Math 261 Calculus I. Test 1 Study Guide. Name. Decide whether the limit exists. If it exists, find its value. 1) lim x 1. f(x) 2) lim x -1/2 f(x)

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Math 261 Calculus I Test 1 Study Guide Name Decide whether the it exists. If it exists, find its value. 1) x 1 f(x) 2) x -1/2 f(x) Complete the table and use the result to find the indicated it. 3) If f(x) = x 4-1, find x - 1 x 1 f(x). x 0.9 0.99 0.999 1.001 1.01 1.1 f(x) 1

Use the properties of its to help decide whether the it exists. If the it exists, find its value. 1 x + 6-1 6 4) x 5) x 9 x - 3 x - 9 Find all points where the function is discontinuous. 6) 7) 8) 9) 10) 2

Find all values x = a where the function is discontinuous. 0 if x< 0 11) g(x) = x 2-2x if 0 x 2 2 if x > 2 12) f(x) = ln x - 8 x + 3 Find the average rate of change for the function over the given interval. 13) y = 5x3-8x2 + 1 between x = -3 and x = 2 Suppose the position of an object moving in a straight line is given by the specified function. Find the instantaneous velocity at time t. 14) s(t) = t 2 + 6t + 3, t = 4 Find the instantaneous rate of change for the function at the given value. 15) s(t) = 3t2 + 5t - 7 at t = -2 Find the equation of the tangent line of the curve for the given value of x. 16) y = x2 + 5x, x = 4 Solve the problem. 17) Given - f(x) = L l, + f(x) = L r, and Ll Lr, which of the following statements is true? I. f(x) = Ll II. f(x) = Lr III. f(x) does not exist. A) I B) none C) III D) II 18) What conditions, when present, are sufficient to conclude that a function f(x) has a it as x approaches some value of a? A) The it of f(x) as from the left exists, the it of f(x) as from the right exists, and at least one of these its is the same as f(a). B) The it of f(x) as from the left exists, the it of f(x) as from the right exists, and these two its are the same. C) Either the it of f(x) as from the left exists or the it of f(x) as from the right exists D) f(a) exists, the it of f(x) as from the left exists, and the it of f(x) as from the right exists. 3

Use the graph to evaluate the it. 19) f(x) Provide an appropriate response. 20) It can be shown that the inequalities 1 - x2 6 < x sin(x) < 1 hold for all values of x close to zero. What, if 2-2 cos(x) anything, does this tell you about x sin(x) 2-2 cos(x)? Explain. 21) Write the formal notation for the principle "the it of a quotient is the quotient of the its" and include a statement of any restrictions on the principle. g(x) A) If B) C) If D) g(x) = M and f(x) = L, then g(x) f(x) = g(a), provided that f(a) 0. f(a) g(x) = M and g(x) f(x) = g(a) f(a). f(x) = L, then g(x) f(x) = g(x) f(x) = g(x) f(x) = M, provided that L 0. L f(x) = M, provided that f(a) 0. L 22) Provide a short sentence that summarizes the general it principle given by the formal notation [f(x) ± g(x)] = f(x) ± g(x) = L ± M, given that f(x) = L and g(x) = M. A) The sum or the difference of two functions is the sum of two its. B) The it of a sum or a difference is the sum or the difference of the functions. C) The it of a sum or a difference is the sum or the difference of the its. D) The sum or the difference of two functions is continuous. 4

23) The statement "the it of a constant times a function is the constant times the it" follows from a combination of two fundamental it principles. What are they? A) The it of a product is the product of the its, and the it of a quotient is the quotient of the its. B) The it of a product is the product of the its, and a constant is continuous. C) The it of a constant is the constant, and the it of a product is the product of the its. D) The it of a function is a constant times a it, and the it of a constant is the constant. Determine the it by sketching an appropriate graph. 24) f(x), where f(x) = x2 + 2 for x -4 x -4 + 0 for x = -4 Find the it, if it exists. 7 - x 25) x 7 7 - x Provide an appropriate response. 26) It can be shown that the inequalities -x x cos 1 x x hold for all values of x 0. Find x cos 1 x if it exists. 27) The inequality 1- x 2 2 < sin x < 1 holds when x is measured in radians and x < 1. x Find sin x x if it exists. Find the it. 28) x -2 1 x + 2 29) x -9 + 1 x + 9 Find all vertical asymptotes of the given function. 30) f(x) = x + 9 x 2-36 x - 1 31) R(x) = x 3 + 2x 2-80x A) x = -10, x = 8 B) x = -10, x = 0, x = 8 C) x = -8, x = 0, x = 10 D) x = -8, x = -30, x = 10 5

Find the it, if it exists. 32) (x 2-16 + 3 x 2-36) x -3 33) x A) -4 B) 4 C) Does not exist D) -10 5x - 1 x 3 A) B) 64 C) 125 D) Does not exist Provide an appropriate response. 34) Is f continuous on (-2, 4]? f(x) = x 3, -4x, 4, 0, -2 < x 0 0 x < 2 2 < x 4 x = 2 A) Yes B) No Find the it, if it exists. 6 (x - 35) 3)3 x 3 + x - 3 Provide an appropriate response. 36) Use the Intermediate Value Theorem to prove that 9x 3-5x 2 + 10x + 10 = 0 has a solution between -1 and 0. Find numbers a and b, or k, so that f is continuous at every point. 37) f(x) = -4, ax + b, 5, x < -4-4 x -3 x > -3 Solve the problem. 38) Complete the statement for the definition of the it: f(x) = L x x0 means that 6

39) Identify the incorrect statements about its. I. The number L is the it of f(x) as x approaches x0 if f(x) gets closer to L as x approaches x0. II. The number L is the it of f(x) as x approaches x0 if, for any > 0, there corresponds a > 0 such that f(x) - L < whenever 0 < x - x0 <. III. The number L is the it of f(x) as x approaches x0 if, given any > 0, there exists a value of x for which f(x) - L <. Use the graph to find a > 0 such that for all x, 0 < x - x0 < f(x) - L <. 40) 4.2 4 3.8 y = x + 3 f(x) = x + 3 x0 = 1 L = 4 = 0.2 0.8 1 1.2 NOT TO SCALE 41) f(x) = x0 = 2 L = 2 x y = x = 1 4 1.66 1.41 1.16 1.3575 2 2.7675 NOT TO SCALE A function f(x), a point x0, the it of f(x) as x approaches x0, and a positive number is given. Find a number > 0 such that for all x, 0 < x - x0 < f(x) - L <. 42) f(x) = 6x + 5, L = 17, x0 = 2, and = 0.01 7

Prove the it statement 43) (3x - 4) = 2 x 2 44) x 7 x 2-49 x - 7 = 14 8

Answer Key Testname: UNTITLED1 1) 1 2) -1 3) x 0.9 0.99 0.999 1.001 1.01 1.1 f(x) 3.439 3.940 3.994 4.006 4.060 4.641 ; it = 4.0 4) - 1 36 5) 1 6 6) x = 4 7) x = 1 8) x = -2, x = 0, x = 2 9) x = 6 10) None 11) a = 2 12) a = 8, -3 13) 43 14) 14 15) -7 16) slope is 13 17) C 18) B 19) does not exist 20) Answers may vary. One possibility: 1 - x2 6 = 1 = 1. According to the squeeze theorem, the function x sin(x) x2, which is squeezed between 1 - and 1, must also approach 1 as x approaches 0. Thus, 2-2 cos(x) 6 x sin(x) 2-2 cos(x) = 1. 21) A 22) C 23) C 24) 18 25) Does not exist 26) 0 27) 1 28) Does not exist 29) 30) x = -6, x = 6 31) B 32) D 33) C 34) B 35) 0 36) Let f(x) = 9x 3-5x 2 + 10x + 10 and let y0 = 0. f(-1) = -14 and f(0) = 10. Since f is continuous on [-1, 0] and since y0 = 0 is between f(-1) and f(0), by the Intermediate Value Theorem, there exists a c in the interval (-1, 0) with the property that f(c) = 0. Such a c is a solution to the equation 9x 3-5x 2 + 10x + 10 = 0. 9

Answer Key Testname: UNTITLED1 37) a = 9, b = 32 38) if given any number > 0, there exists a number > 0, such that for all x, 0 < x - x0 < implies f(x) - L <. 39) I and III 40) 0.2 41) 0.6425 42) 0.001667 43) Let > 0 be given. Choose = /3. Then 0 < x - 2 < implies that (3x - 4) - 2 = 3x - 6 = 3(x - 2) = 3 x - 2 < 3 = Thus, 0 < x - 2 < implies that (3x - 4) - 2 < 44) Let > 0 be given. Choose =. Then 0 < x - 7 < implies that x 2-49 x - 7-14 = (x - 7)(x + 7) x - 7-14 = (x + 7) - 14 for x 7 = x - 7 < = Thus, 0 < x - 7 < implies that x2-49 x - 7-14 < 10