Origin - Calculus - Solved Exam, Exams of Calculus

This is the Solved Exam of Calculus which includes Rambling, Worst Three, Rabbits, Commonly, Proportional, Average Rate, Proper Justiffication, Derivative etc. Key important points are: Origin, Local Maxima, Local Minima, Inflection Points, Values, Continuous, Differentiable, Straight Line, Acceleration, Experiment

Typology: Exams

2012/2013

Uploaded on 03/06/2013

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MATH 105, Sections A&B Final Exam SOLUTIONS 1. (20 points) Find the domain of the following function. f (x) = ¢/U=228-=*) Solution. (ns)fs-s"} ros fae) Lo ene ee) >o= dom f =| (—Vv5, 0) U (0, 1] U [¥75, +0) | = + # - +t x TOT TEE —— TTT -& e ‘ e 2. (20 points) Let f(x) = x — 4r¥, i a) Find local maxima of f(x). b) Find local minima of f(x). c) Find inflection points of f(a). Solution. f’(r) = 42° — 12%? = da? (x—3) => f(x) has | no local maxima and a local minimum at 2: =3} f(x) = 122? — 242 = 12a (x — 2) => f(x) has] inflection points at x = 0 and x = 2 Se Pod. + = + ff MeN 3-7 Ft VIP OV ed F az +80 if >2 bide if2 <2 Find the values of a and 6 such that f («) is continuous and 3. (20 points) Let f (x) = { differentiable at 2 = 2. Solution. The fact that f (x) is continuous means that Linn fiz) = lim f(x), and that f (x) is differentiable == = means that kim (z= Jim | J’ (x). Thus we have to find the derivative and set up a system for a and b; =a 2! tigy J 4008 ifs >2 1a+80=4b+4a _ da+20=b+4 3a+20=b PO)=) oe ite<2 | aa=4b 8a=b 8a=b 3a +20 = 84 => 20 = 5a [a = 4] => 4, (20 points) An object moves along a straight line such that its acceleration is a(t) = 15 + & The experiment shows that at f = 1 we have the following data: the velocity of the object is 19 units per second, and it’s 9 units far from the origin in the positive direction. a) Find the velocity at time t = 4. 'b) Find a function that describes the position of the object at time t. Solution. v(t) = fo dt = if (15vi+ Sat = 1 fePares fae = 1008/2 + gt? +. Cy => o(l) = 104640, =19 = OC, =3 = v(t) = 108? + 6? +3 a}u (4) = 10-497? + 6-44? +3 = 80 +12 43 =[95 units per second] b) p(t) = fvwae= 10 f 8Pa +6 [eras far = 405/? 4 g¢8 4 34 Cy = pQ) =44+4434C2 =9 => Cp =-2 = [p(t} = 40? +4 +3t 5, (20 points) A mold grows at a rate proportional to the amount present. Initially, its weight was 10 grams; after 3 days, it was 10e grams. a) Determine a differential equation that describes the process and a function that expresses the weight of the mold as a function of time. ‘b) What is the weight of the mold after 15 days?