition
of the derivative? What are some interpretations of the derivative?
2. Find the limits. Where appropriate, describe the limit (does it correspond to a hole? a
horizontal/vertical/oblique asymptote?)
tan(5x) sec 3x
(a) lim
x→0
12x
Solution:
sin(5x)
1
tan(5x) sec 3x
1 sin(5x)
1
cos(5x) · cos(3x)
lim
= lim
= lim
· (5x)
x→0
x→0
x→0 12x
12x
12x
5x
cos(5x) · cos(3x)
sin(5x)
1
5
5
5x
lim
lim
=
· (1) · (1) =
= lim
x→0
x→0 cos(5x) cos(3x)
x→0 12x
5x
12
12
8 cos(9x)
x→0 x − 5
(b) lim
Solution:
8 cos(9x)
8 cos 0
8
=
=−
x→0 x − 5
(0 − 5)
5
lim
x2 + x − 56
x→7
x−7
(c) lim
Solution:
x2 + x − 56
(x − 7)(x + 8)
= lim
= lim (x + 8) = 15
x→7
x→7
x→7
x−7
x−7
lim
This limit corresponds to a hole.
x2 + x − 56
x→8
x−7
(d) lim
Solution:
x2 + x − 56
82 + 8 − 56
=
= 16
x→8
x−7
8−7
lim
1
This limit doesn’t correspond to a point of discontinuity.
x−9
(e) lim √
x→9
x−3
Solution:
√
√
√
( x − 3)( x + 3)
x−9
√
= lim
= lim ( x + 3) = 6
lim √
x→9
x→9
x − 3 x→9
x−3
This limit represents a hole.
12x − 8
x→−2 3x + 6
(f) lim
Solution: This limit does not exist. It represents a vertical asymptote of x = −2.
12x − 8
x→∞ 3x + 6
(g) lim
Solution: This limit is 12
3 , since the numerator and denominator have the same degree. It represents a horizontal asymptote of y = 12/3.
12x2 − 8
x→∞ 3x + 6
(h) lim
Solution: This limit is ∞, since the degree of the top is larger than the bottom, and
because for large positive values of x, this number is positive. It represents an oblique
asymptote of y = 4x − 8, by using long division.
(i) lim
x→∞
12
3x + 6
Solution: This limit is 0, since the numerator has smaller degree than the denominator. It represents a horizontal asymptote y = 0.
(j) lim
x→0
1
3+x
−
x
1
3
Solution:
lim
x→0
1
3+x
−
x
1
3
1
1
1
1 3 − (3 + x)
= lim
−
= lim
x→0 x
x→0 x
3+x 3
3(3 + x)
1
−x
−1
1
= lim
= lim
=−
x→0 x
x→0 3(3 + x)
3(3 + x)
9
Solution: We could also notice that this is g 0 (3), where g(t) =
g 0 (t) = −t−2 , we see g 0 (3) = −1/9.
1
t
= t−1 . Then since
3. Calculate each of the following using rules of differentiation (you may leave your answers
unsimplified):
2
(a)
d
dx
1
x2
−
3
x4
Solution: We can rewrite the function as x−2 −3x −4 , so using the power rule, constant
d
1
multiple rule, and sum rule, we have dx
− x34 = −2x−3 + 12x−5 .
x2
(b) g 0 (x), where g(x) = (x5 − 3x3 + 5)(x4 + 5x3 − 4)
5
d
Solution: We have dx
x − 3x3 + 5 = 5x4 −9x2 and
so using the product rule, we get
d
dx
x4 + 5x3 − 4 = 4x3 +15x2 ,
g 0 (x) = (5x4 − 9x2 )(x4 + 5x3 − 4) + (x5 − 3x3 + 5)(4x3 + 15x2 ).
(c)
dy
dx ,
where y =
x2 −1
2x+1
Solution: Rewriting y =
f (x)
g(x)
with f (x) = x2 −1 and g(x) = 2x+1, we see f 0 (x) = 2x
and g 0 (x) = 2, so using the quotient rule,
dy
dx
=
(2x)(2x+1)−(x2 −1)(2)
.
(2x+1)2
(d) f (100) (x), where f (x) = 2145x12 + 7485x9 + 346791x5 + 765x + 6
Solution: Since this is a polynomial of degree 12, we know that by the 12th derivative,
we have a constant, so all higher-order derivatives are just 0. Hence f (100) (x) = 0.
4. Find f 0 (x), where f (x) =
√
2x + 5, DIRECTLY FROM THE DEFINITION.
Solution:
p
√
2(x + h) + 5 − 2x + 5
f (x + h) − f (x)
f (x) = lim
= lim
h→0
h→0
h
h
#
"p
p
√
√
2(x + h) + 5 − 2x + 5
2(x + h) + 5 + 2x + 5
·p
= lim
√
h→0
h
2(x + h) + 5 + 2x + 5
"
!#
"
!#
1
2(x + h) + 5 − (2x + 5)
1
2h
= lim
= lim
· p
· p
√
√
h→0 h
h→0 h
2(x + h) + 5 + 2x + 5
2(x + h) + 5 + 2x + 5
!
2
2
1
= lim p
= √
=√
√
h→0
2 2x + 5
2x + 5
2(x + h) + 5 + 2x + 5
0
so f 0 (x) =
√ 1
.
2x+5
f (5 + h) − f (5)
, given that the line y = 3x − 2 is tangent to the graph of y = f (x)
h→0
h
at the point (5, 13).
5. Find lim
3
Solution: This limit is just f 0 (5), which is the slope of the tangent line at x = 5, which is
3, since y = 3x − 2 is the tangent line.
(
kx2 + 6
6. Let g(x) =
36−x2
x−6
if x = 6
. Find the value of k that makes g(x) continuous on (−∞, ∞).
if x =
6 6
Solution: To be continuous, we need limx→6 g(x) = g(6). Now, g(6) = 36k + 6, and
2
(6−x)(6+x)
= − limx→6 (x−6)(6+x)
= − limx→6 (6 +
limx→6 (g(x)) = limx→6 36−x
x−6 = limx→6
x−6
x−6
x) = 12. So we need to solve 12 = 36k + 6, which gives k = 1/6.
7. In each of the following, sketch a graph for f (x) satisfying the given conditions. (There should
only be one graph for each of (a), (b).) Be sure to label the relevant points.
(a) lim f (x) = 6, lim f (x) = −2, and f (4) = 0.
x→4−
x→4+
(b) lim f (x) = 1, f (−3) = 8, and f 0 (x) is positive on (−∞, −3) ∪ (−3, ∞).
x→−3
8. Let > 0. Find δ > 0 so that |2x − 4| < whenever |x − 2| < δ. What limit is being verified
here?
Solution: |2x−4| < means − < 2x−4 < , so −+4 < 2x < +4, so −/2 < x−2 < /2,
and δ = /2. The limit being verified is limx→2 (2x) = 4.
9. Assume g(x), h(x) are continuous, differentiable functions with values as shown:
x
−1
2
5
8
g(x)
4
8
12
16
h(x)
2
8
32
128
g 0 (x)
10
9
8
7
h0 (x)
3
1
15
4
(a) Find lim (h(x)g(x)).
x→5
Solution: Since both functions are continuous, this is h(5)g(5) = 32 ∗ 12 = 384.
(b) Find
d
dx
[g(x)h(x)]|x=−1 .
Solution: From the product rule, this is g 0 (−1)h(−1) + g(−1)h0 (−1) = (10)(2) +
(4)(3) = 32.
h
i
3h(x) d
(c) Find dx
x·g(x) x=5
Solution: From product and quotient rule:
(3h0 (5))(5g(5)) − (3h(5))(g(5) + 5g 0 (5))
3 · 15 · 5 · 12 − 3 · 32 · (12 + 5 · 8)
−2292
=
=
2
(5g(5))
25 · 144
3600
4
(Note - on the actual exam, I would use less-terrible numbers since you will not have
calculators.)
10. If h(x) represents the size (in cubic meters) of the box needed to hold my sock collection when
I have spent x dollars on socks, what are the units for h0 (x)?
Solution: (cubic meters)/dollar (i.e. m3 /$)
11. After being in the air for t seconds, Harry’s snitch is s feet to the right of the center of the
Quidditch field, where s is given by:
s(t) = 4t2 + 12t − 16.
(a) How far from the center of the field does the snitch start? Does it start to the right of
center or left of center? (Include units)
Solution: The snitch starts s(0) feet to the right of the center, which is s(0) = −16
feet. Since s(0) is negative, we know this is 16 feet to the left of center, since s gives
the distance to the right.
(b) What is the initial velocity of the snitch? Include units.
Solution: The velocity of the snitch is v(t) = s0 (t) = 8t + 12, using the sum, constant
multiple, and power rules. Then the initial velocity is v(0) = 12 feet/second.
(c) What is the velocity of the snitch when it has been in the air for 2 seconds?
Solution: v(2) = 8(2) + 12 = 28 feet/sec.
(d) What direction is the snitch moving when it has been in the air for 2 seconds? Explain
how you know.
Solution: Since v(2) = 28 is positive, we know the snitch is moving to the right at
this time.
(e) At what time(s) is the snitch in the center of the Quidditch field? Find the speed and
direction of travel at these times.
Solution: The snitch is in the center when s(t) = 0, so when 4t2 + 12t − 16 = 0. Then
0 = t2 + 3t − 4 = (t − 1)(t + 4). So at time t = 1 and time t = −4, the snitch is in the
center. Since the domain is t ≥ 0, we see the only time is after 1 second. At this time,
the velocity is v(1) = 8 + 12 = 20 feet/sec, which is positive, so the speed is also 20
feet/sec to the right.
12. Let A be a constant and write r(x) =
√
x+1
−1 < x ≤ 1
x2 + A(x − 1) + 1 1 < x < 3
5
(a) Is r(x) continuous at x = 1? Explain.
Solution: To be continuous, we need limx→1− r(x) = limx→1+ r(x) = r(1). The
√
left-hand limit is limx→1− x + 1 = 2, which is also r(1). The right-hand limit is
limx→1+ (x2 + A(x − 1) + 1) = 1 + A(0) + 1 = 2, so this function is continuous at x = 1.
(b) Find r0 (x). (Write it as a piecewise function).
(
Solution:
r0 (x)
=
1
√
2 x
−1 < x < 1
2x + A 1 < x < 3
(c) Find the value of A so that r0 (1) is defined.
Solution: We want both pieces of r0 (x) to have the same value at 1, so
and A = − 23 .
1
2
= 2 + A,
13. Let f, f 0 be differentiable.
(a) If f 0 (x) is positive, what can you say about f (x)?
Solution: f is increasing
(b) If f 0 (x) is negative, what can you say about f (x)?
Solution: f is decreasing
(c) If f 0 (x) is decreasing, what can you say about f 00 (x)?
Solution: f 00 ≤ 0
(d) If f 00 (x) is positive, what can you say about f 0 (x)?
Solution: f 0 is increasing
2
14. Use calculus to determine all values of x for which f (x) = 73 x3 + 21
2 x − 28x + 12 has horizontal
tangent lines.
Solution: f 0 (x) = 7x2 + 21x − 28 = 7(x2 + 3x − 4) = 7(x − 1)(x + 4). To get a horizontal
tangent line, we need f 0 (x) = 0, so x = 1 or −4.
15. Suppose f (x) is a continuous function on (−∞, ∞) with the following properties:
lim
h→0
f (6 + h) − f (6)
= 9,
h
6
lim f (x) = 5.
x→6
(a) Find an equation for the tangent line to f (x) at x = 6. Where must we use the fact that
f is continuous?
Solution: Since f is continuous, f (6) = limx→6 f (x) = 5. Also, the slope of the
(6)
tangent line is f 0 (6) = limh→0 f (6+h)−f
= 9. So the equation of the tangent line is
h
y = 9(x − 6) + 5.
(b) Is f (x) differentiable at x = 6? Explain your reasoning.
Solution: yes - f (x) is differentiable wherever the limit of the difference quotient
(x0 )
limh→0 f (x0 +h)−f
is defined, so since we have a value for the limit at x0 = 6, we
h
know f 0 exists there.
16. The graph of f (x) is shown below.
(a) Does f (x) appear to be continuous everywhere in its domain? Explain.
Solution: yes - it appears the limit at every point is equal to the functions value.
There are no holes/skips/jumps
(b) Does f (x) appear to be differentiable everywhere in its domain? Explain.
Solution: no - between the points E and F there is a sharp point, where the function
is not differentiable
(c) At what labeled points is f 0 (x) positive? negative? zero?
7
Solution: f 0 is positive at C, E, F, since f is increasing; f 0 is negative at D since f
is decreasing; f 0 = 0 at A, B, since there is a horizontal tangent line
(d) Write the points B, D, E, F in order from least value for f 0 (x) to greatest.
Solution: D,B,F,E
(e) Suppose now that g is a function such that g 0 (x) = f (x). At what labeled points is g(x)
increasing?
Solution: g is increasing at all of the labeled points, since g 0 (x) = f (x) is positive at
all labeled points.
8