Targeted Review Solutions

Determine if the given functions are inverses.
$f (x) = \sqrt[3]{4 x + 2} g (x) = \frac{x^{3} - 2}{4}$

$\begin{matrix} y = \sqrt[3]{4 x + 2} \\ {(x)}^{3} = {(\sqrt[3]{4 y + 2})}^{3} \\ x^{3} = 4 y + 2 \\ x^{3} - 2 = 4 y \\ y = \frac{x^{3} - 2}{4} \end{matrix}$

The functions f and g are inverses because when x and y are switched in f the result is g.

Solve by completing the square.
$5 x^{2} - 10 x + 15 = 0$

$x^{2} - 2 x + 3 = 0 x^{2} - 2 x + = - 3 + x^{2} - 2 x + 1 = - 3 + 1 {(x - 1)}^{2} = - 2 \sqrt{{(x - 1)}^{2}} = \pm \sqrt{- 2} x - 1 = \pm i \sqrt{2}$

$x = 1 \pm i \sqrt{2}$

Write the equation when the solution is $\pm 6 i .$

$\begin{array}{rcl} \begin{matrix} x = - 6 i & x = 6 i \\ (x - (- 6 i)) = 0 & (x - 6 i) = 0 \end{matrix} \\ (x + 6 i) (x - 6 i) & = & 0 \\ x^{2} - 36 (- 1) & = & 0 \end{array}$

$\begin{array}{rcl} x^{2} + 36 & = & 0 \end{array}$

Evaluate.

$\begin{matrix} i = \sqrt{- 1} \\ i^{2} = - 1 \\ i^{3} = - i \\ i^{4} = 1 \end{matrix}$

Name the parent graph and its end behavior.

Cube root, $A s x \to + \infty, f (x) \to + \infty, and as x \to - \infty, f (x) \to - \infty$

Square root, $A s x \to + \infty, f (x) \to + \infty, and as x \to 0, f (x) \to 0$

Quadratic (parabola), $A s x \to + \infty, f (x) \to + \infty, and as x \to - \infty, f (x) \to + \infty$

Linear, $A s x \to + \infty, f (x) \to + \infty, and as x \to - \infty, f (x) \to - \infty$

Multiple Choice

Solve.
${(x - \frac{2}{5})}^{2} = \frac{16}{25}$

$\frac{6}{5}$
$\frac{4}{5}$
$- \frac{2}{5}, \frac{6}{5}$
$- \frac{6}{5}, \frac{2}{5}$

Note

This option does not include both solutions.
This option added the solutions together into one value.
The signs of the solutions are reversed.

$\sqrt{{(x - \frac{2}{5})}^{2}} = \sqrt{\frac{16}{25}} x - \frac{2}{5} = \pm \frac{4}{5} x = \frac{2}{5} \pm \frac{4}{5} x = \frac{2}{5} - \frac{4}{5}, \frac{2}{5} + \frac{4}{5} x = - \frac{2}{5}, \frac{6}{5}$

Name the vertex of the parabola.
$f (x) = x^{2} - 6 x + 13$

(0, 13)
(–3, 4)
(3, 4)
(3, 22)

Note

This option is the y-intercept.
This option has the incorrect sign for the x-coordinate.
This option occurs if 13 is not subtracted from both sides when completing the square.

$x^{2} - 6 x + 13 = 0 x^{2} - 6 x + {(\frac{- 6}{2})}^{2} = - 13 + {(\frac{- 6}{2})}^{2} {(x - 3)}^{2} = - 13 + 9 {(x - 3)}^{2} = - 4 f (x) = {(x - 3)}^{2} + 4 v e r t e x : (3, 4)$

Name the domain and range of the inverse to the function.

$Domain : \{x | x \in ℝ\}, Range : \{y | y \in ℝ\}$
$D omain : \{x | x \in ℝ, x \geq 0\}, R ange : \{y | y \in ℝ, y \geq 3\}$
$D omain : \{x | x \in ℝ, x \geq 3\}, R ange : \{y | y \in ℝ, y \geq 0\}$
$D omain : \{x | x \in ℝ, x \leq 0\}, R ange : \{y | y \in ℝ, y \leq 3\}$

Note

The function is only in the first quadrant, the inverse will also be in Q1. All real numbers include negative values.
This option is the domain and range for the function, not the inverse.
The direction of the inequality symbols does not change for this inverse.

A student graphed a function through $(0, 3)$ and knows that as $x \to + \infty, y \to + \infty$ . Select all true statements about the function.

The graph represents a cube root function.
The graph represents a cubic function.
$As x \to - \infty, y \to - \infty$
$As x \to - \infty, y \to + \infty$

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L19	L24	L23	L15	L17	L17	L17	L17	L13	L27	L21	L18

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics