Calculus Early Transcendentals (for AP) by John Rogawski, Ray Cannon

By John Rogawski, Ray Cannon

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Example text

Represent 1/7 and 4/27 as repeating decimals. x (A) (B) y 36. Plot each pair of points and compute the distance between them: (a) (1, 4) and (3, 2) (b) (2, 1) and (2, 4) (c) (0, 0) and (−2, 3) 11 39. Determine the domain and range of the function (ii) a lies between 1 and 7. (iii) The distance from a to 5 is less than 13 . 24. Describe x : Real Numbers, Functions, and Graphs y x x (d) (−3, −3) and (−2, 3) (C) 37. Find the equation of the circle with center (2, 4): (a) with radius r = 3. (b) that passes through (1, −1).

3x + 5| < 1 In Exercises 19–22, describe the set as a union of finite or infinite intervals. 19. {x : |x − 4| > 2} 20. {x : |2x + 4| > 3} 21. {x : |x 2 − 1| > 2} 22. {x : |x 2 + 2x| > 2} 23. Match (a)–(f) with (i)–(vi). 1 (i) a lies to the right of 3. (iv) The distance from a to 3 is at most 2. (v) a is less than 5 units from 13 . (vi) a lies either to the left of −5 or to the right of 5. x < 0 as an interval. x+1 f : {r, s, t, u} → {A, B, C, D, E} defined by f (r) = A, f (s) = B, f (t) = B, f (u) = E.

For example, quadrant III consists of points (x, y) such that x < 0 and y < 0. The distance d between two points P1 = (x1 , y1 ) and P2 = (x2 , y2 ) is computed using the Pythagorean Theorem. In Figure 11, we see that P1 P2 is the hypotenuse of a right triangle with sides a = |x2 − x1 | and b = |y2 − y1 |. Therefore, d 2 = a 2 + b2 = (x2 − x1 )2 + (y2 − y1 )2 We obtain the distance formula by taking square roots. 4 CHAPTER 1 PRECALCULUS REVIEW y y 2 b 1 −2 P = (a, b) II (−, +) x −1 1 2 x a −1 III (−, −) −2 (A) FIGURE 10 Rectangular coordinate system.

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