Section 1.2
| 12. |
|
Section 1.3
| 2. |
(a) Stretch vertically by a factor of 5. (b) Shift 5 units to the right. (c) Reflect across the x-axis. (d) Reflect across the x-axis and stretch vertically by a factor of 5. (e) Compress horizontally by a factor of 5. (f) Stretch vertically by a factor of 5 and shift downward 3 units. |
Section 2.2
| 2. | We can make f(x) arbitrarily close to 3 by taking x sufficintly close to 1 and less than 1; we can make f(x) arbitrarily close to 7 by taking x sufficiently close to 1 and greater than 1. The limit of f(x) as x approaches 1 does not exist in this situation, because no matter how close to 1 we move x, the function f(x) will take on values close to 3 and close to 7. So f(x) cannot be made to stay close to a single number by taking x close to 1. |
| 4. |
(a) 3 (b) 4 (c) 2 (d) Does not exist; left- and right-hand limits are different (e) 3 |
| 8. |
(a) 0 (b) (infinity) (c) -(infinity) (d) -(infinity) (e) x=-5; x=0; x=4 |
Section 2.3
| 10. |
|
Section 2.5
| 6. |
Here's one simple solution.
|
| 16. |
There is an infinite discontinuity at x=1. The limit of f(x) as x
approaches 1 from the left is -(infinity); the limit of f(x) as x
approaches 1 from the right is (infinity).
|
| 20. |
There is a jump discontinuity at x = 2. The limit of f(x) as x
approaches 2 from the left is -1, and the limit of f(x) as x
approaches 2 from the right is 0.
|
| 34. |
The function has jump discontinuities at x=-1 and x=1.
It is continuous from the left at -1 and continuous from the right at
1.
|
Section 2.6
| 8. |
The slope is equal to limh -> 0(1/sqrt(1+h)-1)/h = limh -> 0(1-sqrt(1+h))/(h sqrt(1+h)) = limh -> 0((1 - (1+h))/(h sqrt(1+h) (1 + sqrt(1+h))) = limh -> 0-h/(h sqrt(1+h) (1 + sqrt(1+h))) = limh -> 0-1/(sqrt(1+h) (1+sqrt(1+h))) = -1/(sqrt(1) (1 + sqrt(1)) = -1/2. Using the point-slope form, the equation of the line is y - 1 = (-1/2)(x - 1). |
Section 3.2
| 4. |
(a) II. The slope is negative, then positive, then negative, and has
two zeros. (b) IV. The slope is a positive constant, then a negative constant, then a positive constant. (c) I. The slope starts close to zero, then is increasingly negative, then swings through to a positive number, and then decreases toward zero. (d) III. There are three points where the tangent slope is zero. |
Section 3.4
| 8. |
(a) Maximum height is 180 feet. (b) On the way up, 16 ft/s; on the way down, -16 ft/s. |
| 16. |
(a) 218.75 gallons/minute (b) 187.5 gallons/minute (c) 125 gallons/minute (d) 0 gallons/minute The outflow is fastest at t=0, and slows at a constant rate to zero gallons per minute at t=40. |
The proof
Because HTML doesn't handle mathematical notation well, I've supplied the proof separately as a PDF document. If you have the Adobe Acrobat Reader, click here to see the proof.
| 44. | The equation of the line is y - 1 = (-sqrt(3)/2)(x - pi/6). |
Section 3.9
| 14. | The boat approaches the dock at sqrt(65)/8 meters per second. |
Section 3.10
| 8. | L(x) = -2 + (1/12)(x+8) |
| 42. | dV = 0.625*pi cubic meters, or about 1,963 liters. |
Section 4.7
| 2. | The numbers are 50 and -50. |
| 12. |
The minimum cost is about $163.54. The exact value is 20*(9/2)^(2/3)+180/(9/2)^(1/3). |
| 22. |
The dimensions of the largest rectangle are Width: 2*(8/3)^(1/2) Height: 16/3. |
| 30. |
The dimensions that give the largest printed area are Width = (120)^(1/2) Height = (3/2)*(120)^(1/2). |
| 34. |
The length of the shortest ladder is ((256^(1/3)+4)^2 + (64/(256^(2/3)))(256^(1/3)+4)^2)^(1/2) feet. This is about 16.65 feet. |
Section 7.2*
| 2. | (1/2) ln (a) + (1/2)ln(b2 + c2) |
| 6. | ln(6) |
| 48. | One root is approximately 1.05800640109; the other is approximately -1.96463559749. |
Section 7.3*
| 12. | The equation can be rearranged to read (ex - 3) (ex - 4) = 0, so the roots are x = ln(3) and x = ln(4). |