Calculus I Practice Exam 1A

Transcription

1 Calculus I Practice Exam A Calculus I Practice Exam A Tis practice exam empasizes conceptual connections and understanding to a greater degree tan te exams tat are usually administered in introductory single-variable calculus courses. It is designed to guide students wo are taking suc courses to a deeper mastery of te material. Wile a number of questions ere are fairly typical for actual examinations, you sould not infer from te expression practice exam tat exams encountered in introductory single-variable calculus courses will ask te same types of questions. Multiple coice: 2 x 4. Evaluate te limit algebraically: lim x x A) 0 B) 7 C) 7 D) E) None of tese coices. 2. Evaluate te limit algebraically: lim x 3 3x 9 7(x 2 9) A) 0 B) 2 7 C) 7 D) 7 90 E) None of tese coices. x Evaluate te limit algebraically: lim. DNE means does not exist. x 47 x 47 A) 4 B) C) 7 43 D) DNE E) None of tese coices.

2 Calculus I Practice Exam A 4. Consider te following grap of a function f. Ten wic one of te following statements describes te beavior of te function near x = a? A) lim f(x) exists, lim f(x) exists, lim + B) lim f(x) exists, lim f(x) does not exist, lim + C) lim f(x) exists, lim f(x) exists, lim + D) lim f(x) exists, lim f(x) does not exist, lim + f(x) does not exist, f(a) exists. f(x) does not exist, f(a) exists. f(x) does not exist, f(a) does not exist. E) None of te oters. f(x) does not exist, f(a) does not exist. 5. For wic value of te constant c is te function f continuous? f(x) = { cx2 + for x x c for x > A) c=- B) c=0 C) c= D) c=2 E) None of te oters

3 Calculus I Practice Exam A 6. Explain wy te following function is discontinuous at x=a: A) Bot limit and function value exist, but are not equal to eac oter. B) Neiter limit nor function value exist. C) Even toug te limit exists, te function value does not. D) Even toug te function value exists, te limit does not. E) Bot limit and function value exist and are equal to eac oter. 7. Evaluate te following limit in terms of te constant c: lim x ( 9x 2 + cx 3x) A) 0 B) c 9 C) c 3 D) c 6 E) c 2 2x 8. Evaluate te following limit: lim 0 +x x 2x 5 + A) - B) C) 2 D) 2 E) None of te oters

4 Calculus I Practice Exam A 9. Suppose f(x) = 2g(x) + sin x cos x + 2 x and g (π) =. Evaluate f (π). A) f (π) = 4 π 3 B) f (π) = 3 4 π 3 C) f (π) = + π D) f (π) = 3 + π E) None of te oters 0. Te following table sows values of te position function of an object tat is moving on a straigt line. Time in seconds Position in meters Wat was te average velocity of tis object from t= to t=9 seconds in meters per second? A) B) C) D) 4.35 E) None of te above.

5 Calculus I Practice Exam A Free response. At wat point does te normal to y = x 2 x + at (,) intersect te parabola a second time? 2. Based on te grap of f given below, determine at wic values of x te quantity lim f(x+) f(x) does not exist, and explain wy for eac value. 3. Let f(x) =. Use te limit definition of te derivative to find f '( x ). x+2

6 Answers: Calculus I Practice Exam A Multiple Coice: C 2B 3A 4A 5B 6A 7D 8C 9B 0C Free Response:. To find te normal, we take te derivative of y = x 2 x + : y = 2x and evaluate tat at x=: y =. Terefore, te slope of te tangent of te function at (,) is, wic means tat te slope of te normal (te line perpendicular to te tangent) is -. To find te equation of te normal, we use te fact tat te normal also intersects (,) and plug tat point into te point-slope form: y y = m(x x ) y = (x ) Simplifying tat, we get y = 2 x as te equation of te normal. Tis line intersects te given parabola at (,) by design, and supposedly intersects it a second time. To find tis intersection point, we set te normal and te parabola equal to eac oter: x 2 x + = 2 x Simplifying tis, we get x 2 = wic means x = ±. Te solution x = corresponds to te intersection point (,) tat we already know. Terefore, x = is te location of te intersection we are seeking. We find te y coordinate of te intersection point by plugging x = into y = 2 x and find y = 3. For verification purposes,we may confirm tat we get te same y value by plugging te x value into x 2 x +.

7 Calculus I Practice Exam A Terefore, te second intersection point of te normal wit te quadratic is (-,3). f(x+) f(x) 2. Te quantity lim is te derivative of f at x. Tis does not exist at x= because f as a corner tere (different limit of te difference quotient from te left and te rigt.) x=2 because f as a jump discontinuity tere (te derivative of f does not exist were f is discontinuous.) x=3 because f as a vertical tangent tere, so te limit of te different quotient is infinite, i.e. does not exist. 3. We use te limit definition of derivative: f (x) f(x + ) f(x) x x + 2 We simplify te double fraction by creating a common denominator in te numerator and combining te fractions tere. Ten we cancel te common factor, wic removes te removable discontinuity at = 0 and evaluate te limit by substituting = 0: lim x x + 2 x + 2 (x + + 2)(x + 2) x (x + + 2)(x + 2) = (x + + 2)(x + 2) (x + + 2)(x + 2) = (x + 2) 2 Note: wile tis guide is being made freely available to ASU students and te general public for personal use, it is not to be uploaded to tird-party websites, especially not ones tat profit from suc content. If you found tis document on a tird-party website suc as Course Hero or Cegg, te document is being served to you in violation of copyrigt law.