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1. A certain radioactive element has a half-life of 5 hours. If its initial mass is at

6,470 grams, how many grams are left after 2 days?

2. Carlo’s online investment is worth Php870,000 on its 3rd year and Php650, 000

on its 7th year. What is the worth of his initial investment?

3. The pressure exerted on us by the atmosphere decreases exponentially as you

go up. The pressure at ground level is 1,013 hPa and decreases to 965 hPa at

381 meters. What is the pressure at the summit of Mt. Apo which is at 2,954

4. A strain of bacteria growing on the palm of your hands is 9 bacteria after 6

minutes. If you start with only one bacterium and follows an exponential growth

model, how many bacteria could be present at the end of 1 hour?

5. A certain breed of rats doubles its population every 2 months. Assuming there

were only 6 rats initially at a certain area, how many months will it take for the

population to grow to 68 rats?

Compute the area of the plane region bounded by the curve x=y 2 -2 and the line y=-x using integration along the y-axis.

Consider the R²-R function defined by f(x,y)=x²-2x-y and let c be the contour curve of f at level 0.

1.find a Cartesian equation for the tangent line L to C at the point (-1,3).

2.sketch the contour curve C together with the line L in R².show all intersections with the axes.

3.find an equation for the tangent plane V to the graph of f in R³ at the point (-1,3,0).

A delivery company needs the measurements of a rectangular box such that the length plus twice the width plus twice the heights should not exceed 150cm.what is the maximum volume of such a box?

Consider the R²-R function f defined by f(x,y)=x²+2y²-x²y.

Show that f has two saddle points

Consider the R²-R function f defined by f(x,y)=e^xIn(1+y).

Find the third order Taylor polynomial of f about the point (0,0)

Convert (√3,-1)into polar coordinates (r,0) so that r≥0 and 0≤0<2π

Find implicit differentiation of 2y+cot(xy^2)=3xy

A rock is thrown horizontally from the top of a cliff 88 m high, with a horizontal speed of 25 m/s. with what velocity does the rock hit

Derive an equation  x ( t ) for the instantaneous position of the car as a function of time. Identify the

●      value  x  when  t  = 0 s

●      asymptote of this function as  t  → ∞

t(0-28 m/s) (s) t(400m)(s) t (Maxspeed)(s)

2.5 9.75 8.0

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lim x → 1 1 x − 1 x − 1 = −1 lim x → 1 1 x − 1 x − 1 = −1

lim x → 2 h ( x ) = −1 . lim x → 2 h ( x ) = −1 .

lim x → 2 | x 2 − 4 | x − 2 lim x → 2 | x 2 − 4 | x − 2 does not exist.

a. lim x → 2 − | x 2 − 4 | x − 2 = −4 ; lim x → 2 − | x 2 − 4 | x − 2 = −4 ; b. lim x → 2 + | x 2 − 4 | x − 2 = 4 lim x → 2 + | x 2 − 4 | x − 2 = 4

a. lim x → 0 − 1 x 2 = + ∞ ; lim x → 0 − 1 x 2 = + ∞ ; b. lim x → 0 + 1 x 2 = + ∞ ; lim x → 0 + 1 x 2 = + ∞ ; c. lim x → 0 1 x 2 = + ∞ lim x → 0 1 x 2 = + ∞

a. lim x → 2 − 1 ( x − 2 ) 3 = − ∞ ; lim x → 2 − 1 ( x − 2 ) 3 = − ∞ ; b. lim x → 2 + 1 ( x − 2 ) 3 = + ∞ ; lim x → 2 + 1 ( x − 2 ) 3 = + ∞ ; c. lim x → 2 1 ( x − 2 ) 3 lim x → 2 1 ( x − 2 ) 3 DNE. The line x = 2 x = 2 is the vertical asymptote of f ( x ) = 1 / ( x − 2 ) 3 . f ( x ) = 1 / ( x − 2 ) 3 .

Does not exist.

11 10 11 10

lim x → −1 − f ( x ) = −1 lim x → −1 − f ( x ) = −1

f is not continuous at 1 because f ( 1 ) = 2 ≠ 3 = lim x → 1 f ( x ) . f ( 1 ) = 2 ≠ 3 = lim x → 1 f ( x ) .

f ( x ) f ( x ) is continuous at every real number.

Discontinuous at 1; removable

[ −3 , + ∞ ) [ −3 , + ∞ )

f ( 0 ) = 1 > 0 , f ( 1 ) = −2 < 0 ; f ( x ) f ( 0 ) = 1 > 0 , f ( 1 ) = −2 < 0 ; f ( x ) is continuous over [ 0 , 1 ] . [ 0 , 1 ] . It must have a zero on this interval.

Let ε > 0 ; ε > 0 ; choose δ = ε 3 ; δ = ε 3 ; assume 0 < | x − 2 | < δ . 0 < | x − 2 | < δ .

Thus, | ( 3 x − 2 ) − 4 | = | 3 x − 6 | = | 3 | · | x − 2 | < 3 · δ = 3 · ( ε / 3 ) = ε . | ( 3 x − 2 ) − 4 | = | 3 x − 6 | = | 3 | · | x − 2 | < 3 · δ = 3 · ( ε / 3 ) = ε .

Therefore, lim x → 2 3 x − 2 = 4 . lim x → 2 3 x − 2 = 4 .

Choose δ = min { 9 − ( 3 − ε ) 2 , ( 3 + ε ) 2 − 9 } . δ = min { 9 − ( 3 − ε ) 2 , ( 3 + ε ) 2 − 9 } .

| x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε | x 2 − 1 | = | x − 1 | · | x + 1 | < ε / 3 · 3 = ε

δ = ε 2 δ = ε 2

Section 2.1 Exercises

a. 2.2100000; b. 2.0201000; c. 2.0020010; d. 2.0002000; e. (1.1000000, 2.2100000); f. (1.0100000, 2.0201000); g. (1.0010000, 2.0020010); h. (1.0001000, 2.0002000); i. 2.1000000; j. 2.0100000; k. 2.0010000; l. 2.0001000

y = 2 x y = 2 x

a. 2.0248457; b. 2.0024984; c. 2.0002500; d. 2.0000250; e. (4.1000000,2.0248457); f. (4.0100000,2.0024984); g. (4.0010000,2.0002500); h. (4.00010000,2.0000250); i. 0.24845673; j. 0.24984395; k. 0.24998438; l. 0.24999844

y = x 4 + 1 y = x 4 + 1

a. −0.95238095; b. −0.99009901; c. −0.99502488; d. −0.99900100; e. (−1;.0500000,−0;.95238095); f. (−1;.0100000,−0;.9909901); g. (−1;.0050000,−0;.99502488); h. (1.0010000,−0;.99900100); i. −0.95238095; j. −0.99009901; k. −0.99502488; l. −0.99900100

y = − x − 2 y = − x − 2

−49 m/sec (velocity of the ball is 49 m/sec downward)

Under, 1 unit 2 ; over: 4 unit 2 . The exact area of the two triangles is 1 2 ( 1 ) ( 1 ) + 1 2 ( 2 ) ( 2 ) = 2.5 units 2 . 1 2 ( 1 ) ( 1 ) + 1 2 ( 2 ) ( 2 ) = 2.5 units 2 .

Under, 0.96 unit 2 ; over, 1.92 unit 2 . The exact area of the semicircle with radius 1 is π ( 1 ) 2 2 = π 2 π ( 1 ) 2 2 = π 2 unit 2 .

Approximately 1.3333333 unit 2

Section 2.2 Exercises

lim x → 1 f ( x ) lim x → 1 f ( x ) does not exist because lim x → 1 − f ( x ) = −2 ≠ lim x → 1 + f ( x ) = 2 . lim x → 1 − f ( x ) = −2 ≠ lim x → 1 + f ( x ) = 2 .

lim x → 0 ( 1 + x ) 1 / x = 2.7183 lim x → 0 ( 1 + x ) 1 / x = 2.7183

a. 1.98669331; b. 1.99986667; c. 1.99999867; d. 1.99999999; e. 1.98669331; f. 1.99986667; g. 1.99999867; h. 1.99999999; lim x → 0 sin 2 x x = 2 lim x → 0 sin 2 x x = 2

lim x → 0 sin a x x = a lim x → 0 sin a x x = a

a. −0.80000000; b. −0.98000000; c. −0.99800000; d. −0.99980000; e. −1.2000000; f. −1.0200000; g. −1.0020000; h. −1.0002000; lim x → 1 ( 1 − 2 x ) = −1 lim x → 1 ( 1 − 2 x ) = −1

a. −37.931934; b. −3377.9264; c. −333,777.93; d. −33,337,778; e. −29.032258; f. −3289.0365; g. −332,889.04; h. −33,328,889 lim x → 0 z − 1 z 2 ( z + 3 ) = − ∞ lim x → 0 z − 1 z 2 ( z + 3 ) = − ∞

a. 0.13495277; b. 0.12594300; c. 0.12509381; d. 0.12500938; e. 0.11614402; f. 0.12406794; g. 0.12490631; h. 0.12499063; ∴ lim x → 2 1 − 2 x x 2 − 4 = 0.1250 = 1 8 ∴ lim x → 2 1 − 2 x x 2 − 4 = 0.1250 = 1 8

a. 10.00000; b. 100.00000; c. 1000.0000; d. 10,000.000; Guess: lim α → 0 + 1 α cos ( π α ) = ∞ , lim α → 0 + 1 α cos ( π α ) = ∞ , actual: DNE

False; lim x → −2 + f ( x ) = + ∞ lim x → −2 + f ( x ) = + ∞

False; lim x → 6 f ( x ) lim x → 6 f ( x ) DNE since lim x → 6 − f ( x ) = 2 lim x → 6 − f ( x ) = 2 and lim x → 6 + f ( x ) = 5 . lim x → 6 + f ( x ) = 5 .

Answers may vary.

a. ρ 2 ρ 2 b. ρ 1 ρ 1 c. DNE unless ρ 1 = ρ 2 . ρ 1 = ρ 2 . As you approach x SF x SF from the left, you are in the high-density area of the shock. When you approach from the right, you have not experienced the “shock” yet and are at a lower density.

Section 2.3 Exercises

Use constant multiple law and difference law: lim x → 0 ( 4 x 2 − 2 x + 3 ) = 4 lim x → 0 x 2 − 2 lim x → 0 x + lim x → 0 3 = 3 lim x → 0 ( 4 x 2 − 2 x + 3 ) = 4 lim x → 0 x 2 − 2 lim x → 0 x + lim x → 0 3 = 3

Use root law: lim x → −2 x 2 − 6 x + 3 = lim x → −2 ( x 2 − 6 x + 3 ) = 19 lim x → −2 x 2 − 6 x + 3 = lim x → −2 ( x 2 − 6 x + 3 ) = 19

− 5 7 − 5 7

lim x → 4 x 2 − 16 x − 4 = 16 − 16 4 − 4 = 0 0 ; lim x → 4 x 2 − 16 x − 4 = 16 − 16 4 − 4 = 0 0 ; then, lim x → 4 x 2 − 16 x − 4 = lim x → 4 ( x + 4 ) ( x − 4 ) x − 4 = 8 lim x → 4 x 2 − 16 x − 4 = lim x → 4 ( x + 4 ) ( x − 4 ) x − 4 = 8

lim x → 6 3 x − 18 2 x − 12 = 18 − 18 12 − 12 = 0 0 ; lim x → 6 3 x − 18 2 x − 12 = 18 − 18 12 − 12 = 0 0 ; then, lim x → 6 3 x − 18 2 x − 12 = lim x → 6 3 ( x − 6 ) 2 ( x − 6 ) = 3 2 lim x → 6 3 x − 18 2 x − 12 = lim x → 6 3 ( x − 6 ) 2 ( x − 6 ) = 3 2

lim x → 9 t − 9 t − 3 = 9 − 9 3 − 3 = 0 0 ; lim x → 9 t − 9 t − 3 = 9 − 9 3 − 3 = 0 0 ; then, lim t → 9 t − 9 t − 3 = lim t → 9 t − 9 t − 3 t + 3 t + 3 = lim t → 9 ( t + 3 ) = 6 lim t → 9 t − 9 t − 3 = lim t → 9 t − 9 t − 3 t + 3 t + 3 = lim t → 9 ( t + 3 ) = 6

lim θ → π sin θ tan θ = sin π tan π = 0 0 ; lim θ → π sin θ tan θ = sin π tan π = 0 0 ; then, lim θ → π sin θ tan θ = lim θ → π sin θ sin θ cos θ = lim θ → π cos θ = −1 lim θ → π sin θ tan θ = lim θ → π sin θ sin θ cos θ = lim θ → π cos θ = −1

lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1 = 1 2 + 3 2 − 2 1 − 1 = 0 0 ; lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1 = 1 2 + 3 2 − 2 1 − 1 = 0 0 ; then, lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1 = lim x → 1 / 2 ( 2 x − 1 ) ( x + 2 ) 2 x − 1 = 5 2 lim x → 1 / 2 2 x 2 + 3 x − 2 2 x − 1 = lim x → 1 / 2 ( 2 x − 1 ) ( x + 2 ) 2 x − 1 = 5 2

lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72 lim x → 6 2 f ( x ) g ( x ) = 2 lim x → 6 f ( x ) lim x → 6 g ( x ) = 72

lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7 lim x → 6 ( f ( x ) + 1 3 g ( x ) ) = lim x → 6 f ( x ) + 1 3 lim x → 6 g ( x ) = 7

lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5 lim x → 6 g ( x ) − f ( x ) = lim x → 6 g ( x ) − lim x → 6 f ( x ) = 5

lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28 lim x → 6 [ ( x + 1 ) f ( x ) ] = ( lim x → 6 ( x + 1 ) ) ( lim x → 6 f ( x ) ) = 28

lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6 lim x → −3 − ( f ( x ) − 3 g ( x ) ) = lim x → −3 − f ( x ) − 3 lim x → −3 − g ( x ) = 0 + 6 = 6

lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1 lim x → −5 2 + g ( x ) f ( x ) = 2 + ( lim x → −5 g ( x ) ) lim x → −5 f ( x ) = 2 + 0 2 = 1

lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3 lim x → 1 f ( x ) − g ( x ) 3 = lim x → 1 f ( x ) − lim x → 1 g ( x ) 3 = 2 + 5 3 = 7 3

lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46 lim x → −9 ( x f ( x ) + 2 g ( x ) ) = ( lim x → −9 x ) ( lim x → −9 f ( x ) ) + 2 lim x → −9 ( g ( x ) ) = ( −9 ) ( 6 ) + 2 ( 4 ) = −46

The limit is zero.

b. ∞. The magnitude of the electric field as you approach the particle q becomes infinite. It does not make physical sense to evaluate negative distance.

Section 2.4 Exercises

The function is defined for all x in the interval ( 0 , ∞ ) . ( 0 , ∞ ) .

Removable discontinuity at x = 0 ; x = 0 ; infinite discontinuity at x = 1 x = 1

Infinite discontinuity at x = ln 2 x = ln 2

Infinite discontinuities at x = ( 2 k + 1 ) π 4 , x = ( 2 k + 1 ) π 4 , for k = 0 , ± 1 , ± 2 , ± 3 ,… k = 0 , ± 1 , ± 2 , ± 3 ,…

No. It is a removable discontinuity.

Yes. It is continuous.

k = −5 k = −5

k = −1 k = −1

k = 16 3 k = 16 3

Since both s and y = t y = t are continuous everywhere, then h ( t ) = s ( t ) − t h ( t ) = s ( t ) − t is continuous everywhere and, in particular, it is continuous over the closed interval [ 2 , 5 ] . [ 2 , 5 ] . Also, h ( 2 ) = 3 > 0 h ( 2 ) = 3 > 0 and h ( 5 ) = −3 < 0 . h ( 5 ) = −3 < 0 . Therefore, by the IVT, there is a value x = c x = c such that h ( c ) = 0 . h ( c ) = 0 .

The function f ( x ) = 2 x − x 3 f ( x ) = 2 x − x 3 is continuous over the interval [ 1.25 , 1.375 ] [ 1.25 , 1.375 ] and has opposite signs at the endpoints.

b. It is not possible to redefine f ( 1 ) f ( 1 ) since the discontinuity is a jump discontinuity.

Answers may vary; see the following example:

False. It is continuous over ( − ∞ , 0 ) ∪ ( 0 , ∞ ) . ( − ∞ , 0 ) ∪ ( 0 , ∞ ) .

False. Consider f ( x ) = { x if x ≠ 0 4 if x = 0 . f ( x ) = { x if x ≠ 0 4 if x = 0 .

False. IVT only says that there is at least one solution; it does not guarantee that there is exactly one. Consider f ( x ) = cos ( x ) f ( x ) = cos ( x ) on [ − π , 2 π ] . [ − π , 2 π ] .

False. The IVT does not work in reverse! Consider ( x − 1 ) 2 ( x − 1 ) 2 over the interval [ −2 , 2 ] . [ −2 , 2 ] .

R = 0.0001519 m R = 0.0001519 m

D = 345,826 km D = 345,826 km

For all values of a , f ( a ) a , f ( a ) is defined, lim θ → a f ( θ ) lim θ → a f ( θ ) exists, and lim θ → a f ( θ ) = f ( a ) . lim θ → a f ( θ ) = f ( a ) . Therefore, f ( θ ) f ( θ ) is continuous everywhere.

Section 2.5 Exercises

For every ε > 0 , ε > 0 , there exists a δ > 0 , δ > 0 , so that if 0 < | t − b | < δ , 0 < | t − b | < δ , then | g ( t ) − M | < ε | g ( t ) − M | < ε

For every ε > 0 , ε > 0 , there exists a δ > 0 , δ > 0 , so that if 0 < | x − a | < δ , 0 < | x − a | < δ , then | φ ( x ) − A | < ε | φ ( x ) − A | < ε

δ ≤ 0.25 δ ≤ 0.25

δ ≤ 2 δ ≤ 2

δ ≤ 1 δ ≤ 1

δ < 0.3900 δ < 0.3900

Let δ = ε . δ = ε . If 0 < | x − 3 | < ε , 0 < | x − 3 | < ε , then | x + 3 − 6 | = | x − 3 | < ε . | x + 3 − 6 | = | x − 3 | < ε .

Let δ = ε 4 . δ = ε 4 . If 0 < | x | < ε 4 , 0 < | x | < ε 4 , then | x 4 | = x 4 < ε . | x 4 | = x 4 < ε .

Let δ = ε 2 . δ = ε 2 . If 5 − ε 2 < x < 5 , 5 − ε 2 < x < 5 , then | 5 − x | = 5 − x < ε . | 5 − x | = 5 − x < ε .

Let δ = ε / 5 . δ = ε / 5 . If 1 − ε / 5 < x < 1 , 1 − ε / 5 < x < 1 , then | f ( x ) − 3 | = 5 x − 5 < ε . | f ( x ) − 3 | = 5 x − 5 < ε .

Let δ = 3 M . δ = 3 M . If 0 < | x + 1 | < 3 M , 0 < | x + 1 | < 3 M , then f ( x ) = 3 ( x + 1 ) 2 > M . f ( x ) = 3 ( x + 1 ) 2 > M .

0.328 cm, ε = 8 , δ = 0.33 , a = 12 , L = 144 ε = 8 , δ = 0.33 , a = 12 , L = 144

lim x → a f x + lim x → a g x = L + M lim x → a f x + lim x → a g x = L + M

Review Exercises

False. A removable discontinuity is possible.

8 / 7 8 / 7

2 / 3 2 / 3

Since −1 ≤ cos ( 2 π x ) ≤ 1 , −1 ≤ cos ( 2 π x ) ≤ 1 , then − x 2 ≤ x 2 cos ( 2 π x ) ≤ x 2 . − x 2 ≤ x 2 cos ( 2 π x ) ≤ x 2 . Since lim x → 0 x 2 = 0 = lim x → 0 − x 2 , lim x → 0 x 2 = 0 = lim x → 0 − x 2 , it follows that lim x → 0 x 2 cos ( 2 π x ) = 0 . lim x → 0 x 2 cos ( 2 π x ) = 0 .

[ 2 , ∞ ] [ 2 , ∞ ]

c = −1 c = −1

δ = ε 3 δ = ε 3

0 m / sec 0 m / sec

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• Authors: Gilbert Strang, Edwin “Jed” Herman
• Publisher/website: OpenStax
• Book title: Calculus Volume 1
• Publication date: Mar 30, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/calculus-volume-1/pages/1-introduction
• Section URL: https://openstax.org/books/calculus-volume-1/pages/chapter-2

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