Solução_Calculo_Stewart_6e

F.
222 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
where sin x + C>0 ⇔ sin x>−C ⇔ sin −1 (−C) 0 implies that f is increasing on R, so there is exactly one zero of f, and hence, exactly one real
root of the equation 3x +2cosx +5=0.
47. Since f is continuous on [32, 33] and differentiable on (32, 33), then by the Mean Value Theorem there exists a number c in
(32, 33) such that f 0 (c) = 1 5 c−4/5 =
5√
33 −
5 √ 32
33 − 32
= 5√ 33 − 2,but 1 5 c−4/5 > 0 ⇒ 5√ 33 − 2 > 0 ⇒ 5√ 33 > 2. Also
f 0 is decreasing, so that f 0 (c) f 0 (c) = 5√ 33 − 2 ⇒ 5√ 33 < 2.0125.
Therefore, 2 < 5√ 33 < 2.0125.
49. (a) g(x) =f(x 2 ) ⇒ g 0 (x) =2xf 0 (x 2 ) by the Chain Rule. Since f 0 (x) > 0 for all x 6= 0,wemusthavef 0 (x 2 ) > 0 for
x 6= 0,sog 0 (x) =0 ⇔ x =0.Nowg 0 (x) changes sign (from negative to positive) at x =0, since one of its factors,
f 0 (x 2 ), is positive for all x, and its other factor, 2x, changes from negative to positive at this point, so by the First
Derivative Test, f has a local and absolute minimum at x =0.
(b) g 0 (x) =2xf 0 (x 2 ) ⇒ g 00 (x) =2[xf 00 (x 2 )(2x)+f 0 (x 2 )] = 4x 2 f 00 (x 2 )+2f 0 (x 2 ) by the Product Rule and the Chain
Rule. But x 2 > 0 for all x 6= 0, f 00 (x 2 ) > 0 [since f is CU for x>0], and f 0 (x 2 ) > 0 for all x 6= 0, so since all of its
factors are positive, g 00 (x) > 0 for x 6= 0.Whetherg 00 (0) is positive or 0 doesn’t matter [since the sign of g 00 does not
change there]; g is concave upward on R.
51. If B =0, the line is vertical and the distance from x = − C A to (x 1,y 1 ) is
x 1 + C A = |Ax 1 + By 1 + C|
√ , so assume
A2 + B 2
B 6= 0. The square of the distance from (x 1,y 1) to the line is f(x) =(x − x 1) 2 +(y − y 1) 2 where Ax + By + C =0,so
we minimize f(x) =(x − x 1) 2 + − A B x − C 2
B − y1 ⇒ f 0 (x) =2(x − x 1)+2
− A B x − C
B − y1 − A
.
B
f 0 (x) =0 ⇒ x = B2 x 1 − ABy 1 − AC
and this gives a minimum since f 00 (x) =2
1+ A2
> 0. Substituting
A 2 + B 2 B 2
this value of x into f(x) and simplifying gives f(x) = (Ax 1 + By 1 + C) 2
, so the minimum distance is
A 2 + B 2
f(x) =
|Ax 1 + By 1 + C|
√
A2 + B 2 .