Practice 1 Solutions

For problems 1–3, use the graph and equations.

$f (x) = 2 \sqrt{x + 4} g (x) = \frac{1}{2} {(x - 1)}^{2}$

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

$\begin{array}{rcl} [f \circ f] (- 4) & = & f [f (- 4)] \\ f (- 4) & = & 0 \\ f (0) & = & 4 \end{array}$

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

Note

The values were determined by reading the graph.

$g [f (- 4)]$

$\begin{array}{rcl} g [f (- 4)] & = & g (0) \\ = & \frac{1}{2} \end{array}$

$g [f (- 4)] = \frac{1}{2}$

Note

Read f(–4) from the graph, and then check g(0) with the equation.

$[f \circ g] (- 3)$

$\begin{array}{rcl} [f \circ g] (- 3) & = & f [g (- 3)] \\ = & f (\frac{1}{2} {(- 3 - 1)}^{2}) \\ = & f (\frac{1}{2} {(- 4)}^{2}) \\ = & f (\frac{1}{2} (16)) \\ = & f (8) \\ = & 2 \sqrt{8 + 4} \\ = & 2 \sqrt{12} \\ = & 4 \sqrt{3} \end{array}$

$[f \circ g] (- 3) \begin{array}{rcl} = \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} \sqrt{3} \end{array}$

Note

g(–3) can be determined from the graph. f(8) needs to be determined algebraically because the function continues past what is viewable (extrapolation).

State the domain and range of $[g \circ f] (x)$ .
$f (x) = \{(- 13.4, 1.56), (8.73, - 9.08), (6.54, - 6.79), (- 23.4, 12.1)\} g (x) = \{(- 6.79, 1.2), (1.56, - 9.67), (- 13.4, 2.43), (12.1, - 10.21)\}$

$g [f (- 13.4)] = g (1.56) = - 9.67 g [f (8.73)] = g (- 9.08) = undefined g [f (6.54)] = g (- 6.79) = 1.2 g [f (- 23.4)] = g (12.1) = - 10.21$

$Domain : \{- 23.4, - 13.4, 6.54\} R ange : \{- 10.21, - 9.67, 1.2\}$

For problems 5–9, use the following functions to determine the given composite function. List all domain restrictions for the composite function.

$k (x) = x^{2} - 3 x + 2 j (x) = 2 x - 5 p (x) = \frac{1}{x - 2}$

$[k \circ j] (x)$

$\begin{array}{rcl} [k \circ j] (x) & = & k [j (x)] \\ = & k (2 x - 5) D {omain}_{j} : \{x | x \in ℝ\} \\ = & {(2 x - 5)}^{2} - 3 (2 x - 5) + 2 \\ = & 4 x^{2} - 20 x + 25 - 6 x + 15 + 2 \\ = & 4 x^{2} - 26 x + 42 \end{array}$

$[k \circ j] (x) = 4 x^{2} - 26 x + 42$

$[j \circ k] (x)$

$\begin{array}{rcl} [j \circ k] (x) & = & j [k (x)] \\ = & j (x^{2} - 3 x + 2) D {omain}_{k} : \{x | x \in ℝ\} \\ = & 2 (x^{2} - 3 x + 2) - 5 \\ = & 2 x^{2} - 6 x + 4 - 5 \end{array}$

$[j \circ k] (x) = 2 x^{2} - 6 x - 1$

$[p \circ j] (x)$

$\begin{array}{rcl} [p \circ j] (x) & = & p [j (x)] \\ = & p (2 x - 5) D {omain}_{p \circ j} : \{x | x \in ℝ, x \neq \frac{7}{2}\} \\ = & \frac{1}{2 x - 5 - 2} \end{array}$

$[p \circ j] (x) = \frac{1}{2 x - 7}, x \neq \frac{7}{2}$

$[p \circ k] (x)$

$\begin{array}{rcl} [p \circ k] (x) & = & p [k (x)] \\ = & p (x^{2} - 3 x + 2) D {omain}_{p \circ k} : \{x | x \in ℝ, x \neq 0, 3\} \\ = & \frac{1}{x^{2} - 3 x + 2 - 2} = \frac{1}{x^{2} - 3 x} \\ = & \frac{1}{x (x - 3)} \end{array}$

$[p \circ k] (x) = \frac{1}{x^{2} - 3 x}, x \neq 0, 3$

Note

The denominator needs to be factored to determine all of the values excluded from the domain.

$[j \circ p] (x)$

$\begin{array}{rcl} [j \circ p] (x) & = & j [p (x)] \\ = & j (\frac{1}{x - 2}) D {omain}_{p} : \{x | x \in ℝ, x \neq 2\} \\ = & 2 (\frac{1}{x - 2}) - 5 \\ = & \frac{2}{x - 2} - 5 \end{array}$

$[j \circ p] (x) = \frac{2}{x - 2} - 5, x \neq 2$

Determine the composite function.

$[g \circ f] (\frac{1}{x}), if f (x) = 3 x and g (x) = x^{2}$

$\begin{array}{rcl} [g \circ f] (\frac{1}{x}) & = & g [f (\frac{1}{x})] \\ = & g [3 (\frac{1}{x})] = g (\frac{3}{x}) \\ = & {(\frac{3}{x})}^{2} \\ = & \frac{9}{x^{2}}, x \neq 0 \end{array}$

$\begin{array}{rcl} [g \circ f] \end{array} (\frac{1}{x}) \begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{9}{x^{2}} \end{array} \begin{array}{rcl} , \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} x \end{array} \begin{array}{rcl} \neq \end{array} \begin{array}{rcl} 0 \end{array}$

$[h \circ k] (2 x - 1), if h (x) = - x^{2} and k (x) = 4 x + 5$

$\begin{array}{rcl} [h \circ k] (2 x - 1) & = & h [k (2 x - 1)] \\ = & h [4 (2 x - 1) + 5] = h [8 x - 4 + 5] \\ = & h (8 x + 1) \\ = & - {(8 x + 1)}^{2} = - (64 x^{2} + 16 x + 1) \\ = & - 64 x^{2} - 16 x - 1 \end{array}$

$[h \circ k] (2 x - 1) = - 64 x^{2} - 16 x - 1$

$[b \circ g] (\frac{4}{c}), if b (x) = \frac{1}{x} and g (x) = \sqrt{x}$

$\begin{array}{rcl} [b \circ g] (\frac{4}{c}) & = & b [g (\frac{4}{c})] \\ = & b [\sqrt{\frac{4}{c}}] = b (\frac{2}{\sqrt{c}}) \\ = & (\frac{1}{\frac{2}{\sqrt{c}}}) = 1 \div \frac{2}{\sqrt{c}} = 1 \cdot \frac{\sqrt{c}}{2} \\ = & \frac{\sqrt{c}}{2} \end{array}$

$[b \circ g] (\frac{4}{c}) = \frac{\sqrt{c}}{2}$

Note

You don’t need to clear the square root from the denominator in the second step because when you divide in the next step, it flips to the numerator.

Determine f and g such that $[f \circ g] (x) = h (x)$ and $h (x) = 2 {(x - 7)}^{3} + 5 .$ Select all that apply.

$f (x) = x - 7 g (x) = 2 x^{3} - 5$
$f (x) = 2 x^{3} + 5 g (x) = x - 7$
$f (x) = 2 x + 5 g (x) = {(x - 7)}^{3}$
$f (x) = {(x - 7)}^{3} g (x) = 2 x + 5$

Note

The first and fourth options have f(x) and g(x) reversed.

Determine j and k such that $[j \circ k] (x) = m (x)$ and $m (x) = \frac{1}{2} | x - 6 | + 3 .$ Select all that apply.

$j (x) = \frac{1}{2} x + 3 k (x) = | x - 6 |$
$j (x) = | x + 3 | k (x) = \frac{1}{2} x - 6$
$j (x) = x - 6 k (x) = \frac{1}{2} | x | + 3$
$j (x) = \frac{1}{2} | x | + 3 k (x) = x - 6$

Note

The second option has the constants reversed.

The third option has j(x) and k(x) reversed.

Determine a and g such that $[a \circ g] (x) = z (x)$ and $z (x) = \sqrt[3]{{(x - 2)}^{2}} - 1 .$ Select all that apply.

$a (x) = \sqrt[3]{x - 1} g (x) = x - 2$
$a (x) = \sqrt[3]{x^{2}} - 1 g (x) = x - 2$
$a (x) = \sqrt[3]{x} - 1 g (x) = {(x - 2)}^{2}$
$a (x) = x - 1 g (x) = \sqrt[3]{{(x - 2)}^{2}}$

Note

The first option does not square $(x - 2)$ and includes –1 under the radical.

The regular price of a new gaming system is x dollars. A store allows customers to apply a $200 off coupon as well as a 15% off coupon. Determine which coupon should be applied first to get the lowest price.

$f (x) = x - 200 g (x) = 0.85 x [f \circ g] (x) = 0.85 x - 200 [g \circ f] (x) = 0.85 (x - 200)$

$[f \circ g] (x) = 0.85 x - 200$ allows for the greatest discount. 15% off coupon should be applied first, then the $200 off coupon.

Note

Use a specific dollar amount for the gaming system if you are struggling to determine the best discount.

For problems 17–20, determine if f(x) and g(x) are inverses.

$f (x) = \sqrt[3]{x - 2} + 5 g (x) = {(x - 5)}^{3} + 2$

$\begin{array}{rcl} [f \circ g] (x) & = & \sqrt[3]{({(x - 5)}^{3} + 2) - 2} + 5 \\ = & \sqrt[3]{{(x - 5)}^{3}} + 5 \\ = & x - 5 + 5 \\ = & x \end{array}$

$\begin{array}{rcl} [g \circ f] (x) & = & {(\sqrt[3]{x - 2} + 5 - 5)}^{3} + 2 \\ = & {(\sqrt[3]{x - 2})}^{3} + 2 \\ = & x - 2 + 2 \\ = & x \end{array}$

f(x) and g(x) are inverses

$f (x) = \frac{1}{2} x + 4 g (x) = 2 x - 4$

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

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

$f (x) = \frac{\sqrt{x - 3}}{4} g (x) = 16 x^{2} + 3, x \geq 0$

$\begin{array}{rcl} [f \circ g] (x) & = & f [g (x)] \\ = & \frac{\sqrt{16 x^{2} + 3 - 3}}{4}, x \geq 0 \\ = & \frac{\sqrt{16 x^{2}}}{4} \\ = & \frac{4 x}{4} \\ = & x \end{array}$

$\begin{array}{rcl} [g \circ f] (x) & = & g [f (x)] \\ = & 16 {(\frac{\sqrt{x - 3}}{4})}^{2} + 3 \\ = & 16 (\frac{x - 3}{16}) + 3 \\ = & x - 3 + 3 \\ = & x \end{array}$

f(x) and g(x) are inverses

$f (x) = \sqrt{x - 1} + 5 g (x) = {(x - 5)}^{2} + 1, x \geq 5$

$\begin{array}{rcl} [f \circ g] (x) & = & \sqrt{{(x - 5)}^{2} + 1 - 1} + 5, x \geq 5 \\ = & \sqrt{{(x - 5)}^{2}} + 5 \\ = & x - 5 + 5 \\ = & x \end{array}$

$\begin{array}{rcl} [g \circ f] (x) & = & {(\sqrt{x - 1} + 5 - 5)}^{2} + 1 \\ = & {(\sqrt{x - 1})}^{2} + 1 \\ = & x - 1 + 1 \\ = & x \end{array}$

f(x) and g(x) are inverses