Section 3.2 Working with Derivatives

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1 Section 3.2 Working with Derivatives Problem (a) If f 0 (2) exists, then (i) lim f(x) must exist, but lim f(x) 6= f(2) (ii) lim f(x) =f(2). (iii) lim f(x) =f 0 (2) (iv) lim f(x) need not exist. The correct answer is (ii): sin x (b) Assuming that lim =, we can conclude x!0 x (i) 0 0 = f 0 (2) exists () f di erentiable at x =2 =) f continuous at x =2 () lim f(x) =f(2) (ii) the tangent line to y =sinx at (0, 0) has slope. (iii) you can cancel the x s. (iv) for all x near 0, sin x = x. (v) for all x near 0, sin x x. Statement (ii) is true: This problem has two correct answers: (ii) and (v). f 0 sin(h + 0) sin(0) (0) = lim h!0 h sin(h) h!0 h = =) slope of tangent line at (0, 0) is Statement (v) is true: When x is near 0 the tangent line y = x is a good approximation to f. Problem 2 Define the function f by f(x) =x /3 and consider the graph of this function:

2 Which of the following two statements are true: (a) The graph of f has a tangent line at x =0. This statement is true! The function f has a vertical tangent at x =0: (b) The derivative f 0 (0) is defined. This statement is false! f(0 + h) f(0) h /3 lim = lim h!0 + h h!0 + h = lim h!0 + h 2 3 = because the limit is of the form: pos 0 + =) f 0 (0) is undefined 2

3 Problem 3 Suppose we are given the graph of a function f: (a) Use this graph to find the following: (Assume all values will be integers or + or ) (i) all x where f(x) =0, f(x) is zero when the function crosses the x-axis. Therefore f(x) =0when x = 5,x=, and x =5. (ii) all x where f(x) > 0, f(x) is positive when the graph of the function is above the x-axis. Therefore f(x) > 0 on ( 5, ) [ (5, ). (iii) all x where f(x) < 0, and f(x) is negative when the graph of the function is below the x-axis. Therefore f(x) < 0 on (, 5) [ (, 5). (iv) all x where f(x) attains a local maximum and all x where f attains a local minimum. f(x) has a local maximum at x = 3. f(x) has a local minimum at x =2. Without sketching the graph of f 0 find (b) (i) all x where f 0 (x) =0, f 0 (x) is zero when the tangent line has a slope of zero, which is approximately at x = 3 and x =2. Note, for this question, these are the same answers as the (local) highest and lowest point for the graph of f. (ii) all x where f 0 (x) > 0, f 0 (x) is positive when the slope of the tangent line is positive. Observe that f is increasing on (, 3), (2, ) and this same set of intervals is where the tangent lines have positive slope. Therefore f 0 (x) > 0 on (, 3), (2, ). 3

4 (iii) all x where f 0 (x) < 0, and f 0 (x) is negative when the slope of the tangent line is negative. Observe that f is decreasing on ( 3, 2) and this interval is where the tangent lines have negative slope. Therefore f 0 (x) < 0 on ( 3, 2). (iv) On the following intervals, does f 0 (x) seem to be increasing or decreasing? i. (, 0) ii. (0, ) (c) Sketch a graph of f 0. decreasing increasing The graph of f 0 is approximately Problem 4 Use the graph of g 4

5 (a) Find the values of t in (0, 4) at which g is not continuous. g is not continuous at t =. (b) Find the values of t in (0, 4) at which g is not di erentiable. g is not di erentiable at t =and t =2. Problem 5 Given the following graph of a function h sketch a graph of the derivative h 0. The graph of the derivative is in red. Important Note: Despite being drawn on the same graph, the units for f and f 0 are not the same! 5

6 Problem 6 (a) Fill in the blanks if the limit exists. if the limit exists. f 0 (x)??????? h f 0 (x) = lim h!0 f(x + h) f(x) h (b) Let f(x) = x +4. Use the (limit) definition of derivative in (a) to find f 0 (x). f 0 f(x + h) f(x) (x) = lim h!0 h h!0 x+4 (x+h+4) (x+h+4)(x+4) h h!0 x+h+4 h!0 h h (x + h + 4)(x + 4) h!0 (x + h + 4)(x + 4) = (x + 4) 2 h x+4 6

7 Problem 7 Let f(x) = 5 x. (a) For a<5, findf 0 (a). Recall that f 0 f(x) (a) = lim x!a x f(a). a When a<5, 0 < 5 a and so f(a) = 5 a =5 a. Since we are considering the limit as x approaches a, (and therefore interested in values of x really close to a), we may also assume that x<5 and therefore f(x) =5 x. Thus (b) For a>5, findf 0 (a). f 0 f(x) f(a) (a) = lim (5 x) (5 a) (x a) =. x!a When a>5, 0 > 5 a and so f(a) = 5 a = (5 a) =a 5. Since we are considering the limit as x approaches a, we may also assume that x>5 and therefore f(x) =x 5. Thus (c) Determine whether f 0 (5) exists. f 0 f(x) f(a) (a) = lim (x 5) (a 5) x a =. x!a If f 0 f(x) f(5) (5) exists then lim exists. But when x is close to 5 it may be that x>5 x!5 x 5 or x<5. So in order to find this limit, we have to check whether the two one-sided limits are equal. f(x) f(5) f(x) f(5) This means, lim = lim x!5 x 5 x!5 + x 5 f(x) f(5) lim x!5 x 5 f(x) f(5) lim x!5 + x 5 f(x) f(5) lim x!5 x 5 = lim x!5 (x 5) (0) x 5 (x 5) (0) = lim = x!5 + x 5 6= lim x!5 + f(x) f(5) x 5 = 7

8 Therefore, f 0 (5) does not exist. 8 if a>5 >< f 0 (a) = undefined if a =5 >: if a<5 (d) Sketch a graph of the function f(x) and its derivative f 0 (x) 8

9 Problem 8 An object is moving along a horizontal line. Let s (t) be the position function of the object. The graph of s (t) is given in the figure below. Using the graph, answer the questions below. Here s is in meters, and t in seconds. y y = s ( t ) t NOTE : The point (2, -) is on the graph. (a) Find s (0). Solution : s (0) = (b) At what time (or times) is the object at the origin? Solution : s () = 0 and s (4) = 0, so at t = and t = 4. (c) At what time (or times) is the object farthest from the origin? Solution : At t = 0, and t = 2. (d) Find the average velocity, v AV, of the object during the time interval [0, 2]. s (2) - s (0) Solution : v AV = = - - = -2 m / sec In problems (e) - (g) remember : instantaneous velocity = the slope of the tangent line. So compare velocities, by comparing the slopes of corresponding tangent lines. (e) Compare instantaneous velocities : circle the bigger of the two given choices. Solution : i v (2.5), v (3.5) ii v (0.2), v (0.8) iii v (.3), v (.8).

10 2 MIDTERM - MORE PROBLEMS.nb f Circle the interval or intervals where the instantaneous veolcity, is positive. v (t), (0, ), (, 2), (2, 4). (g) Circle the interval or intervals where v (t), the instantaneous veolcity, is increasing. (0, ), (, 2), (2, 4).