Homework 6

# Homework 6 - vu(tv2894 – Homework 6(Quest – miner...

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Unformatted text preview: vu (tv2894) – Homework 6 (Quest) – miner – (55096) This print-out should have 5 questions. Multiple-choice questions may continue on the next column or page – ﬁnd all choices before answering. 001 10.0 points 10 9 8 3. 7 6 5 4 3 2 1 0 -1 -2 -3 -4 1 8 6 4 2 −2 −2 246 8 If f is a function having 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 8 6 4 2 −2 −2 2 4 6 8 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 as its graph, which of the following could be the graph of f ′ ? 8 1. 7 9 9 -4-3-2-10 1 2 3 4 5 6 7 8 9 8 4. 8 7 6 6 5 4 4 3 2 2 1 0 -1 2468 -2 −2 −2 -3 -4 10 9 -4-3-2-10 1 2 3 4 5 6 7 8 9 8 5. 8 7 6 6 5 4 4 3 2 2 1 0 -1 2468 -2 −2 −2 -3 -4 -4-3-2-10 1 2 3 4 5 6 7 8 9 8 6 4 2 −2 −2 2 46 8 6 5 4 3 2 1 0 -1 -2 -3 -4 Explanation: The slope of the tangent to the graph is zero at x = 3, is positive to the left of x = 3 and is negative to the right of x = 3. In addition, the slope of the tangent is decreasing as x increases, so the graph of f ′ is 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -4-3-2-10 1 2 3 4 5 6 7 8 9 9 8 2. 7 6 5 4 3 2 1 0 -1 -2 -3 -4 8 6 4 2 −2 −2 2 4 6 8 8 6 4 2 −2 −2 2 46 8 correct -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 -4-3-2-10 1 2 3 4 5 6 7 8 9 002 10.0 points vu (tv2894) – Homework 6 (Quest) – miner – (55096) If f is a function deﬁned on (−2, 2) whose graph is 2 1 2 1 −2 −1 −1 −2 −2 −1 −1 −2 1 2 5. 2 1 −2 −1 −1 −2 2 1 1 2 −2 −1 −1 −2 1 2 1 2 1 2 2 4. which of the following is the graph of its derivative f ′ ? 1. 2 1 −2 −1 −1 −2 2. 2 1 −2 −1 −1 −2 3. 2 1 −2 −1 −1 −2 1 2 1 2 correct 6. Explanation: Since the graph on (−2, 2) consists of straight lines joined at x = −1 and x = 1, the derivative of f will exist at all points in (−2, 2) except x = −1 and x = 1, eliminating the answer whose graph contain ﬁlled dots at x = −1 and x = 1. On the other hand, the graph of f (i) has slope −1 on (−2, −1), (ii) has slope 1 on (−1, 1), and (iii) has slope −1 on (1, 2). Consequently, the graph of f ′ consists of the horizontal lines and ‘holes’ in vu (tv2894) – Homework 6 (Quest) – miner – (55096) 2 1 3 −2 −1 −1 −2 1 2 The graph shows that f has a removable discontinuity at x = −3 and a jump discontiinuity at x = 3, so f will not be diﬀerentiable at these points. On the other hand, at x = 1 the graph is continuous but has a ‘corner’, so it will not be diﬀerentiable at this point also. Thus, on (−6, 6) the function f will fail to be diﬀerentiable at the points x = −3, 1, 3 . 004 10.0 points Below is the graph of a function f . keywords: graph, derivative, piecewise linear function, graph of derivative from graph of function, discontinuous derivative 003 10 9 8 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 7 6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -7 -8 6 4 2 −6 −4 −2 −2 −4 −6 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 10.0 points Below is the graph of a function f . 8 6 4 2 −6 −4 −2 −2 −4 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 6 2 4 6 Use this graph to determine all the values of x on (−7, 7) at which f is continuous but not diﬀerentiable. 1. x = −1, 3 2. x = 5 correct 3. x = −1, 3, 5 4. x = −1, 5 5. x = 3, 5 Explanation: Since f is not deﬁned at x = −1, it is not continuous at x = −1. Thus the only possible Use the graph to determine all the values of x in (−6, 6) at which f is not diﬀerentiable. 1. x = −3, 1 2. x = −3, 3 3. x = −3, 1, 3 correct 4. x = −3 5. x = 1, 3 Explanation: vu (tv2894) – Homework 6 (Quest) – miner – (55096) values for x are x = 3 and x = 5. However, f is not continuous at x = 3 because the left and right hand limits exist but do not coincide at x = 3. On the other hand, f is continuous at x = 5, but it is not diﬀerentiable at x = 5. Consequently, f is continuous but not diﬀerentiable only at x = 5. 4 6. s : v: a: 005 10.0 points The ﬁgure below shows the graphs of three functions of time t: Explanation: Experience tells us that the car is (i) moving forwards when its velocity is positive, (ii) moving backwards when its velocity is negative, (iii) speeding up when its velocity is positive and increasing, i.e., when both velocity and acceleration are positive, (iv) slowing down when its velocity is positive but decreasing, i.e., when its velocity is positive but its acceleration is negative. Now ds dv v= a= , dt dt so we need to look at the slope of the tangent line to three graphs to determine which graph is that of position, that of velocity and that of acceleration. Inspection of the graphs thus shows that s: v: a: . t One is the graph of the position function s of a car, one is its velocity v , and one is its acceleration a. Identify which graph goes with which function. 1. s : 2. s : 3. s : 4. s : 5. s : v: v: v: v: v: a: a: a: a: a: correct ...
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