Practice 2 Solutions

For problems 1–10, use the functions: $f (x) = \frac{1}{x - 5} g (x) = x^{3} - 10 x^{2} + 25 x h (x) = x^{2} - 4 x - 5$

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

$(f g) (x)$

$\begin{array}{rcl} (f g) (x) & = & f (x) \cdot g (x) \\ (f g) (x) & = & \frac{1}{x - 5} \cdot (x^{3} - 10 x^{2} + 25 x), x \neq 5 \\ (f g) (x) & = & \frac{x^{3} - 10 x^{2} + 25 x}{x - 5} \\ (f g) (x) & = & \frac{x (x^{2} - 10 x + 25)}{x - 5} \\ (f g) (x) & = & \frac{x (x - 5) (x - 5)}{x - 5} \\ (f g) (x) & = & \frac{x (x - 5) (x - 5)}{x - 5}, x \neq 5 \end{array}$

${Domain}_{f} : {x | x \in ℝ, x \neq 5} {Domain}_{g} : {x | x \in ℝ} {Domain}_{f g} : {x | x \in ℝ, x \neq 5}$

$(f g) (x) = x (x - 5) D {omain}_{f g} : \{x | x \in ℝ, x \neq 5\}$

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

$\begin{array}{rcl} (\frac{h}{g}) (x) & = & \frac{h (x)}{g (x)}, g (x) \neq 0 \\ (\frac{h}{g}) (x) & = & \frac{x^{2} - 4 x - 5}{x^{3} - 10 x^{2} + 25 x}, x \neq 0, 5 \\ (\frac{h}{g}) (x) & = & \frac{(x - 5) (x + 1)}{x (x - 5) (x - 5)} \\ (\frac{h}{g}) (x) & = & \frac{(x - 5) (x + 1)}{x (x - 5) (x - 5)}, x \neq 0, 5 \end{array}$

$D {omain}_{h} : \{x | x \in ℝ\} D {omain}_{g} : \{x | x \in ℝ\} D {omain}_{\frac{h}{g}} : \{x | x \in ℝ, x \neq 0, 5\}$

$(\frac{h}{g}) (x) = \frac{x + 1}{x (x - 5)} D {omain}_{\frac{h}{g}} : \{x | x \in ℝ, x \neq 0, 5\}$

$(h - g) (x)$

$(h - g) (x) = h (x) - g (x) (h - g) (x) = (x^{2} - 4 x - 5) - (x^{3} - 10 x^{2} + 25 x) (h - g) (x) = x^{2} - 4 x - 5 - x^{3} + 10 x^{2} - 25 x$

$D {omain}_{g} : \{x | x \in ℝ\} D {omain}_{h} : \{x | x \in ℝ\} D {omain}_{h - g} : \{x | x \in ℝ\}$

$(h - g) (x) = - x^{3} + 11 x^{2} - 29 x - 5 D {omain}_{h - g} : \{x | x \in ℝ\}$

Find $f (x^{2} - 4)$

$f (x^{2} - 4) = \frac{1}{x^{2} - 4 - 5} f (x^{2} - 4) = \frac{1}{x^{2} - 9} x^{2} = 9 x = \pm \sqrt{3}$

$D {omain}_{f} : \{x | x \in ℝ\} D {omain}_{f (x^{2} - 4)} : \{x | x \in ℝ, x \neq \pm \sqrt{3}\}$

$f (x^{2} - 4) = \frac{1}{x^{2} - 9} D {omain}_{f (x^{2} - 4)} : \{x | x \in ℝ, x \neq \pm \sqrt{3}\}$

Evaluate.

$h (x - 2)$

$h (x - 2) = {(x - 2)}^{2} - 4 (x - 2) - 5 h (x - 2) = x^{2} - 4 x + 4 - 4 x + 8 - 5$

$h (x - 2) = x^{2} - 8 x + 7$

$(f + g) (- 1)$

$(f + g) (- 1) = f (- 1) + g (- 1) (f + g) (- 1) = \frac{1}{- 1 - 5} + {(- 1)}^{3} - 10 {(- 1)}^{2} + 25 (- 1) (f + g) (- 1) = - \frac{1}{6} - 1 - 10 - 25$

$(f + g) (- 1) = - 36 \frac{1}{6}$

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

$(\frac{f g}{h}) (x) = \frac{f (x) \cdot g (x)}{h (x)} (\frac{f g}{h}) (x) = \frac{\frac{1}{x - 5} \cdot (x^{3} - 10 x^{2} + 25 x)}{x^{2} - 4 x - 5}, x \neq - 1, 5 (\frac{f g}{h}) (x) = \frac{\frac{1}{x - 5} \cdot (x (x - 5) (x - 5))}{(x - 5) (x + 1)} (\frac{f g}{h}) (x) = \frac{\frac{1}{x - 5} \cdot (x (x - 5) (x - 5))}{(x - 5) (x + 1)}$

$(\frac{f g}{h}) (x) = \frac{x}{x + 1}, x \neq - 1, 5$

$(g - h) (- 2)$

$(g - h) (- 2) = g (- 2) - h (- 2) (g - h) (- 2) = {(- 2)}^{3} - 10 {(- 2)}^{2} + 25 (- 2) - ({(- 2)}^{2} - 4 (- 2) - 5) (g - h) (- 2) = - 8 - 40 - 50 - (4 + 8 - 5)$

$(g - h) (- 2) = - 105$

$(f + h) (0)$

$(f + h) (0) = f (0) + h (0) (f + h) (0) = \frac{1}{0 - 5} + {(0)}^{2} - 4 (0) - 5 (f + h) (0) = - \frac{1}{5} - 5$

$(f + h) (0) = - 5 \frac{1}{5}$

$h (x^{2})$

$h (x^{2}) = {(x^{2})}^{2} - 4 (x^{2}) - 5$

$h (x^{2}) = x^{4} - 4 x^{2} - 5$

For problems 11–14, determine the value of a using the functions: $f (x) = a x^{2} - 10 g (x) = \frac{3}{a x + 2}$

$f (\sqrt{5}) = 0$

$a {(\sqrt{5})}^{2} - 10 = 0 5 a = 10$

$a = 2$

$g (- 2) = - \frac{3}{4}$

$\frac{3}{a (- 2) + 2} = - \frac{3}{4} \frac{3}{- 2 a + 2} = - \frac{3}{4} - 3 (- 2 a + 2) = 12 - 2 a + 2 = - 4 - 2 a = - 6$

$a = 3$

$f (- 1) = - 13$

$a {(- 1)}^{2} - 10 = - 13 a - 10 = - 13$

$a = - 3$

$g (\frac{1}{2}) = \frac{3}{4}$

$\frac{3}{a (\frac{1}{2}) + 2} = \frac{3}{4} \frac{3}{\frac{1}{2} a + 2} = \frac{3}{4} 3 (\frac{1}{2} a + 2) = 12 \frac{1}{2} a + 2 = 4 \frac{1}{2} a = 2$

$a = 4$

For problems 15–20, evaluate using the graph.

Note

If you want to review these problems using technology, use the equations: $f (x) = x^{2} - 6 x + 8 and g (x) = 2 x + 1$ .

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

$(\frac{f}{g}) (4) = \frac{f (4)}{g (4)} (\frac{f}{g}) (4) = \frac{0}{9}$

$(g - f) (0)$

$(g - f) (0) = g (0) - f (0) (g - f) (0) = 1 - 8$

–7

$(f g) (- 1)$

$(f g) (- 1) = f (- 1) \cdot g (- 1) (f g) (- 1) = 15 \cdot - 1$

–15

$(f + g) (3)$

$(f + g) (3) = f (3) + g (3) (f + g) (3) = - 1 + 7$

$(f - g) (7)$

$(f - g) (7) = f (7) - g (7) (f - g) (7) = 15 - 15$

$(\frac{g}{f}) (2)$

$(\frac{g}{f}) (2) = \frac{g (2)}{f (2)} (\frac{g}{f}) (2) = \frac{5}{0}$

Undefined