Solutions Worksheet 1

Jens Christian Jorgensen

Spring 2013

1) a) False

b) False

c) False

d) True

e) True

2) Sketches omitted - use graphical calculator to check your answer.

3) a) D = R, R = {15}b) D = R, R = [0,∞)

c) D = R \ (−1, 3), R = [0,∞)

d) D = R \ {0}, R = {±1}

4) a) Means that we can get f(x) arbitrarily close to L for x sufficiently close to a.

b) Not necessarily. Only if f is continuous at x = 1.

c) Nothing. The concept of limit is agnostic to the value of f at the given point.

5) a) 3

b) π

c) |a|d) 8

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Solutions Worksheet 2

Jens Christian Jorgensen

Spring 2013

1) a) ∞b) ∞c) DNE

d) −∞

2) k = 1, a = −2.

a) 8

b) 5

c) 33

d) 2√

2

3) a) False

b) False

c) True

d) True

e) True

f) False

4) limh→0±|h|h = ±1. Use graphical calculator or similar to check your result.

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Worksheet 3 - Solution

Spring 2013

1) Comprehension check:

a) See Course textbook.

b) f(x) is continuous at x = c if limx→c f(x) = f(c). The function f : [a, b] → R is continuous if f iscontinuous at every point x ∈ [a, b].

c) A function f can fail to be continuous at a point c ∈ R if: 1) the limit limx→c f(x) exists but is notequal to f(c) (removable discontinuity), 2) the left and right limits exists, but are not equal; that is,limx→c− f(x) 6= limx→c+ f(x) or 3) the limit limx→c f(x) does not exist.

d) False

e) True. Follows immediately from the limit laws.

f) limx→a+ f(x) = f(a).

2) Put f(x) = 2x3 + x− 5. Then f(1) = −2 < 0 and f(2) = 13 > 0, so a root must exist in (1, 2).

3) Limit laws:

a) 3

b) 25/4

4) Limits at ∞:

a) ∞b) 2

c) 0

5) Vertical asymptotes: We have cot(x) = cos(2x)sin(2x) . Potential vertical asymptotes are found by solving

sin(2x) = 0, which gives x = π2n, for n = 0, 1,−1, 2,−3, . . . . Since the numerator is cos(2π2n) = cos(πn) =

±1, it follows that |f(x)| → ∞ as x → π2n, and hence all the listed points correspond to vertical

asymptotes.

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Worksheet 4 - Solution

Spring 2013

1) Comprehension check:

a) Continuity at x = a: limx→a f(x) = f(a). Differentiability at x = a:limx→af(x)−f(a)

x−a exists.

b) cont 6=⇒ diff, diff =⇒ cont

c) Kink example: |x| . Cusp example:√|x| . Jump example: x/|x|

d) g(1) = 6, g′(1) = 5.

e) f ′(c) = limx→cf(x)−f(c)

x−c = limh→0f(c+h)−f(c)

h

f) The slope of the graph of f at x = a.

g) False

h) f(x) = 1, g(x) = x.

i) Yes: diff =⇒ cont

2) (a) f ′(x) = 6x− 2

(b) f ′(x) = − 1(x+3)2

(c) f ′(x) = 1√x

3) a = 0, b = 2.

4) f ′(x) = 2x + 3. f ′ < 0 for x < 3/2. f ′ > 0 for x > −3/2. f ′(−3/2) = 0.

5) Put f ′(a) = 32

√3. f(3) =

√27 + 2. Tg line is: y − f(a) = f ′(a)(x− a).

6) f(−2) < 0, f(3) > 0. By IVT there exists a c ∈ (−2, 3), such that f(c) = 0.

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Worksheet 5 - Solution

Spring 2013

1) Use f(x) = 1 and g(x) = x.

2) See course text book.

3) a) f ′(x) = 12(x−1)

√x−

√x

(x−1)2

b) f ′(x) = (6x + 1) cos(x)−(3x2 + x

)sin(x)

c) f ′(x) =2(x3+2)

x4 +3(x2+1)

x3 − 5(x3+2)(x2+1)x6

d) f(x) = sec2(x)− cot(x) csc(x)

e) f ′(x) = 2x2 sin(x) cos(x) + 2x sin2(x)

4) (a) h′(2) = −2

(b) h′(2) = 8

(c) h′(2) = −4

(d) h′(2) = 5/4

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Worksheet 6 - Solution

Spring 2013

1) See course text book.

2) h′(1) = 60

3) sin(θ) = 3/5, cos(θ) = 4/5, cot(θ) = 4/3

4) a) f ′(x) = 6x2+73(2x3+7x+3)2/3

b) g′(x) = cos(x) sec2(sin(x))

c) h′(x) = 4 sec(x)2 tan(x)

d) f ′(x) = ex+3x2(1 + 6x)

e) g′(x) = cos(x) cos(sin(x)) cos(sin(sin(x)))

5) x ∈ π2 + πZ

6) y = −x+ 32

7) (a) sin(θ) = 3/5

(b) cos(θ) = 4/5

(c) cot(θ) = 4/3

8) (a) h′(2) = −2/3

(b) h′(2) = 40f ′(16) (there appears to be a misprint in the problem)

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Worksheet 7 - Solution

Spring 2013

1) dydx = −x

2

y2

2) dydx = y sin(x)+y cos(xy)

cos(x)−x cos(xy)

3) dydx = −3x2−y2

2(x−2)y

4) The exact volume of paint needed is V (r) = 4π3

[(r + d)3 − r3

], d = 0.2. Linearize in r = 0: (r + d)3 '

3r2d+ r3, so V (r) ' 4πr2 · d. For r = 200, d = 0.2 we get V (r) = 8000π [cm3] = pi125 [liters].

5) Linear approx: l(x) = 18 + 4. Approx.:

√15.75 = f(−.25) ' l(−.25) = 127

32

6) (a) 3/2

(b) −3

(c) 3

7) e = 2.71828 . . .

8) f−1(x) = (x− 1)2 − 5, Df−1 = [1,∞), Rf−1 = [−5,∞).

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Worksheet 9 - Solution

Spring 2013

1) See course text book

2) a) limx→1x5−1x3−1 =

[00

] H= limx→15x4

3x2 = 53 .

b) limx→0sin(4x)tan(5x) =

[00

] H= limx→04 cos(4x)5 sec2(5x) = 4

5 .

c) limx→−∞ x2ex = limx→−∞x2

e−x =[00

] H= limx→−∞2x−e−x =

[00

] H= limx→−∞ limx→−∞2

e−x = 0

3) 0 = limx→0sin(3x)+ax+bx3

x3 =[00

] H= limx→0cos(3x)+a+3bx2

3x2 =[3+a0

]. Must have a = −3. Then,

...H= lim

x→0

−9 sin(3x) + 6bx

6x=

[00] H= lim

x→0

−27 cos(3x) + 6b

6=−92

+ b. (1)

Thus b = −9/2.

4) a) Point where f ′(c) = 0 or where f ′ does not exist.

b) Point c such that f(x) ≤ f(c) for x near c.

c) Point c such that f(x) ≥ f(c) for all x.

5) Critical points

(a) c = 0,−2/3

(b) c = ±√

3

(c) c = 15

6) (a) True

(b) False

(c) True

(d) True. f(x) = 1/x.

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Worksheet 10 - Solution

Spring 2013

1) See course text book

2) No

3) f(x) = 7 for all x.

4) a) c = 12 ln( 6

1−e−6 ).

b) c = 3√

2− 2.

5) He drove at least 159/2 > 65 mile/hour at some point.

a) False

b) True

c) False

d) True

6) Critical points

(a) c = −3, 4

(b) Increase: (−∞,−3] and [4,∞). Decrease: [−3, 4].

(c) Local max c = −3. Local min: c = 4.

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Worksheet 11 - Solution

Spring 2013

1) a) See Calculus Cheat sheet under 1st derivative test.

b) See Calculus Cheat sheet under 2nd derivative test. It does not always work; if f ′′(c) = 0 in a criticalpoint it is inconclusive.

2) (a) Critical points: c = 3,−2

(b) Increase: (−∞,−2] and [3,∞). Decrease: [−2, 3].

(c) Local max: (−2, 48). Local min: (3, 77)

(a) Convex: (−π, 0] and [π, 2π). Concave: (0, π).

(b) Inflection points: c = 0, π.

3) Optimal price = 11.5 $

4) All the rope should be used for the circle.

5) The equation is f ′(x) = 0, where f(x) = 2x+ 3√

(10− x)2 + 1 and x is km of land pipeline.

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Worksheet 12 - Solution

Spring 2013

1) a) False

b) True

c) True

2) 54

3) (a) limn→∞∑n

i=14n ( 4i

n )2.

(b) A ' limn→∞∑4

i=1(i−12 )2 = 21. (True value A = 21.33)

4) (a) R =∑4

i=1 f(1 + i) = 7312 .

(b) f ′(x) = −1(x−1)2 ≤ 0, so f is decreasing.

(c) R >∫ 4

0f(x) dx

5) (a) 5

(b) −2

(c) 0

(d) −1

6) limn→∞∑n

i=12n

(2 + 2i

n

)2 =∫ 2

0f(x) dx, where f(x) = (2 + x)2. The Riemann sum is obtained by using

an equipartition with right endpoints on [0, 2]. The solution is not unique.

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Worksheet 13 - Solution

Spring 2013

1) a) See textbook.

b) f ′(x) =√

x5 − 1 > 0.

c) g′(x) = 3x2√

x15 − 1

2) (a) f ′(x) = − cos(x5).

(b) g′(x) = 2 sin( 1x2 )x−3.

(c) h′(x) = (2 + x4)5

3) (a) 63

(b) e2 − 1

(c) 1

4) Same as WS 12 # 3.

5) (a) 15 log(x) + c

(b) 16 (x3 − 2x3/2

(c) ex + sin(x)− cos(x) + c

(d) tan(x) + c

6) (a) 1543

(b) 6

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