Test 16 (Lessons 31–32): Polynomial Functions Solutions

For problems 1–4, find each function. Name any restriction(s) on the domain.

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

$(f \cdot g) (x)$

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

$[f \circ g] (x)$

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

$[g \circ f] (x)$

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

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

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

When a stone is dropped into water the circular ripples create a wave that increases at a rate of 3 inches per second, 3t. If $A (r) = \frac{22}{7} r^{2}$ , what will the area of the wave be in the water after 7 seconds? (Hint: Determine $[A \circ r] (t)$ )

$\begin{array}{rcl} [A \circ r] (t) & = & A [r (t)] \\ = & \frac{22}{7} {(3 t)}^{2} \\ = & \frac{22}{7} (9 t^{2}) \end{array}$

$\begin{array}{rcl} A [r (7)] & = & \frac{22}{7} (9) {(7)}^{2} \\ = & 22 (9) (7) \\ = & 1386 \end{array}$

The area of the wave after 7 seconds is 1386 square inches.

For problems 6–8, use the graph and the functions $a (x) = | x - 2 | - 3$ and $b (x) = x^{3} - 1$ to evaluate.

Note

The points marked are those that students may mark to help determine answers. It is not required that students mark the graph.

$(a - b) (2)$

$\begin{array}{rcl} (a - b) (2) & = & a (2) - b (2) \\ = & - 3 - 7 \\ = & - 10 \end{array}$

$(a + b) (- 3)$

$\begin{array}{rcl} (a + b) (- 3) & = & a (- 3) + b (- 3) \\ = & 2 + ({(- 3)}^{3} - 1) \\ = & 2 + - 28 \\ = & - 26 \end{array}$

Note

Recall that the equation for b(x) can be determined using skills learned in previous lessons.

$[b \circ a] (1)$

$\begin{array}{rcl} [b \circ a] (1) & = & b [a (1)] \\ = & b (- 2) \\ = & - 9 \end{array}$

Determine which pair of functions are inverses.
$h (x) = \frac{1}{x + 2} - 3 j (x) = - \frac{3}{x} - 2 k (x) = \frac{1}{x + 3} - 2$

$\begin{array}{rcl} [h \circ j] (x) & = & \frac{1}{(- \frac{3}{x} - 2) + 2} - 3 \\ = & \frac{1}{- \frac{3}{x}} - 3 \\ = & - \frac{x}{3} - 3 \\ [h \circ j] (x) & \neq & x, not inverses \end{array}$

$\begin{array}{rcl} [h \circ k] (x) & = & \frac{1}{(\frac{1}{x + 3} - 2) + 2} - 3 \\ = & \frac{1}{\frac{1}{x + 3}} - 3 \\ = & x + 3 - 3 \\ = & x \end{array}$

$\begin{array}{rcl} [k \circ h] (x) & = & \frac{1}{(\frac{1}{x + 2} - 3) + 3} - 2 \\ = & \frac{1}{\frac{1}{x + 2}} - 2 \\ = & x + 2 - 2 \\ = & x \end{array}$

The functions h and k are inverses.

Determine c and d when $[d \circ c] = r (x)$ .
$r (x) = \sqrt[3]{{(x - 1)}^{2}} + 5$

$\begin{matrix} Option 1 c (x) = {(x - 1)}^{2} d (x) = \sqrt[3]{x} + 5 & Option 2 c (x) = \sqrt[3]{{(x - 1)}^{2}} d (x) = x + 5 & Option 3 c (x) = x - 1 d (x) = \sqrt[3]{x^{2}} + 5 \end{matrix}$