Question 2.8.3: Evaluating Combinations of Functions If possible, use the gi...
Evaluating Combinations of Functions
If possible, use the given representations of functions ƒ and g to evaluate
(ƒ + g)(4), (ƒ – g)(-2), (ƒg)(1), and ( \frac{ƒ}{g} )(0).
(a) 
(b)
| x | ƒ(x) | g(x) |
| -2 | -3 | undefined |
| 0 | 1 | 0 |
| 1 | 3 | 1 |
| 4 | 9 | 2 |
(c) ƒ(x) = 2x + 1, g(x) = \sqrt{x}
(a) From the figure, ƒ(4) = 9 and g(4) = 2.
(ƒ + g)(4)
= ƒ(4) + g(4) (ƒ + g)(x)= ƒ(x) + g(x)
= 9 + 2 Substitute.
= 11 Add.
For (ƒ – g)(-2), although ƒ(-2) = -3, g(-2) is undefined because -2 is not in the domain of g. Thus (ƒ – g)(-2) is undefined.
he domains of ƒ and g both include 1.
(ƒg)(1)
= ƒ(1) • g(1) (ƒg)(x) = ƒ(x) • g(x)
= 3 • 1 Substitute.
= 3 Multiply.
The graph of g includes the origin, so g(0) = 0. Thus ( \frac{ƒ}{g} )(0) is undefined.
(b) From the table, ƒ(4) = 9 and g(4) = 2.
(ƒ + g)(4)
= ƒ(4) + g(4) (ƒ + g)(x) = ƒ(x) + g(x)
= 9 + 2 Substitute.
= 11 Add.
In the table, g(-2) is undefined, and thus (ƒ – g)(-2) is also undefined.
(ƒg)(1)
= ƒ(1) • g(1) (fg)(x) = ƒ(x) • g(x)
= 3 • 1 ƒ(1) = 3 and g(1) = 1
= 3 Multiply.
The quotient function value ( \frac{ƒ}{g} )(0) is undefined because the denominator, g(0), equals 0.
(c) Using ƒ(x) = 2x + 1 and g(x) =\sqrt{x}, we can find (ƒ + g)(4) and (ƒg)(1). Because -2 is not in the domain of g, (ƒ – g)(-2) is not defined.
\left.\begin{matrix}(ƒ + g)(4) \\ \\ = ƒ(4) + g(4)\\ \\ = (2 • 4 + 1) + \sqrt{4}\\ \\= 9 + 2 \\ \\ = 11\end{matrix} \right| \begin{matrix} (ƒg)(1) \\ \\ = ƒ(1) • g(1) \\ \\ = (2 • 1 + 1) • \sqrt{1}\\ \\ = 3(1)\\ \\ = 3 \end{matrix}( \frac{ƒ}{g} )(0) is undefined since g(0) = 0.