108 D CHAPTER 15
2(cos
2 x - sin
2 x) - cos x = 2(2 cos
2 * - 1) - cos x = 4 cos
2 x - cos x - 2. Since f(x) has a period of 27r, we need
only consider, say, [-ir, ir]. Moreover, since f(-x) = f(x), we only have to consider [0, ir], and then reflect
in the _y-axis. The critical numbers are the solutions in [0, TT] of sin x = 0 or 2 cos x - 1 = 0. The equation
sin x = 0 has the solutions 0, TT. The equation 2 cos x — 1=0 is equivalent to cos x = \, having the
solution x = ir/3. /"(0) = 1>0, so there is a relative minimum at x = 0, y = 0. /"(7r) = 3>0, so there
is a relative minimum at x = TT, y = -2. f(ir/3) = - § < 0, so there is a relative maximum at x = ir/3,
y = 1. There are inflection points between 0 and 7r/3 and between ir/3 and TT; they can be approximated by
using the quadratic formula to solve f"(x) = 4 cos
2 x - cos x - 2 = 0 for cos x, and then using a cosine table to
approximate x. See Fig. 15-24.
Fig. 15-25
15.36
f(jc) = |sinjt:l.
I Since f(x) is even, has a period of TT, and coincides with sin A; on [0, Tr/2], we get the curve of Fig. 15-25.
15.37 f(x) = sin x + x.
I f'(x) = cosx + l. f"(x) = —sinx. The critical numbers are the solutions of cos;t=-l, x = (2n + l)-rr.
The first derivative test yields the case { + , +}, and, therefore, we obtain only inflection points at x = (2n + I)TT,
y = (2n + l)TT. See Fig. 15-26.
15.39
Fig. 15-26
Fig. 15-27
15.38 f(x) - sin x + sin |jc|.
I Case 1. x>0. Then /(*) = 2sin*. Case 2. *<0. Then f(x) = 0 [since sin (-x) = -sin*]. See
Fig. 15-27.
Fie. 15-24