Page 288 - Calculus Demystified

Chapter 1
(d) θ =−2π/3 radians 275
2
(e) θ = π /180 radians
(f) θ = 157π/9000 radians
2 2
2
2
17. (a) f ◦ g(x) =[(x − 1) ] + 2[(x − 1) ]+ 3; g ◦ f(x) = ([x + 2x +
2
3]− 1) .
# #
3 √
3
2
2
(b) f ◦ g(x) = x − 2 + 1; g ◦ f(x) = [ x + 1] − 2.
2
2
2
(c) f ◦ g(x) = sin([cos(x − x)]+ 3[cos(x − x)] ); g ◦ f(x) =
2
2 2
cos([sin(x + 3x )] −[sin(x + 3x )]).
(d) f ◦ g(x) = e ln(x−5)+2 ; g ◦ f(x) = ln(e x+2 − 5).
2
2
2
2
(e) f ◦g(x) = sin([ln(x −x)] +[ln(x −x)]); g◦f(x) = ln([sin(x +
2
2
x)] −[sin(x + x)]).
2
x
(f) f ◦ g(x) = e [e −x 2 ] 2 ; g ◦ f(x) = e −[e ] 2 .
(g) f ◦g(x) =[(2x −3)(x +4)]·[(2x −3)(x +4)+1]·[(2x −3)(x +
4)+2]; g◦f(x) = (2[(x(x+1)(x+2)]−3)([(x(x+1)(x+2)]+4).
18. (a) f is invertible, with f −1 (t) = (t − 5) 1/3 .
(b) g is not invertible since g(0) = g(1) = 0.
2
(c) h is invertible, with h −1 (t) = sgn t · t .
(d) f is invertible, with f −1 (t) = (t − 8) 1/5 .
(e) g is invertible, with g −1 (t) =−[ln t]/3.
√
(f) h is not invertible, since sin π/4 = sin 9π/4 = 2/2.
(g) f is not invertible, since tan π/4 = tan 9π/4 = 1.
√
(h) g is invertible, with g −1 (x) = sgn x · |x|.
19. We will do (a), (c), (e), and (g).

Fig. S1.19(a)