Math 121 โ Section 2.1 Solutions
19. The relation {(2,6),(โ3,6),(4,9),(2,10)}does not represent a function since the input 2 gives two
different outputs 6,10.
27. y=x2defines yas a function of x
29. y=1
xdefines yas a function of x
30. y=|x|defines yas a function of x
35. y= 2x2โ3x+ 4 defines yas a function of x
47. The domain of f(x) = โ5x+ 4 is all real numbers.
49. The domain of f(x) = x
x2+ 1 is all real numbers . Note that there are no real roots of the denomi-
nator.
51. The domain of f(x) = x
x2โ16 is all real numbers except x=ยฑ4 . Using interval notation, the
domain is (โโ,โ4) โช(โ4,4) โช(4,โ).
55. For the function h(x) = โ3xโ12 we need:
3xโ12 โฅ0
3xโฅ12
xโฅ4
The domain is xโฅ4 . Using interval notation, the domain is [4,โ).
61. For the functions f(x) = 3x+ 4 and g(x) = 2xโ3 we have:
(a) (f+g)(x) = (3x+ 4) + (2xโ3) = 5x+ 1
(b) (fโg)(x) = (3x+ 4) โ(2xโ3) = x+ 7
(c) (fg)(x) = (3x+ 4)(2xโ3) = 6x2โxโ12
(d) ๎f
g๎(x) = 3x+ 4
2xโ3
(e) (f+g)(3) = 5(3) + 1 = 16
(f) (fโg)(4) = 4 + 7 = 11
(g) (fg)(2) = 6(2)2โ2โ12 = 10
(h) ๎f
g๎(1) = 3(1) + 4
2(1) โ3=โ7
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