Math 106: Calculus I, Spring 2018: Midterm Exam II Monday, April Give your name, TA and section number:

Size: px

Start display at page:

Download "Math 106: Calculus I, Spring 2018: Midterm Exam II Monday, April Give your name, TA and section number:"

Vincent McDaniel
5 years ago
Views:

1 Math 106: Calculus I, Spring 2018: Midterm Exam II Monday, April Give your name, TA and section number: Name: TA: Section number: 1. There are 6 questions for a total of 100 points. The value of each part of each question is stated. 2. Do not open your booklet until told to begin. The exam will be 50 minutes long. 3. You may not use phones, calculators, books, notes or any other paper. Write all your answers on this booklet. If you need more space, you can use the back of the pages. 4. Unless specified otherwise, you must show ALL your work and explain your answers clearly to obtain full credit! 5. Read the questions carefully! Make sure you understand what each question asks of you. Please read the following statement and then sign and date it: I agree to complete this exam without unauthorized assistance from any person, materials, or device. Signature: Date: Question Points Score Total: 100 i

2 1. (a) (4 points) Write the formal definition (i.e. the it definition) of derivative f (x) of a real valued function f(x) at a fixed value x. The definition of f at a fixed point x is f f(x + h) f(x) (x) =. (b) (8 points) Let f(x) = x 2. Show from the definition that f (x) = 2x. Show all your work. NOTE: In this question, do not use any rules for differentiation. Work straight from the definition, and compute the necessary it that shows up in the definition to show it is equal to 2x. f (x) = (x + h) 2 x 2 = x 2 + 2xh + h 2 x 2 = x 2 + 2xh + h 2 x 2 = h(2x + h) = 2x + h h 0 = 2x 1

3 2. In the following questions, you can use rules for differentiation. Feel free to consult the last page of the exam, which contains some of the rules in case you forgot them. Show all your work and reasoning for why you can apply the laws you are using. Do not skip steps. (a) (8 points) Let Compute f (x). f(x) = cos 1 + 4x 4. We will need two iterations of the chain rule to compute this. f (x) = sin 1 + 4x 4 = sin 1 + 4x 4 = sin 1 + 4x 4 d 1 + 4x 4 dx x x 4 16x3 d dx (1 + 4x4 ) (b) (8 points) Let Compute f (x). f(x) = ln x x 2 for x > 0 f (x) = = 1 x x2 ln x 2x x 4 x 2x ln x = x 4 = 1 2 ln x x 3 by the quotient rule 2

4 In the following questions, you can use rules for differentiation. Feel free to consult the last page of the exam, which contains some of the rules in case you forgot them. Show all your work and reasoning for why you can apply the laws you are using. Do not skip steps. (c) (8 points) Let Compute f (0). f(x) = (2x 1)ex2 (x 1) 3 (x 2 + 5) for x 1. If furthermore x 1 2, so that f(x) or any or the factors are not zero, we can take ln: ln f(x) = ln 2x 1 + ln e x2 3 ln x 1 ln x = ln 2x 1 + x 2 3 ln x 1 ln x and differentiating both sides we get f (x) f(x) = 2 2x 1 + 2x 3 1 x 1 2x x Thus f (0) = f(0) ( ) = = 1 5 (d) (8 points) Find an equation of the tangent line to the curve y = 2x sin x at the point ( π 2, π). Differentiating both sides with respect to x: y = 2 sin x + 2x cos x Evaluating at the point x = π 2 : y ( π 2 ) = 2 sin π 2 + 2π 2 cos π 2 = π 0 = 2. Thus the equation of the line tangent to the curve at the point ( π 2, π) is y π = 2(x π 2 ) so so y π = 2x π y = 2x. 3

5 3. (10 points) You are inflating a spherical balloon at the rate of 7 cm 3 /sec. How fast is its radius increasing when the radius is 4 cm? NOTE: The volume of a sphere of radius r is given by the formula V = 4πr3 3. Note that r and V are both functions of t, and of course, V depends on r (it is a composite of r(t) and the function 4πr3 3.) We know dv dt dr = 7 and we want to know dt at the time t when r(t) = 4. By the chain rule, dv dt = 4πr2 dr dt. So at the time t when r(t) = 4, so at this time t, 7 = 4π4 2 dr dt dr dt = 7 64π cm/sec 4

6 4. (a) (8 points) Let F (x) = f(g(x)). Suppose we know the following values f( 2) = 8 f ( 2) = 4 f (5) = 3 g(5) = 2 g (5) = 6 Find F (5). Note: You might not need all the given values in your computation of F (5). By the chain rule, F (x) = f (g(x))g (x). Thus F (5) = f (g(5))g (5) = f ( 2) 6 = 4 6 = 24. (b) (8 points) Give an example of a degree 2 polynomial p(x) with p (0) = 3 and p (0) = 5. A degree 2 polynomial has the form p(x) = ax 2 + bx + c. We have p (x) = 2ax + b, and we need p (0) = 3, so we need b = 3. We also have p (x) = 2a and we need p (0) = 5, so we need a = 5 2. The choice of c does not matter, as long as b = 3 and a = 5 2, the required conditions of p will be satisfied. So you can choose any real number c and get a valid example. For example, or or p(x) = 5 2 x2 + 3x + 6 p(x) = 5 2 x2 + 3x p(x) = 5 2 x2 + 3x + π These are all valid examples and any other choice of c would also give a valid example. You just had to choose one value for c to construct your example. 5

7 5. (20 points) Which of the following statements are true or false? Write True or False on the line accordingly. No justification is needed! (a) If f(x) exists, then f is differentiable at a. FALSE the it at a existing says nothing x a about differentiability at a. It doesn t even imply that f is continuous at a (unless that it were equal to f(a)), and even continuity at a would not imply differentiability at a, see next question. (b) If f is continuous at a, then f is differentiable at a. class and on the practice. FALSE we saw counterexamples of this in f(a + h) f(a) (c) If exists, then f is differentiable at a. definition what it means to be differentiable at a. TRUE that it existing is by (d) If f is differentiable at a, then f(x) = f(a). TRUE we proved exactly this implication in x a class, and what it says is that if f is differentiable at a then it s continuous at a. (e) If f is defined on [a, b], and we have that f(a) < 0 and f(b) > 0, then f has a 0 on (a, b). FALSE We would also need f continuous to apply IVT. 6

8 6. (10 points) Show that ( 1 ) x cos = 0 x Explain your reasoning and show all your work. Note that cos 1 x theorem. does not exist, so we cannot apply the product rule. We will use the sandwich Note that for all x 0, 1 cos 1 x 1 REMARK: When we multiply through by x, depending on whether x is positive or negative, the inequalities get preserved or reversed. For x > 0: x x cos 1 x x holds for all x > 0, so we can apply the sandwich theorem for right its Since x = x = 0 (and in particular the right its are 0), we can conclude that For x < 0: x cos 1 + x = 0 x x cos 1 x x holds for all x < 0, so we can apply the sandwich theorem for left its Since x = x = 0 (and in particular the left its are 0), we can conclude that x cos 1 x = 0 ( 1 ) Since the left and right its agree, we can conclude that x cos = 0. x COMMENT: If you used only one set of inequalities x x cos 1 x x or x x cos 1 x x to conclude ( 1 ) ( 1 ) that x cos = 0, that is wrong: the sandwich theorem does not apply to compute x cos x x because neither one of those sets of inequalities holds for all points around 0, which was the hypothesis on the sandwich theorem. One of the sets of inequalities holds to the right, and one to the left of 0. However, if you noted that x x cos 1 x x for ALL x that is correct, and you can compute the its x = 0 and x = 0 to conclude that the middle one is 0. 7

Calculus I: Practice Midterm II

Calculus I: Practice Midterm II Calculus I: Practice Mierm II April 3, 2015 Name: Write your solutions in the space provided. Continue on the back for more space. Show your work unless asked otherwise. Partial credit will be given for