Solução_Calculo_Stewart_6e

F.
156 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
5. (a) Since f 0 (x) > 0 on (1, 5), f is increasing on this interval. Since f 0 (x) < 0 on (0, 1) and (5, 6), f is decreasing on these
intervals.
(b) Since f 0 (x) =0at x =1and f 0 changes from negative to positive there, f changes from decreasing to increasing and has
a local minimum at x =1.Sincef 0 (x) =0at x =5and f 0 changes from positive to negative there, f changes from
increasing to decreasing and has a local maximum at x =5.
7. There is an inflection point at x =1because f 00 (x) changes from negative to positive there, and so the graph of f changes
from concave downward to concave upward. There is an inflection point at x =7because f 00 (x) changes from positive to
negative there, and so the graph of f changes from concave upward to concave downward.
9. (a) f(x) =2x 3 +3x 2 − 36x ⇒ f 0 (x) =6x 2 +6x − 36 = 6(x 2 + x − 6) = 6(x +3)(x − 2).
We don’t need to include the “6”inthecharttodeterminethesignoff 0 (x).
Interval x +3 x − 2 f 0 (x) f
x− 1 2 ,and
f 00 (x) < 0 ⇔ x