Math 131 Exam 1 October 4, :00-9:00 p.m.

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1 Name (Last, First) My Solutions ID # Signature Lecturer Section (01, 02, 03, etc.) university of massachusetts amherst department of mathematics and statistics Math 131 Exam 1 October 4, :00-9:00 p.m. Instructions Turn off all cell phones and watch alarms! Put away ipods, etc. There are seven (7) questions. Do all work in this exam booklet. You may continue work to the backs of pages and the blank page at the end, but if you do so indicate where. You may use a calculator. If you do, be sure to show the set-up for what you are calculating and do not round intermediate results. Otherwise, this is a closed-book exam: do not use any books or paper except this exam booklet. Organize your work in an unambiguous order. Show all necessary steps. Answers given without supporting work may receive 0 credit! Be ready to show your UMass ID card when you hand in your exam booklet. QUESTION POINTS SCORE TOTAL 100

2 #1. A particle moves along the x-axis with equation of motion s(t) = 3t t where s is measured in meters and t is measured in seconds. (a) (5 points) Compute the average velocity of the particle on the interval [1, 3]. s(1) = 7 s(3) = 5 Therefore, the average velocity on the interval [1, 3] is: Average Velocity = s(3) s(1) 3 1 = = 1 m/s (b) (5 points) Use a it to compute the instantaneous velocity of the particle at time t = 3. We compute: s(t + h) s(t) h 0 h = h 0 [3(t + h) (t + h) 2 + 5] [3t t 2 + 5] h = h 0 3h 2th + h 2 h = h 0 3 2t + h = 3 2t Thus, the instantaneous velocity at t = 3 is: 3 2(3) = 3 m/s (c) (5 points) Does the particle ever come to a stop? (That is, is its instantaneous velocity ever exactly zero?) If so, at what time does this happen? Using the previous part the particle stops when: i.e. at t = 3 s the particle stops t = 0 1

3 #2. Find the exact value of each it without relying on a graph or table of values. If the value does not exist, indicate this by writing DNE. Make sure to show all of your work. (a) (5 points) t 1 3 t + 3 t (t+1) t t 1 t + 1 = t t 1 t + 1 = 3 t 1 t = 3 (b) (5 points) x x 10 x x 5 x x 10 x x 5 = x x 10 x 1 x 5 (10 + 3x 5 ) 1 x 5 = x 1 1 x = 1 3 x 5 (c) (5 points) x 0 x 2 + 2x x 2 (x 1)(2x + 4) x 0 + x 2 + 2x x 2 (x 1)(2x + 4) = x 0 + x + 2 x(x 1)(2x + 4) = x x(x 1) = and x 0 x 2 + 2x x 2 (x 1)(2x + 4) = x 0 x + 2 x(x 1)(2x + 4) = x 0 1 2x(x 1) = So, x 0 x 2 + 2x x 2 (x 1)(2x + 4) = DNE 2

4 #3. (10 points) The it of f(x) = x as x approaches 9 is x = 3. x 9 Demonstrate this by finding the greatest possible value of δ corresponding to ε = 0.01 in the definition of the it. Note: a calculator is recommended for this problem, though not required. If you do use a calculator, you must carefully describe and explain your methods in order to receive full credit. Recall the we say that x a f(x) = L if for every ɛ > 0 there exists some δ > 0 such that 0 < x a < δ = f(x) L < ɛ For this particular Limit and the given ɛ = 0.01 we have that: whenever, f(x) L = x 3 < < x 3 < < x < < x < Now since, and = = means that the greatest possible value of δ corresponding to ɛ = 0.01 is δ = min{0.0599, } =

5 #4. Use the graph of a function y = f(x) shown below to answer both parts of this question. 4 y 5 5 x (a) (6 points) Find the value of each expression below, or write DNE if it does not exist. You do not need to show your work. x 4 f(x) = 1 x 2 f(x) = 3 x 1 f(x) = 4 x 1 + f(x) = 1 x 5 f(x) = DNE f(5) = 2 (b) (14 points) Use the graph to determine whether each statement is True or, and circle the correct choice. You do not need to show your work. At x = 4, f(x) is continuous from the left. At x = 4, f(x) is continuous from the right. At x = 4, f(x) is continuous. At x = 4, f(x) is differentiable. At x = 1, f(x) is continuous from the left. At x = 1, f(x) is continuous from the right. At x = 5, f(x) is continuous from the left. At x = 5, f(x) is continuous from the right. f(x) is continuous on the interval ( 4, 1). f(x) is differentiable on the interval ( 4, 1). f(x) is differentiable on the interval (1, 2). f(x) is differentiable on the interval (0, 2). f(x) is differentiable on the interval (1, 5). f(x) is differentiable on the interval (3, 6). True True True 4

6 #5. (15 points) Without relying on a graph, locate exactly all horizontal and vertical asymptotes of the function below. Explain (in terms of its) why the asymptotes occur at these places. y = f(x) = x 3 + x 2 (6x + 6)(x 2 5) = x 2 (x + 1) 6(x + 1)(x + 5)(x 5). Vertical asymptotes: Our possible candidates to have vertical asymptotes are x = 1, 5, 5. After a little bit of work we find that: x 1 f(x) = 1 24 x ( 5) + f(x) = x ( f(x) = 5) + Therefore, we have vertical asymptotes at x = 5 and at x = 5. Horizontal asymptotes: Using the trick of dividing by the highest power (see question 4.b) we find that: f(x) = f(x) = 1 x x 6 Therefore, y = 1 6 is a horizontal asymptote to f(x). 5

7 #6. (10 points) Use ideas and concepts from Calculus I to explain why we can be certain that the equation f(x) = x 4 4x 1 = 0 has at least two solutions. Note: if you use any theorems, make sure to explain why any such theorems apply in this situation! We use the Intermediate Value Theorem. First note that since f(x) is a polynomial it is continuous on the entire real line and hence it is continuous on every closed interval [a, b]. We compute: f(0) = 1 f(2) = = 7 f( 2) = = 23 Since, f(0) < 0 < f(2) and f(0) < 0 < f( 2) by the intermediate value theorem we have know that there exists real numbers c in ( 2, 0) and d in (0, 2) such that f(c) = f(d) = 0 i.e. the equation has at least two solutions. 6

8 #7. Both parts of this problem concern the function f(x) = 1 x 2. (a) (10 points) Use the it definition of the derivative to find a formula for the derivative f (x). So, For, a 0 we have that: f f(x) f(a) (a) = x a x a = x a 1 x 2 1 a 2 x a a 2 x 2 a = 2 x 2 x a x a (x a)(a + x) = x a a 2 x 2 (x a) (a + x) = x a ax = 2a a 4 = 2 a 3 f (x) = 2 x 3 (b) (5 points) Write an equation for the tangent line to y = f(x) at x = 2. The slope of the tangent line at x = 2 is Since, f(2) = 1, we have that: 4 m = f (2) = = 1 4 y = 1 4 (x 2) = x

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