Solução_Calculo_Stewart_6e

F.
190 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
25.
From the graph of f(x) =sin(x +sin3x) in the viewing rectangle [0,π] by [−1.2, 1.2], it looks like f has two maxima
and two minima. If we calculate and graph f 0 (x) =[cos(x +sin3x)] (1 + 3 cos 3x) on [0, 2π], we see that
the graph of f 0 appears to be almost tangent to the x-axis at about x =0.7. The graph of
f 00 = − [sin(x +sin3x)] (1 + 3 cos 3x) 2 +cos(x +sin3x)(−9sin3x) is even more interesting near this x-value:
it seems to just touch the x-axis.
If we zoom in on this place on the graph of f 00 ,weseethatf 00 actually does cross the axis twice near x =0.65,
indicating a change in concavity for a very short interval. If we look at the graph of f 0 onthesameinterval,weseethatit
changes sign three times near x =0.65, indicating that what we had thought was a broad extremum at about x =0.7 actually
consists of three extrema (two maxima and a minimum). These maximum values are roughly f(0.59) = 1 and f(0.68) = 1,
and the minimum value is roughly f(0.64) = 0.99996. There are also a maximum value of about f(1.96) = 1 and minimum
values of about f(1.46) = 0.49 and f(2.73) = −0.51. The points of inflection on (0,π) are about (0.61, 0.99998),
(0.66, 0.99998), (1.17, 0.72), (1.75, 0.77), and(2.28, 0.34). On(π, 2π), they are about (4.01, −0.34), (4.54, −0.77),
(5.11, −0.72), (5.62, −0.99998),and(5.67, −0.99998). TherearealsoIPat(0, 0) and (π, 0). Note that the function is odd
and periodic with period 2π, and it is also rotationally symmetric about all points of the form ((2n +1)π, 0), n an integer.
27. f(x) =x 4 + cx 2 = x 2 x 2 + c . Note that f is an even function. For c ≥ 0, the only x-intercept is the point (0, 0). We
calculate f 0 (x) =4x 3 +2cx =4x x 2 + 1 c ⇒ f 00 (x) =12x 2 +2c. Ifc ≥ 0, x =0is the only critical point and there
2
is no inflection point. As we can see from the examples, there is no change in the basic shape of the graph for c ≥ 0;itmerely