advertisement

Version PREVIEW – HW 03 – hoffman – (57225) This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 List the five slopes in decreasing order. 1. A, B, D, E, C 2. C, A, E, B, D correct CalC2f01a 001 10.0 points 3. C, A, D, B, E Find an expression for the slope of the secant line through the points P (5, h(5)) and Q(x, h(x)) on the graph of y = h(x). h(5) + h(x) 5+x 1. slope = Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is h(x) − h(5) 5+x C, A, E, B, D . h(5) − h(x) 4. slope = x−5 h(x) − h(5) correct x−5 5. slope = 5. D, B, E, A, C 6. A, C, B, E, D h(x) + h(5) 2. slope = 5−x 3. slope = 4. D, C, B, A, E Explanation: Since the slope of the line through the points P (5, h(5)) and Q(x, h(x)) is the ratio of the rise divided by the run, we see that the h(x) − h(5) slope = . x−5 CalC2f05a 003 10.0 points Find an equation for the tangent line at the point P (−1, f (−1)) on the graph of f when f is defined by f (x) = x2 − 3x + 4 . CalC2f03s 002 10.0 points 1. y − 5x − 3 = 0 Consider the slope of the given curve at the five points shown. 2. y − 5x + 3 = 0 B A 3. y + 5x − 3 = 0 correct 4. y + 5x + 5 = 0 C D E 5. y + 5x − 5 = 0 6. y − 5x + 5 = 0 Explanation: Version PREVIEW – HW 03 – hoffman – (57225) The slope of the tangent line P (−1, f (−1)) is the value of the limit at f (−1 + h) − f (−1) . h h→0 lim 2 What is her speed after 9 minutes, and in what direction is she heading at that time? 1. away from RLM at 20 yds/min. Now 2. towards RLM at 25 yds/min. correct f (−1 + h) = (−1 + h)2 − 3(−1 + h) + 4 3. away from RLM at 30 yds/min. 2 = h − 5h + 8 , 4. towards RLM at 30 yds/min. while f (−1) = 8. Thus f (−1 + h) − f (−1) h 5. towards RLM at 20 yds/min. h2 − 5h = h− 5. h As h approaches 0, therefore, = f (−1 + h) − f (−1) → −3 . h Consequently, by the point-slope formula, an equation for the tangent line at P (−1, f (−1)) is y − 8 = −5(x + 1) , i.e., y + 5x − 3 = 0 . 6. away from RLM at 25 yds/min. Explanation: The graph is linear on [8, 10], so the student’s speed at time t = 9 is the (absolute value of the) slope of this line. Hence slope = 100 − 150 = −25 . 10 − 8 The fact that her distance from RLM is decreasing at t = 9 indicates that she is walking towards RLM at that time. CalC2f15c 004 10.0 points A Calculus student leaves the RLM building and walks in a straight line to the PCL Library. Her distance from RLM after t minutes is given by the graph CalC3a00Ex2 005 10.0 points If P (a, f (a)) is the point on the graph of f (x) = 3x2 + 2x + 1 yards 500 at which the tangent line is parallel to the line 400 y = 6x + 2 , 300 determine a. 200 100 2 4 6 8 10 mins 12 1. a = 1 6 2. a = 1 2 Version PREVIEW – HW 03 – hoffman – (57225) f (x) − f (a) x→a x−a f (x + h) − f (x) III. lim x→a h II. lim 2 3. a = correct 3 4. a = 3 1 3 1. II only 5. a = 0 2. I and II only correct Explanation: The slope of the tangent line at the point P (a, f (a)) on the graph of f is the value f (a + h) − f (a) f ′ (a) = lim h→0 h of the derivative of f at x = a. To compute the value of f ′ (a), note that 3. I only 4. I and III only 5. I, II, and III Explanation: Both of f (a + h) = 3(a + h)2 + 2(a + h) + 1 2 f ′ (a) = lim h→0 2 = 3a + h(6a + 2) + 3h + 2a + 1 , f (a + h) − f (a) h and while f ′ (a) = lim 2 f (a) = 3a + 2a + 1 . x→a f (x) − f (a) x−a are valid definitions of f ′ (a). By contrast, Thus f (a + h) − f (a) = h{(6a + 2) + 3h} , in which case f ′ (a) = lim {(6a + 2) + 3h} = 6a + 2 . h→0 lim x→a f (a + h) − f (a) f (x + h) − f (x) = h h because f is continuous. Consequently, f ′ (a) is given only by If the tangent line at P is parallel to the line I and II . y = 6x + 2 , then they have the same slopes, so CalC3a05 007 10.0 points f ′ (a) = 6a + 2 = 6 . If f is a differentiable function, find the value of the limit Consequently, a = 2 . 3 CalC3a01a 006 10.0 points If f is a differentiable function, then f ′ (a) is given by which of the following? I. lim h→0 f (a + h) − f (a) h lim h→0 5f (x + h) − 5f (x − h) . h 1. limit doesn’t exist 2. limit = 5f ′ (x) 3. limit = 10f ′ (x) correct Version PREVIEW – HW 03 – hoffman – (57225) Explanation: By definition 4. limit = 0 f ′ (a) = lim 5. limit = −10f ′ (x) h→0 Explanation: Since f is differentiable, f (x + h) − f (x) f (x) = lim h→0 h ′ f (x) − f (x − h) = lim . h h→0 But then 5f (x + h) − 5f (x − h) h h→0 n f (x + h) − f (x) = 5 lim h h→0 f (x) − f (x − h) o , + h lim and so by the properties of limits, limit = 10f ′ (x). CalC3a20s 008 10.0 points For what function f and number a is the limit √ 4+h−2 lim h h→0 the value of f ′ (a)? 1. f (x) = x2 , a = 1 4 1 , x a = 2 3. f (x) = x, a = 2 2. f (x) = 4. f (x) = x1/2 , a = 2 5. f (x) = x1/2 , a = 4 correct 6. f (x) = 1 , x2 4 a = 4 f (a + h) − f (a) . h When f (a + h) − f (a) = h √ 4+h−2 , h therefore, inspection shows that f (x) = x1/2 , a = 4 .