Question 2.8.7: Determining Composite Functions and Their Domains Given that...
Determining Composite Functions and Their Domains
Given that ƒ(x) = \frac{6}{x – 3} \text{and} g(x) = \frac{1}{x} , find each of the following.
(a) (ƒ ∘ g)(x) and its domain (b) (g ∘ ƒ)(x)and its domain
(a) (ƒ ∘ g)(x)
= ƒ(g(x)) By definition
= ƒ(\frac{1}{x}) g(x) =\frac{1}{x}
= \frac{6}{\frac{1}{x} – 3} ƒ(x) = \frac{6}{x – 3}
= \frac{6x}{1 – 3x} Multiply the numerator
and denominator by x.
The domain of g is the set of all real numbers except 0, which makes g(x) undefined. The domain of ƒ is the set of all real numbers except 3. The expression for g(x), therefore, cannot equal 3. We determine the value that makes g(x) = 3 and exclude it from the domain of ƒ ∘ g.
\frac{1}{x} = 3 The solution must be excluded.
1 = 3x Multiply by x.
x =\frac{1}{3} Divide by 3.
Therefore, the domain of ƒ ∘ g is the set of all real numbers except 0 and \frac{1}{3}, written in interval notation as
(-∞, 0) ∪ (0, \frac{1}{3}) ∪(\frac{1}{3}, ∞).
(b) (g ∘ ƒ)(x)
= g(ƒ(x)) By definition
= g ( \frac{6}{x – 3} ) ƒ(x) = \frac{6}{x – 3}
= \frac{1}{\frac{6}{x – 3}} Note that this is meaningless if x = 3; g(x) =\frac{1}{x} .
= \frac{x – 3}{6} \frac{1}{\frac{a}{b}} = 1÷\frac{ a}{b} = 1 • \frac{b}{a} = \frac{b}{a}
The domain of ƒ is the set of all real numbers except 3. The domain of g is the set of all real numbers except 0. The expression for ƒ(x), which is \frac{6}{x – 3} , is never zero because the numerator is the nonzero number 6. The domain of g ∘ ƒ is the set of all real numbers except 3, written (-∞, 3) ∪ (3, ∞).