MATH 2250 Exam 1 Solutions

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1 MATH 2250 Exam 1 Solutions Name Answer every question on the exam there is no penalty for guessing. Calculators and similar aids are not allowed. There are a total of 60 points possible: 20 in Part 1, and 40 in Part True/False questions Mark the following statements true (T) or false (F). In this section (and only in this section) you are not required to justify your answers. Each question in this section is worth 2 points. False 1. If f is a function and x 1 f(x) and x 1 + f(x) both exist, then it must be true that x 1 f(x) exists. (The one-sided its need to be equal in order for the it to exist.) True 2. If f is a function and if x 2 f(x) =, then it must be true that x 2 f(x) does not exist. True 3. If f and g are functions and if x 3 f(x) = 2 and x 3 g(x) = 4 then it must be true that (f(x)g(x)) = 8. x 3 False 4. If f is a function such that x 4 f(x) = 0, and if g is some other function, then it must be true that (f(x)g(x)) = 0. x 4 (This would be true if we knew that x 4 g(x) exists, but as stated it s false. For example, we could have f(x) = x 4 and g(x) = 1 x 4 ). True 5. Let f(x) = x cos x. The Intermediate Value Theorem guarantees that there is a number c such that 0 < c < π 2 and f(c) = 0. (Note that f(0) = 1 and f(1) = π 2, so since 0 is between 1 and π/2 the Intermediate Value Theorem says that f(c) = 0 for some c with 0 < c < π/2.) 1

2 2 True 6. The graph of the function f(x) = 15x2 + 28x x has a slanted asymptote. (Note: You do not need to do any calculations to correctly answer this question.) (The easy way to see this is to note that the degree of the numerator is one larger than the degree of the denominator.) False 7. The graph of the function f(x) = 894x x x x + 93 has a slanted asymptote. (Note: You do not need to do any calculations to correctly answer this question.) (Since the degree of the numerator is equal to the degree of the denominator, there will be a horizontal asymptote, but no slanted one.) False 8. If f is a function which is continuous at x = a, then it must be true that f is differentiable at x = a. (As we discussed in class, the example f(x) = x shows that this is false). True 9. If f is a function which is differentiable at x = a, then it must be true that f is continuous at x = a. False 10. If f and g are continuous at x = 10 then it must be true that f g is continuous at x = 10. (You d need to know that g(10) 0 in order for this to be true. For example the statement is false if f(x) = 1 and g(x) = x 10.)

3 3 2. Extended answer questions For the rest of the exam, you must show your work in order to receive any credit. 1. An object is dropped from a tall building, but air is rapidly blowing upward, so the object falls more slowly than it would otherwise. As a result, the downward distance that the object travels in time t seconds is given, in feet, by d(t) = 16t 2 10t. (a) (4 points) Find the average speed of the object from time t = 1 to time t = 3. and We have so the average speed is d(1) = = 6 d(3) = = 114, d(3) d(1) 3 1 = = 54ft/sec. (b) (8 points) Find the instantaneous speed of the object at t = 1 (i.e., find the derivative d (1)). Note: In order to receive credit, you must directly use the definition of the derivative in terms of its (and not any tricks that you may have learned elsewhere for computing derivatives; we ll discuss these kinds of tricks in class soon, but the point of this problem is to test whether you can use the actual definition). d (1) = h 0 d(1 + h) d(1) h (16(1 + h) 2 10(1 + h)) (16(1) 2 10(1)) = h 0 h ( h + 16h h) 6 = h 0 h 22h + 16h 2 = h 0 h = 22 ft/sec = h 0 ( h)

4 4 2. Evaluate the following its (or write does not exist if they do not exist): (a) (4 points) x 2 + 3x + 2 x + 2 Since (x 2 + 3x + 2) = 2 and (x + 2) = 2, we have (by the it law saying that the it of the quotient is the quotient of the its as long as the it of the denominator is nonzero): x 2 + 3x + 2 = 2 x = 1. (b) (4 points) x 2 + 3x + 2 x 2 x + 2 x 2 + 3x + 2 (x + 1)(x + 2) = x 2 x + 2 x 2 (x + 2) = (x + 1) = 1. x 2 (c) (4 points) 2x 2 + 8x + 1 x x x 2 + 8x + 1 x x 2 1/x /x 2 = 2 + 8/x + 1/x 2 x 1 + 4/x 2 = 2 1 = 2.

5 5 (d) (4 points) x x + 2 x When x > 0 we have x = x. Therefore + x x + 2 x = + x x + 2x = x + 3x = 1 3. On the other hand, when x < 0 we have x = x, and so x x + 2 x = + x x 2x = x + x = 1. Since the left- and right-sided its disagree, the it does not exist. (e) (4 points) sin 2 x (Hint: Remember that sin 2 x + cos 2 x = 1.) sin 2 x = = 1 cos 2 x ()(1 + cos x) = cos x = = 1 2.

6 6 3 (8 points). Where a and b are numbers, let 2x + 3 x < 0 f(x) = x 2 + ax + b 0 x < 3 x x 3. If this function f is continuous at every point, what must be the values of a and b? f(x) = = 3, while f(x) = a 0 + b = b. + If f is continuous at x = 0, then in particular f(x) must exist, so the two one-sided its must agree and we have Also, b = 3. f(x) = a + b = 9 + 3a + b, while f(x) = 3. x 3 x 3 + So if f is continuous at x = 3 we have 9 + 3a + b = 3. Since we already determined that b = 3, solving for a then gives 3a = 15, so a = 5.

Math 106 Answers to Test #1 11 Feb 08

Math 106 Answers to Test #1 11 Feb 08 Math 06 Answers to Test # Feb 08.. A projectile is launched vertically. Its height above the ground is given by y = 9t 6t, where y is the height in feet and t is the time since the launch, in seconds.