Question

1. For a function f(x) (not shown), we have the graphs of f' (x), f ''(x) and f '''(x). Use these graphs to answer the following questions. The difference between a level 1 and 2 pass will be in the number of correct answers and your attention to detail.

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1a. Identify all intervals, approximately, where the original function, f(x) is increasing

1b. Identify all intervals, approximately, where the original function f(x) is decreasing.

1c. Identify all intervals, approximately, where the original function f(x) is concave up

1d. Identify all intervals, approximately, where the original function f(x) is concave down

2. Identify all x-coordinates of extrema (mins/maxes/etc) of the following function: f(x) = x³ − 9x² + 24x

You do NOT need to classify them as local or global.

3. Briefly explain the process of using Calculus to find global maxes and global minimums. Justify the steps in your process by providing sketches of functions that demonstrate all of the significantly different possibilities for locations of global maxes/mins.

4. Suppose g(x) = 4x³ − 3x + 2. On the interval [−2, 2], use calculus to identify x-coordinate of each local / global minimum / maximum value of g(x).

5. The graph below gives G' (t) which is the rate, in questions per hour, that Sarah graded for a recent Celebration of Learning. Use it to answer the following question:

After which hour will Sarah have graded the fewest questions? Use Calculus to briefly defend your answer.

Answer 1

1. For a function for the graph of f'wa) is shown from graph of f'wc we find that graph is always above ac-axis or we can say that f'x) >0 for all values of x on the graph (-5.5) (a) f(x) is increasing on the abbroximate interval (-5,5)) (6) flac) is decreasing [NONE (c) Information of graph f" (ac) shown here f"C0 = 0 at 3c = fabbroximately) -4,-1.6,0,1.6, 4 f" (c) <o on the intervals (-5,-4) U(-1.6,0) 0 (1.634) f (x) > 0 on the intervals (-4,-1.6) U (0,1.6) 0 (4,5) Number line for f"(x) Down Up Down Up Down up. -5 Negative 4 positive to negative positives Negative positives fax) is concare up on the intervals (-4,-1.6) (0.1.6) U (4,5) (d) fox) is concave down on the intervals (-5,-4) 0 (-1.60) U (1.6,4) 2. fac) = xc3-gxc² +242 The critical number of fox) is f'(x) = 322-182 +24 = 0 = 3672-68+8)=0 → 32-62+8=0 = 32 - 23-4x+8 =0 = x(x-2) - 4(x-2) = (x-2)(x-4)=0 - x=2,4 critical numbers x = 2,4 Second derivative test flac) =0 a f (C) = 6x-18 0 f" (2) = 6(2)-18 = -blo f"(4) = 664)-18=670 Thonofore fis maximum at a= 2 and minimum at 3=4