Practice 1 Solutions

For problems 1–10, use the functions: $f (x) = x^{2} - 9 g (x) = x + 3 h (x) = \frac{1}{2 x - 6}$

Find the sum, difference, product, or quotient of the named functions. Determine the domain for each function.

$(\frac{f}{g}) (x)$

$(\frac{f}{g}) (x) = \frac{f (x)}{g (x)}, g (x) \neq 0 (\frac{f}{g}) (x) = \frac{x^{2} - 9}{x + 3}, x \neq - 3 (\frac{f}{g}) (x) = \frac{(x + 3) (x - 3)}{x + 3} (\frac{f}{g}) (x) = \frac{(x + 3) (x - 3)}{x + 3}$

${Domain}_{f} : \{x | x \in ℝ\} D {omain}_{g} : \{x | x \in ℝ\} D {omain}_{\frac{f}{g}} : \{x | x \in ℝ, x \neq - 3\}$

$(\frac{f}{g}) (x) = x - 3 D {omain}_{\frac{f}{g}} : \{x | x \in ℝ, x \neq - 3\}$

$(f h) (x)$

$(f h) (x) = f (x) \cdot h (x) (f h) (x) = (x^{2} - 9) (\frac{1}{2 x - 6}), x \neq 3 (f h) (x) = \frac{x^{2} - 9}{2 x - 6} (f h) (x) = \frac{(x + 3) (x - 3)}{2 (x - 3)} (f h) (x) = \frac{(x + 3) (x - 3)}{2 (x - 3)} (f h) (x) = \frac{x + 3}{2}$

$D {omain}_{f} : \{x | x \in ℝ\} D {omain}_{h} : \{x | x \in ℝ, x \neq 3\} D {omain}_{f h} : \{x | x \in ℝ, x \neq 3\}$

$(f h) (x) = \frac{x + 3}{2} D {omain}_{f h} : \{x | x \in ℝ, x \neq 3\}$

$(g + h) (x)$

$(g + h) (x) = g (x) + h (x) (g + h) (x) = x + 3 + \frac{1}{2 x - 6}, x \neq 3 L C D : 2 (x - 3) (g + h) (x) = \frac{2 (x - 3) (x + 3)}{2 (x - 3)} + \frac{1}{2 (x - 3)} (g + h) (x) = \frac{2 (x^{2} - 9) + 1}{2 (x - 3)} (g + h) (x) = \frac{2 x^{2} - 17}{2 (x - 3)}$

$D {omain}_{g} : \{x | x \in ℝ\} D {omain}_{h} : \{x | x \in ℝ, x \neq 3\} D {omain}_{g + h} : \{x | x \in ℝ, x \neq 3\}$

$(g + h) (x) = \frac{2 x^{2} - 17}{2 (x - 3)} D {omain}_{g + h} : \{x | x \in ℝ, x \neq 3\}$

$(f - g) (x)$

$(f - g) (x) = f (x) - g (x) (f - g) (x) = x^{2} - 9 - (x + 3) (f - g) (x) = x^{2} - x - 12$

${Domain}_{f} : {x | x \in ℝ} {Domain}_{g} : {x | x \in ℝ} {Domain}_{f - g} : {x | x \in ℝ}$

$(f - g) (x) = x^{2} - x - 12 D {omain}_{f - g} : \{x | x \in ℝ\}$

Evaluate.

$(g - h) (2)$

$(g - h) (2) = g (2) - h (2) g (2) = 2 + 3 = 5 h (2) = \frac{1}{2 (2) - 6} = - \frac{1}{2} (g - h) (2) = 5 - \frac{1}{2}$

$(g - h) (2) = \frac{9}{2}$

$f (2 x - 1)$

$f (2 x - 1) = {(2 x - 1)}^{2} - 9 f (2 x - 1) = 4 x^{2} - 4 x + 1 - 9$

$f (2 x - 1) = 4 x^{2} - 4 x - 8$

$(\frac{f h}{g}) (0)$

$(\frac{f h}{g}) (0) = \frac{f (0) \cdot h (0)}{g (0)} (\frac{f h}{g}) (0) = \frac{(0^{2} - 9) (0 + 3)}{2 (0) - 6} (\frac{f h}{g}) (0) = \frac{(- 9) (3)}{- 6} (\frac{f h}{g}) (0) = \frac{- 27}{- 6}$

$(\frac{f h}{g}) (0) = \frac{9}{2}$

$g (\frac{1}{4 x})$

$g (\frac{1}{4 x}) = \frac{1}{4 x} + 3, x \neq 0 LCD : 4 x g (\frac{1}{4 x}) = \frac{1}{4 x} + \frac{12 x}{4 x}$

$g (\frac{1}{4 x}) = \frac{12 x + 1}{4 x}, x \neq 0$

$h (x^{2} + 1)$

$\begin{array}{rcl} h (x^{2} + 1) & = & \frac{1}{2 (x^{2} + 1) - 6} \\ = & \frac{1}{2 x^{2} + 2 - 6} \\ = & \frac{1}{2 x^{2} - 4} = \frac{1}{2 (x^{2} - 2)} \\ x^{2} & = & 2 \\ x & = & \pm \sqrt{2} \end{array}$

$h (x^{2} + 1) = \frac{1}{2 (x^{2} - 2)}, x \neq \pm \sqrt{2}$

Note

Q: When there is a variable in the denominator, what do you need to include as part of your answer?

A: The excluded values

Remember to include all excluded values when taking the square root of a number.

$(g - f) (- 1)$

$(g - f) (- 1) = g (- 1) - f (- 1) f (- 1) = {(- 1)}^{2} - 9 = - 8 g (- 1) = - 1 + 3 = 2 (g - f) (- 1) = 2 - (- 8)$

$(g - f) (- 1) = 10$

For problems 11–14, determine the value of a using the functions: $f (x) = a x^{2} + 4 x + 5 g (x) = \frac{1}{a x - 1}$

$f (- 2) = 25$

$a {(- 2)}^{2} + 4 (- 2) + 5 = 25 4 a - 8 + 5 = 25 4 a = 28$

$a = 7$

$g (\frac{2}{3}) = 1$

$\frac{1}{a (\frac{2}{3}) - 1} = 1 1 = 1 (\frac{2}{3} a - 1) 2 = \frac{2}{3} a$

$a = 3$

$f (1) = - 1$

$a {(1)}^{2} + 4 (1) + 5 = - 1 a + 9 = - 1$

$a = - 10$

$g (- 3) = 3$

$\frac{1}{a (- 3) - 1} = 3 1 = 3 (- 3 a - 1) 1 = - 9 a - 3 4 = - 9 a$

$a = - \frac{4}{9}$

For problems 15–20, evaluate using the graph.

$(\frac{f}{g}) (- 4)$

$(\frac{f}{g}) (- 4) = \frac{f (- 4)}{g (- 4)} (\frac{f}{g}) (- 4) = \frac{1}{4}$

$\frac{1}{4}$

Note

Remember the scale of the graph is one unless otherwise indicated.

$(f - g) (5)$

$(f - g) (5) = f (5) - g (5) (f - g) (5) = 10 - 1$

$(f g) (0)$

$(f g) (0) = f (0) \cdot g (0) (f g) (0) = 2 \cdot 2$

$(\frac{g}{f}) (- 8)$

$(\frac{g}{f}) (- 8) = \frac{g (- 8)}{f (- 8)} (\frac{g}{f}) (- 8) = \frac{6}{0}$

undefined

$(g + f) (- 8)$

$(g + f) (- 8) = g (- 8) + f (- 8) (g + f) (- 8) = 6 + 0$

$(g f) (6)$

$(g f) (6) = g (6) \cdot f (6) (g f) (6) = 0 \cdot 10$