Question 2.8.2: Using Operations on Functions and Determining Domains Let ƒ(...
Using Operations on Functions and Determining Domains
Let ƒ(x) = 8x – 9 and g(x) = \sqrt{2x – 1}. Find each function in (a)–(d).
(a) (ƒ + g)(x) (b) (ƒ – g)(x) (c) (ƒg)(x) (d) (\frac{ƒ}{g})(x)
(e) Give the domains of the functions in parts (a)–(d).
(a) (ƒ + g)(x) (b) (ƒ – g)(x)
= ƒ(x) + g(x) = ƒ(x) – g(x)
= 8x – 9 + \sqrt{2x – 1} = 8x – 9 – \sqrt{2x – 1}
(c) (ƒg)(x) (d) (\frac{ƒ}{g})(x)
= ƒ(x) • g(x) =\frac{ƒ(x)}{g(x)}
= (8x – 9)\sqrt{2x – 1} = \frac{8x – 9}{\sqrt{2x – 1} }
(e) To find the domains of the functions in parts (a)–(d), we first find the domains of ƒ and g.
The domain of ƒ is the set of all real numbers (-∞, ∞).
Because g is defined by a square root radical, the radicand must be nonnegative (that is, greater than or equal to 0).
g(x) = \sqrt{2x – 1} Rule for g(x)
2x – 1 ≥ 0 2x – 1 must be nonnegative.
x ≥ \frac{1}{2} Add 1 and divide by 2.
Thus, the domain of g is [\frac{1}{2} , ∞).
The domains of ƒ + g, ƒ – g, and ƒg are the intersection of the domains of ƒ and g, which is
\begin{matrix}(-∞, ∞) ∩ [\frac{1}{2} , ∞) = [\frac{1}{2} , ∞). & \text{The intersection of two sets}\\ & \text{is the set of all elements} \\ & \text{common to both sets.} \end{matrix}
The domain of \frac{ƒ}{ g} includes those real numbers in the intersection above for which g(x) = \sqrt{2x – 1}≠ 0—that is, the domain of \frac{ƒ}{ g} \text{is} (\frac{1}{2}, ∞).