Practice 2 Solutions

For problems 1–5, solve with the graph and algebraically.

$[f \circ g] (0)$

$\begin{array}{rcl} [f \circ g] (0) & = & f [g (0)] \\ = & f (2) \\ = & 10 \end{array}$

$[f \circ g] (0) = 10$

$[g \circ f] (- 8)$

$\begin{array}{rcl} [g \circ f] (- 8) & = & g [f (- 8)] \\ = & g (0) \\ = & 2 \end{array}$

$[g \circ f] (- 8) = 2$

$[f \circ g] (2)$

$\begin{array}{rcl} [f \circ g] (2) & = & f [g (2)] \\ = & f (2) \\ = & 10 \end{array}$

$[f \circ g] (2) = 10$

$[g \circ f] (- 4)$

$\begin{array}{rcl} [g \circ f] (- 4) & = & g [f (- 4)] \\ = & g (1) \\ = & 2 \end{array}$

$[g \circ f] (- 4) = 2$

$[f \circ g] (- 4)$

$\begin{array}{rcl} [f \circ g] (- 4) & = & f [g (- 4)] \\ = & f (4) \\ = & 10 \end{array}$

$[f \circ g] (- 4) = 10$

For problems 6–10, use the following functions to determine the given composite function. List all domain restrictions for the composite function.

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

$[g \circ g] (x)$

$\begin{array}{rcl} [g \circ g] (x) & = & (x - 2) - 2 {Domain}_{g} : \{x | x \in ℝ\} \\ = & x - 4 \end{array}$

$[g \circ g] (x) = x - 4$

$[g \circ h] (x)$

$\begin{array}{rcl} [g \circ h] (x) & = & g [h (x)] \\ = & g (\frac{1}{2 x + 1}) D omain : \{x | x \in ℝ, x \neq - \frac{1}{2}\} \\ = & \frac{1}{2 x + 1} - 2 \end{array}$

$[g \circ h] (x) = \frac{1}{2 x + 1} - 2, x \neq - \frac{1}{2}$

$[g \circ f] (x)$

$\begin{array}{rcl} [g \circ f] (x) & = & (3 x^{2} - 5) - 2 \\ = & 3 x^{2} - 7 D {omain}_{f} : \{x | x \in ℝ\} \end{array}$

$[g \circ f] (x) = 3 x^{2} - 7$

$[h \circ f] (x)$

$\begin{array}{rcl} [h \circ f] (x) & = & h [f (x)] \\ = & h (3 x^{2} - 5) 6 x^{2} = 9 \\ = & \frac{1}{2 (3 x^{2} - 5) + 1} \sqrt{(x^{2})} = \sqrt{(\frac{3}{2})} \\ = & \frac{1}{6 x^{2} - 9} D {omain}_{h \circ f} : \{x | x \in ℝ, x \neq \frac{\pm 3 \sqrt{2}}{2}\} \end{array}$

$[h \circ f] (x) = \frac{1}{6 x^{2} - 9}, x \neq \frac{\pm 3 \sqrt{2}}{2}$

$[f \circ f] (x)$

$\begin{array}{rcl} [f \circ f] (x) & = & f [f (x)] \\ = & 3 {(3 x^{2} - 5)}^{2} - 5 D omain : \{x | x \in ℝ\} \\ = & 3 (9 x^{4} - 30 x^{2} + 25) - 5 \\ = & 27 x^{4} - 90 x^{2} + 75 - 5 \\ = & 27 x^{4} - 90 x^{2} + 70 \end{array}$

$[f \circ f] (x) = 27 x^{4} - 90 x^{2} + 70$

Determine the composite function.

$[v \circ z] (3 n^{2}), if v (x) = \frac{1}{4} x^{2}, x \geq 0 and z (x) = 2 \sqrt{x}$

$\begin{array}{rcl} [v \circ z] (3 n^{2}) & = & v [z (3 n^{2})] \\ = & v [2 \sqrt{(3 n^{2})}] D {omain}_{z} : \{x | x \in ℝ, x \geq 0\} \\ = & v (2 n \sqrt{3}) \\ = & \frac{1}{4} {(2 n \sqrt{3})}^{2} = \frac{1}{4} (4 n^{2} \cdot 3) \\ = & 3 n^{2} \end{array}$

$[v \circ z] (3 n^{2}) = 3 n^{2}, n \geq 0$

Note

Remember to include the domain restriction in the composition since it is included in the internal function.

Note: v(x) and z(x) are inverses.

$[r \circ m] (5 a), if r (x) = 2 \sqrt[3]{x} + 4 and m (x) = {(x - 2)}^{3}$

$\begin{array}{rcl} [r \circ m] (5 a) & = & r [m (5 a)] \\ = & r [{((5 a) - 2)}^{3}] \\ = & r ({(5 a - 2)}^{3}) \\ = & 2 \sqrt[3]{{(5 a - 2)}^{3}} + 4 = 2 (5 a - 2) + 4 \\ = & 10 a \end{array}$

$[r \circ m] (5 a) = 10 a$

Note

There are no domain restrictions.

$[g \circ h] (7 y), if g (x) = 8 x + 2 and h (x) = \frac{1}{x}$

$\begin{array}{rcl} [g \circ h] (7 y) & = & g [h (7 y)] \\ = & g [\frac{1}{(7 y)}] {Domain}_{h} : \{x | x \in ℝ, x \neq 0\} \\ = & g (\frac{1}{7 y}) \\ = & 8 (\frac{1}{7 y}) + 2 \end{array}$

$[g \circ h] (7 y) = \frac{8}{7 y} + 2, y \neq 0$

Determine f and g such that $[f \circ g] (x) = h (x)$ and $h (x) = \sqrt{5 x^{2} - 7} .$ Select all that apply.

$f (x) = \sqrt{x} g (x) = 5 x^{2} - 7$
$f (x) = \sqrt{x - 7} g (x) = 5 x^{2}$
$f (x) = x g (x) = \sqrt{5 x^{2} - 7}$
$f (x) = \sqrt{5 x - 7} g (x) = x^{2}$

Determine p and r such that $[p \circ r] (x) = w (x)$ and $w (x) = \frac{10 x - 17}{3} .$ Select all that apply.

$p (x) = 10 x - 17 r (x) = \frac{x}{3}$
$p (x) = \frac{x}{3} r (x) = 10 x - 17$
$p (x) = \frac{x}{3} - 17 r (x) = 10 x$
$p (x) = \frac{x - 17}{3} r (x) = 10 x$

Note

The first option reverses p(x) and r(x). The third option did not put –17 in the numerator.

Determine q and c such that $[q \circ c] (x) = h (x)$ and $h (x) = x^{2} - 11 .$ Select all that apply.

$q (x) = \frac{x}{3} - 2 c (x) = 3 x^{2} - 27$
$q (x) = x - 2 c (x) = x^{2} - 9$
$q (x) = x^{2} c (x) = x - 11$
$q (x) = x - 11 c (x) = x^{2}$

Note

The third option has q(x) and c(x) reversed.

Determine if f(x) and g(x) are inverses.

Russell received a 20% off coupon in the mail for a new generator. The store was also having a sale that allowed for 10% off all items. The store determined it was best to use the sale first, then the coupon. Russell asked to use the coupon first, then the sale. Which combination saves Russell the most money?

$f (x) = 0.8 x g (x) = 0.9 x [f \circ g] (x) = 0.8 (0.9 x) = 0.72 x [g \circ f] (x) = 0.9 (0.8 x) = 0.72 x$

$[f \circ g] (x) = [g \circ f] (x)$

Either order in which the coupon and the sale are applied will result in the same outcome.

$f (x) = 8 {(x + 14)}^{3} - 18 g (x) = \frac{\sqrt[3]{x + 18}}{2} - 14$

$\begin{array}{rcl} [f \circ g] (x) & = & f [g (x)] \\ = & 8 {(\frac{\sqrt[3]{x + 18}}{2} - 14 + 14)}^{3} - 18 \\ = & 8 {(\frac{\sqrt[3]{x + 18}}{2})}^{3} - 18 \\ = & 8 (\frac{x + 18}{8}) - 18 \\ = & x + 18 - 18 \\ = & x \end{array}$

$\begin{array}{rcl} [g \circ f] (x) & = & g [f (x)] \\ = & \frac{\sqrt[3]{8 {(x + 14)}^{3} - 18 + 18}}{2} - 14 \\ = & \frac{\sqrt[3]{8 {(x + 14)}^{3}}}{2} - 14 \\ = & \frac{2 (x + 14)}{2} - 14 \\ = & x + 14 - 14 \\ = & x \end{array}$

f(x) and g(x) are inverses

$f (x) = \frac{3}{x - 4} g (x) = \frac{3}{x} + 4$

$\begin{array}{rcl} [f \circ g] (x) & = & f [g (x)] \\ = & \frac{3}{\frac{3}{x} + 4 - 4} \\ = & \frac{3}{\frac{3}{x}} = 3 \div \frac{3}{x} = 3 \cdot \frac{x}{3} \\ = & x \end{array}$

$\begin{array}{rcl} [g \circ f] (x) & = & g [f (x)] \\ = & \frac{3}{\frac{3}{x - 4}} + 4 \\ = & 3 \div \frac{3}{x - 4} + 4 = 3 \cdot \frac{x - 4}{3} + 4 \\ = & x - 4 + 4 \\ = & x \end{array}$

f(x) and g(x) are inverses

$f (x) = \frac{5}{8} x - 7 g (x) = \frac{8}{5} x + 7$

$\begin{array}{rcl} [f \circ g] (x) & = & f [g (x)] \\ = & \frac{5}{8} (\frac{8}{5} x + 7) - 7 \\ = & x + \frac{35}{8} - 7 \\ = & x - \frac{21}{8} \end{array}$

f(x) and g(x) are not inverses