Calculus 2nd Edition Rogawski

24 CHAPTER 1 PRECALCULUS REVIEW
Exercises
In Exercises 1–12, determine the domain of the function.
1. f (x) = x 1/4 2. g(t) = t 2/3
3. f (x) = x 3 + 3x − 4 4. h(z) = z 3 + z −3
5. g(t) = 1
6. f (x) = 1
t + 2
x 2 + 4
7. G(u) = 1
√ x
u 2 8. f (x) =
− 4
x 2 − 9
( )
9. f (x) = x −4 + (x − 1) −3 s
10. F (s) = sin
s + 1
11. g(y) = 10 √ y+y −1 12. f (x) =
x + x −1
(x − 3)(x + 4)
In Exercises 13–24, identify each of the following functions as polynomial,
rational, algebraic, or transcendental.
13. f (x) = 4x 3 + 9x 2 − 8 14. f (x) = x −4
15. f (x) = √ x 16. f (x) = √ 1 − x 2
17. f (x) =
x 2
x + sin x
18. f (x) = 2 x
19. f (x) = 2x3 + 3x
3x − 9x−1/2
9 − 7x 2 20. f (x) =
9 − 7x 2
21. f (x) = sin(x 2 ) 22. f (x) =
x
√ x + 1
23. f (x) = x 2 + 3x −1 24. f (x) = sin(3 x )
25. Is f (x) = 2 x2 a transcendental function?
26. Show that f (x) = x 2 + 3x −1 and g(x) = 3x 3 − 9x + x −2 are rational
functions—that is, quotients of polynomials.
In Exercises 27–34, calculate the composite functions f ◦ g and g ◦ f ,
and determine their domains.
27. f (x) = √ x, g(x) = x + 1
28. f (x) = 1 , g(x) = x−4
x
29. f (x) = 2 x , g(x) = x 2
30. f (x) =|x|, g(θ) = sin θ
31. f(θ) = cos θ, g(x) = x 3 + x 2
32. f (x) = 1
x 2 + 1 ,
g(x) = x−2
33. f (t) = 1 √ t
, g(t) = −t 2
34. f (t) = √ t, g(t) = 1 − t 3
35. The population (in millions) of a country as a function of time t
(years) is P (t) = 30 · 2 0.1t . Show that the population doubles every 10
years. Show more generally that for any positive constants a and k, the
function g(t) = a2 kt doubles after 1/k years.
x + 1
36. Find all values of c such that f (x) =
x 2 has domain R.
+ 2cx + 4
Further Insights and Challenges
In Exercises 37–43, we define the first difference δf of a function f (x)
by δf (x) = f (x + 1) − f (x).
37. Show that if f (x) = x 2 , then δf (x) = 2x + 1. Calculate δf for
f (x) = x and f (x) = x 3 .
38. Show that δ(10 x ) = 9 · 10 x and, more generally, that δ(b x ) =
(b − 1)b x .
39. Show that for any two functions f and g, δ(f + g) = δf + δg and
δ(cf ) = cδ(f ), where c is any constant.
40. Suppose we can find a function P (x) such that δP = (x + 1) k and
P(0) = 0. Prove that P(1) = 1 k , P(2) = 1 k + 2 k , and, more generally,
for every whole number n,
41. First show that
P (n) = 1 k + 2 k +···+n k 1
P (x) =
x(x + 1)
2
satisfies δP = (x + 1). Then apply Exercise 40 to conclude that
1 + 2 + 3 +···+n =
n(n + 1)
2
42. Calculate δ(x 3 ), δ(x 2 ), and δ(x). Then find a polynomial P (x)
of degree 3 such that δP = (x + 1) 2 and P(0) = 0. Conclude that
P (n) = 1 2 + 2 2 +···+n 2 .
43. This exercise combined with Exercise 40 shows that for all whole
numbers k, there exists a polynomial P (x) satisfying Eq. (1). The solution
requires the Binomial Theorem and proof by induction (see Appendix
C).
(a) Show that δ(x k+1 ) = (k + 1)x k +···, where the dots indicate
terms involving smaller powers of x.
(b) Show by induction that there exists a polynomial of degree k + 1
with leading coefficient 1/(k + 1):
P (x) = 1
k + 1 xk+1 +···
such that δP = (x + 1) k and P(0) = 0.