MATH 140A NAMEEXAMINATION I STUDENT NUMBERSEPTEMBER 23, 2002 INSTRUCTORSECTION NUMBERThe examination consists of 15 problems: 11 multiple choice questions followed by 4partial credit problems. For the partial credit problems you must present your workclearly and understandably; no credit will be given for unsupported answers.For this exam calculators are not allowed and are not needed. For multiplechoice problems, please circle the correct answer in each question.The point value for each question is shown next to each question in the left margin. At theend of the examination, the booklet will be collected.THE USE OF CALCULATORS IS NOT PERMITTEDIN THIS EXAMINATION.CHECK THE EXAMINATION BOOKLET BEFOREYOU START. THERE SHOULD BE 15 PROBLEMSON 9 PAGES (INCLUDING THIS ONE).M.C.12.13.14.15.TOTALMATH 140A – EXAMINATION I – – PAGE 21. Find the center and radius of the circle given by the equation5 pts2x2+ 2y2+ 8x − 12y + 8 = 0a) Center: (−2, 3) ; Radius: 3b) Center: (2, 3) ; Radius: 3c) Center: (2, −3) ; Radius: 9d) Center: (4, −6) ; Radius: 3e) Center: (−4, 6) ; Radius: 92. The average rate of change of f (x) = 3x2+ x + 2 on the interval (1, 2) is5 ptsa) −63b) −163c) 6d) 10e) 163. For f (x) = x4, the difference quotientf(x + h) − f(x)his5 ptsa)x4hb)x4+ hhc)x4+ 4hx3+ 6h2x2+ 4h3x + h4hd) 4x3+ 6hx2+ 4h2x + h3e) 1MATH 140A – EXAMINATION I – – PAGE 34. Evaluate limx→2x2(x + 1)2(x − 2)5 ptsa)49b) 1c) ∞d) −∞e) Does not exist.5. Evaluate limx→3x + 3(x − 3)25 ptsa) 0b)13c) 1d) ∞e) Does not exist.6. Evaluate limx→0√x2+ 25 − 5x25 ptsa)110b)15c) 1d) 15e) Does not exist.MATH 140A – EXAMINATION I – – PAGE 47. Evaluate limx→1x2+ 2x − 3x2− 3x + 25 ptsa) −4b) 0c) 1d) ∞e) Does not exist.8. Suppose f (x) is a continuous function, and (1, 3), (2, 0), (3, 5) are points on the graph of5 ptsf(x). What is the value of limx→3−f(x)?a) −3b) 0c) 1d) 5e) Not enough information is known to determine the limit.9. The Intermediate Value Theorem guarantees that x3+ 3x2−12 = 0 has a solution on which5 ptsof the following intervals?a) (−2, −1)b) (−1, 0)c) (0, 1)d) (1, 2)e) (2, 3)MATH 140A – EXAMINATION I – – PAGE 510. If the function5 ptsf(x) =3x2− 2c , x < 2cx − 2x + 2, x ≥ 2is continuous at x = 2, then what is the value of c?a. 5b. 6c. 8d. ∞e. No such value can be found for f (x) to be continuous at x = 2.11. Let f(x) = 2x2− x. Given that f0(1) = 3, what is the equation of the line tangent to the5 ptsgraph of f (x) at x = 1?a. y = 3xb. y = 3x − 1c. y = 3x − 2d. y = 3x − 3e. y = 3x − 4MATH 140A – EXAMINATION I – – PAGE 612. (2 pts each)14 ptsT F a) If both left-hand and right-hand limits of f (x) exist at x = c, then limx→cf(x)must exist.T F b) If f(x) is undefined at x = c, then limx→cf(x) does not exist.T F c) limx→cf(x) may exist if f (x) is discontinuous at x = c.T F d) If limx→cf(x) exists and f (x) is defined at x = c, then f (x) must be continuousat x = c.T F e) If f(x) is continuous at x = c, then limx→c+f(x) = f (c).T F f) If f (x) is differentiable at x = c, then f (x) is also continuous at x = c.T F g) If f(x) is differentiable at x = c, then limx→cf(x) exists.MATH 140A – EXAMINATION I – – PAGE 713. (2 pts each) Suppose f, g, and h are 3 continuous functions, f (x) ≤ g(x) ≤ h(x) for all8 ptsx. If f(1) = 2, f (4) = 5, f (6) = −2, h(1) = 4, h(4) = 5, h(6) = 6. Answer the followingquestions:a) (2 pts) limx→1[f(x) + h(x)] =b) (2 pts) limx→4g(x) =c) (2 pts) What is the name of the theorem you would use to answer part b) above?d) (2 pts) True or False: There is some point x = c on the interval (1, 4) such that f(c) = 3.T FMATH 140A – EXAMINATION I – – PAGE 814. For the function12 ptsf(x) =x + 2x2− 4, x < 31x2− 4x, x > 3Find all points at which f is discontinuous. For each discontinuity, determine the type ofdiscontinuity and justify your answer. (To receive full credit you must show all work verifyingthe type of discontinuity.)MATH 140A – EXAMINATION I – – PAGE 915.11 ptsa) (3 pts) State the limit definition of the derivative.f0(x) =b) (8 pts) Use the limit definition of the derivative to compute the derivative off(x) =√ax , where a is a positive constant.No credit will be given for using any other method to find your