Applied Calculus

Question

Let f(x) = 1/x and g(x) = sin x. Find formulas for the following functions and state their domains.

(a) f g

(b) g f

(c) f f

(d) g g

Answer 1

NOTE: I am assuming "f g" means "f [ g(x) ]"

f(x) = 1/x, g(x) = sin(x)

(A) f [g (x)] = 1/g(x) = 1/sin(x)

Remember that sin(x) has a range of [-1, 1] and 1/x has a domain of (−∞,0)U(0,∞). This means that the domain of our composite function f•g is [-1,0)U(0,1].

(B) g [f (x)] = sin[f (x)] = sin(1/x)

Remember that 1/x has a range of (−∞,0)U(0,∞) and sin(x) has a domain of (−∞,∞). This means that the domain of our composite function g•f is (−∞,0)U(0,∞).

(C) f [f (x)] = 1/f(x) = 1/(1/x) = x

Our composite function is a line and it's domain would be all real numbers, (−∞,∞).

(D) g [g (x)] = sin[g (x)] = sin[sin(x)]

Remember that sin(x) has a range of [-1, 1] and sin(x) has a domain of (−∞,∞). This means that the domain of our composite function g•g is [-1, 1].

Answer 2

It looks like your functions are incomplete. Is it possible that, e.g., when you say "f g" you really mean "f(g)"?

If so, you can solve this by applying the f function to sin(x). The formula would be:

f(g(x)) = f(sin(x)) = 1/sin(x).

Recall that the domain of sin(x) is -1 to 1. You can use that knowledge to determine the range of 1/sin(x). Be careful about the discontinuities at sin(x) = 0.

You can use the same technique with the other functions.