Solução_Calculo_Stewart_6e

F.
184 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
4.6 Graphing with Calculus and Calculators
1. f(x) =4x 4 − 32x 3 +89x 2 − 95x +29 ⇒ f 0 (x) =16x 3 − 96x 2 + 178x − 95 ⇒ f 00 (x) =48x 2 − 192x +178.
f(x) =0 ⇔ x ≈ 0.5, 1.60; f 0 (x) =0 ⇔ x ≈ 0.92, 2.5, 2.58 and f 00 (x) =0 ⇔ x ≈ 1.46, 2.54.
From the graphs of f 0 ,weestimatethatf 0 < 0 and that f is decreasing on (−∞, 0.92) and (2.5, 2.58),andthatf 0 > 0 and f
is increasing on (0.92, 2.5) and (2.58, ∞) with local minimum values f(0.92) ≈−5.12 and f(2.58) ≈ 3.998 and local
maximum value f(2.5) = 4. The graphs of f 0 make it clear that f has a maximum and a minimum near x =2.5,shownmore
clearly in the fourth graph.
From the graph of f 00 ,weestimatethatf 00 > 0 and that f is CU on
(−∞, 1.46) and (2.54, ∞),andthatf 00 < 0 and f is CD on (1.46, 2.54).
There are inflection points at about (1.46, −1.40) and (2.54, 3.999).
3. f(x) =x 6 − 10x 5 − 400x 4 +2500x 3 ⇒ f 0 (x) =6x 5 − 50x 4 − 1600x 3 + 7500x 2 ⇒
f 00 (x) =30x 4 − 200x 3 − 4800x 2 +1500x