Solução_Calculo_Stewart_6e

F.
172 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
(b) Using the Chain Rule and the Quotient Rule we see that f (n) (x) exists for x 6= 0. In fact, we prove by induction that for
each n ≥ 0, there is a polynomial p n and a non-negative integer k n with f (n) (x) =p n (x)f(x)/x kn for x 6= 0.Thisis
true for n =0; suppose it is true for the nth derivative. Then f 0 (x) =f(x)(2/x 3 ),so
which has the desired form.
f (n+1) (x)= x k n
[p 0 n(x) f(x)+p n(x)f 0 (x)] − k nx k n−1 p n(x) f(x) x −2k n
= x k n
p 0 n(x)+p n(x) 2/x 3 − k nx k n−1 p n(x) f(x)x −2k n
= x kn+3 p 0 n(x)+2p n (x) − k n x kn+2 p n (x) f(x)x −(2kn+3)
Now we show by induction that f (n) (0) = 0 for all n.Bypart(a),f 0 (0) = 0. Suppose that f (n) (0) = 0. Then
f (n+1) f (n) (x) − f (n) (0) f (n) (x) p n(x) f(x)/x k n
(0) = lim
= lim =lim
x→0 x − 0
x→0 x x→0 x
f(x)
=limp n (x) lim
x→0 x→0 x = p n(0) · 0=0
kn+1
= lim
x→0
p n(x) f(x)
x k n+1
4.5 Summary of Curve Sketching
1. y = f(x) =x 3 + x = x(x 2 +1) A. f is a polynomial, so D = R.
H.
B. x-intercept =0, y-intercept = f(0) = 0 C. f(−x) =−f(x),sof is
odd; the curve is symmetric about the origin.
D. f is a polynomial, so there is
no asymptote.
E. f 0 (x) =3x 2 +1> 0,sof is increasing on (−∞, ∞).
F. There is no critical number and hence, no local maximum or minimum value.
G. f 00 (x) =6x>0 on (0, ∞) and f 00 (x) < 0 on (−∞, 0),sof is CU on
(0, ∞) and CD on (−∞, 0). Since the concavity changes at x =0,thereisan
inflection point at (0, 0).
3. y = f(x) =2− 15x +9x 2 − x 3 = −(x − 2) x 2 − 7x +1 A. D = R B. y-intercept: f(0) = 2; x-intercepts:
f(x) =0 ⇒ x =2or (by the quadratic formula) x = 7 ± √ 45
2
≈ 0.15, 6.85 C. No symmetry D. No asymptote
E. f 0 (x) =−15 + 18x − 3x 2 = −3(x 2 − 6x +5)
= −3(x − 1)(x − 5) > 0 ⇔ 1