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Name: Class: Date: 8) — F(0 6-4 1 1. a = Sor The values of ¢ which satisfy /"(c) = + seem to be about c = 0.8, 3.2, 4.4, and 6.1L. 1 at it. (b) See the Second Derivative Test and the note that precedes Example 5. 4, (a) See the First Derivative 4. There is an inflection point at « = 1 because ” (r) changes from negative to positive there, and so the graph of f changes from concave downward to concave upward. There is an inflection point atc = 7 because f(r) changes from positive to negative there, and so the graph of f changes from concave upward to concave downward 6. (a) f is increasing on the intervals where f‘(a) > 0, namely, (2,4) and (6,9). (b} f has a local maximum where it changes from increasing to decreasing, that is, where " changes from positive to negative (at = 4). Similarly, where f’ changes from negative to positive, f has a local minimum (at 7 = 2 and at 2 = 6) {c} When /* is increasing, its derivative {” is positive and hence, f is concave upward. This happens on (1. 3), (5,7), and (8,9). Similarly, f is concave downward when J” is decreasing —that is, on (0, 1), (3,5), and (7,8). (d) f has inflection points at» = 1,3, 5, 7, and 8, since the direction of concavity changes at each of these values. 10. (a) flr) = - 7307 f(z) = (P+ 310s) — 2s) ae Ce) _ waa a is determined by the sign of x. Thus, f’(#) > 0 < «> Oand f(x) <0 <= 2 <0. So fis increasing on (0,00) and f is decreasing on (—o0, 0} The denominator is positive so the sign of f"(z) (b) f changes from decreasing to increasing at # = 0. Thus, {(0) = 0 isa local minimum value. ©) f"e) = (x? +.3)7(6) — 6x 20? + 3)Qx) — 6(a? +9) fe? +3—40"] 63-307) —18(e 4+ Ife — 1) : [(2? + 9)°)7 (a? 4 3)1 (a? + 3)* (a? + 3) f(2)>0 @ -Lea< land f"(#) <0 < 2 <—lora# > L Thus, f is concave upward on (—1, 1) and concave downward on (—co, —1) and (1, oc). There are inflection points at (41, 4). PAGE 1 
