Targeted Review Solutions

Simplify.

Note

Problems 1–2

Remember, only terms with identical radicals (degree and radicand) can be combined.

$\sqrt{- 12} + \sqrt{- 75}$

$i \sqrt{2^{2} \cdot 3} + i \sqrt{3 \cdot 5^{2}} 2 i \sqrt{3} + 5 i \sqrt{3}$

$7 i \sqrt{3}$

$\sqrt{40} - \frac{2}{5 \sqrt{6}}$

$\sqrt{2^{3} \cdot 5} - \frac{2}{5 \sqrt{6}} (\frac{\sqrt{6}}{\sqrt{6}}) 2 \sqrt{10} - \frac{2 \sqrt{6}}{5 \cdot 6} 2 \sqrt{10} - \frac{2 \sqrt{6}}{30}$

$2 \sqrt{5} - \frac{\sqrt{6}}{15}$

Rewrite the expression so all terms are in the numerator. Assume all bases are positive.

$\frac{x^{5} y^{4} z^{- 9}}{x^{- 6} y^{13} z^{- 2}}$

$x^{5 - (- 6)} y^{4 - 13} z^{- 9 - (- 2)}$

$x^{11} y^{- 9} z^{- 7}$

${(\frac{3 x^{3} y}{x^{2} y^{5}})}^{- 2}$

${(3 x^{3 - 2} y^{1 - 5})}^{- 2} {(3 x y^{- 4})}^{- 2}$

$3^{- 2} x^{- 2} y^{8}$

Describe the transformation from f(x) to g(x).
$f (x) = {(3)}^{x}, g (x) = - 5 {(3)}^{x + 8} - 2$

g(x) is reflected, vertically stretched, and translated left 8 units and down two units.

Sketch the graph using technology: $g (x) = \frac{1}{2} {(4)}^{x + 1} - 3$

x	y
–1	–2.5
0	-1
1	5

Solve.

$36^{(\frac{1}{2} x)} = 218^{(2 x + 5)}$

$\begin{array}{rcl} 36 = 6^{2} 218 = 6^{3} \\ 6^{2} (^{\frac{1}{2} x}) & = & {6^{3}}^{(2 x + 5)} \\ 6^{x} & = & 6^{(6 x + 15)} \\ x & = & 6 x + 15 \\ - 5 x & = & 15 \end{array}$

x = –3

${(\frac{1}{4})}^{x} < {(\frac{1}{8})}^{x - 2}$

$\begin{array}{rcl} (\frac{1}{4}) & = & 2^{- 2} (\frac{1}{8}) = 2^{- 3} \\ 2^{- 2 (x)} & < & 2^{- 3 (x - 2)} \\ - 2 (x) & < & - 3 (x - 2) \\ - 2 x & < & - 3 x + 6 \end{array}$

x < 6

Multiple Choice

Name the value of b, and if it represents growth or decay for the exponential equation: $y = 6 {(3)}^{- x}$

6, growth
3, growth
$\frac{1}{3},$ decay
$\frac{1}{6},$ decay

$a = 6 b = 3^{- 1} = \frac{1}{3} growth : b > 1 decay : 0 < b < 1$

Note

This option the the a value.
This option is the answer if the negative exponent is ignored.
This option is the reciprocal of the a value.

Find $[g \circ f] (x)$ when $f (x) = 3 x^{2} - 2$ and $g (x) = - x + 6 .$

$- 3 x^{2} + 8$
$3 x^{2} - x + 4$
$3 x^{2} - 36 x + 106$
$- 3 x^{2} + 4$

$[g \circ f] (x) = g [f (x)] [g \circ f] (x) = - (3 x^{2} - 2) + 6 [g \circ f] (x) = - 3 x^{2} + 8$

Note

This expression is the answer to $f (x) + g (x) .$
This expression is the answer to $f [g (x)] .$
This expression did not distribute the negative to all terms in parentheses.

Name the domain and range for the inverse of the function: $g (x) = \frac{3}{x} + 2$

$domain : \{x | x \in ℝ\} range : \{y | y \in ℝ\}$
$domain : \{x | x \in ℝ, x \neq 0\} range : \{y | y \in ℝ, y \neq 2\}$
$domain : \{x | x \in ℝ, x \neq 2\} range : \{y | y \in ℝ, y \neq 0\}$
$domain : \{x | x \in ℝ, x \neq 2\} range : \{y | y \in ℝ\}$

$\begin{array}{rcl} x & = & \frac{3}{y} + 2 \\ x - 2 & = & \frac{3}{y} \\ y (x - 2) & = & 3 \\ y & = & \frac{3}{x - 2} \end{array}$

Note

This option cannot be a domain and range for a reciprocal function. 
This option is the domain and range for the given function, not the inverse.
This option does not have the correct horizontal asymptote.

Write an exponential equation that passes through the points $(3, 128)$ and $(- 1, \frac{1}{32})$ .

$y = 2^{x}$
$y = \frac{1}{2^{11}} {(64)}^{x}$
$y = 8 {(\frac{1}{4})}^{x}$
$y = \frac{1}{4} {(8)}^{x}$

Note

This incorrectly combines a and b before substituting the values into the exponential equation
This is the result if you take the square root instead of the 4th root of b
The values of a and b are switched

$\begin{matrix} y = a b^{x} \\ 128 = a b^{3} & \frac{1}{32} = (\frac{128}{b^{3}}) b^{- 1} & a = \frac{128}{8^{3}} \\ a = \frac{128}{b^{3}} & \frac{1}{32} = \frac{128}{b^{4}} & a = \frac{128}{512} \\ b^{4} = (32) (128) & a = \frac{1}{4} \\ \sqrt[4]{b^{4}} = \sqrt[4]{4096} \\ b = 8 & y = \frac{1}{4} {(8)}^{x} \end{matrix}$

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	11	12	–	–	37	37	38	38	37	32	19	38

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