Practice 2 Solutions

Write the inverse of the given function.

$f (x) = {(\frac{3}{8})}^{- x}$

${(\frac{3}{8})}^{- x} = {(\frac{8}{3})}^{x}$

$f^{- 1} (x) = \log_{\frac{8}{3}} x$

$f (x) = e^{x} + 3$

$y = e^{x} + 3 x = e^{y} + 3 x - 3 = e^{y} \ln (x - 3) = \ln e^{y} y = \ln (x - 3)$

$f^{- 1} (x) = \ln (x - 3)$

$f (x) = 4^{x + 1}$

$y = 4^{x + 1} x = 4^{y + 1} \log_{4} x = \log_{4} 4^{y + 1} \log_{4} x = y + 1 y = \log_{4} x - 1$

$f^{- 1} (x) = \log_{4} x - 1$

$f (x) = 6^{x}$

$f^{- 1} (x) = \log_{6} x$

$r (x) = e^{x - 7}$

$y = e^{x - 7} x = e^{y - 7} \ln x = \ln e^{y - 7} \ln x = y - 7 y = \ln x + 7$

$r^{- 1} (x) = \ln x + 7$

$q (x) = 11^{x} + 9$

$y = 11^{x} + 9 x = 11^{y} + 9 x - 9 = 11^{y} \log_{11} (x - 9) = \log_{11} 11^{y} y = \log_{11} (x - 9)$

$q^{- 1} (x) = \log_{11} (x - 9)$

Graph the inverse without technology. Name the domain and range.

$y = {(\frac{1}{4})}^{x}$

x	y	$y^{- 1} = \log_{\frac{1}{4}} x (y, x)$
–2	16	(16, –2)
–1	4	(4, –1)
0	1	(1, 0)
1	$\frac{1}{4}$	$(\frac{1}{4}, 1)$

$Domain : \{x | x \in ℝ, x > 0\} Range : \{y | y \in ℝ\}$

Note

Q: What does the line $x = 0$ represent in this graph?
A: The vertical asymptote

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

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

x	y	$y^{- 1} = \log_{\frac{3}{5}} x (y, x)$
–1	$\frac{5}{3}$	$(\frac{5}{3}, - 1)$
0	1	(1, 0)
1	$\frac{3}{5}$	$(\frac{3}{5}, 1)$
2	$\frac{9}{25}$	$(\frac{9}{25}, 2)$

$Domain : \{x | x \in ℝ, x > 0\} Range : \{y | y \in ℝ\}$

Graph without technology. Name the end behavior.

$y = \log_{7} x$

$y^{- 1} = 7^{x}$

x	$y^{- 1}$	$y = \log_{7} x (y^{- 1}, x)$
–1	$\frac{1}{7}$	$(\frac{1}{7}, - 1)$
0	1	(1, 0)
1	7	(7, 1)

$As x \to + \infty, f (x) \to + \infty, and as x \to 0, f (x) \to - \infty$

$y = \log_{\frac{1}{2}} x$

$y^{- 1} = {(\frac{1}{2})}^{x}$

x	$y^{- 1}$	$y = \log_{\frac{1}{2}} x (y^{- 1}, x)$
1	$\frac{1}{2}$	$(\frac{1}{2}, 1)$
0	1	(1, 0)
–1	2	(2, –1)

$As x \to + \infty, f (x) \to - \infty, and as x \to 0, f (x) \to + \infty$

Note

Q: Would the inverse of this function represent a growth or decay function?
A: Decay. The base is between 0 and 1.

Describe the transformation from f(x) to g(x).

$f (x) = \log_{7} x g (x) = \log_{7} (x + 4) - 2$

$f (x) : a = 1, h = 0, k = 0 g (x) : a = 1, h = - 4, k = - 2$

From f(x), g(x) was shifted left 4 units and down 2 units.

$f (x) = 2^{x} g (x) = \log_{2} x + 3$

$f (x) : a = 1, h = 0, k = 0 g (x) : a = 1, h = 0, k = 3$

From f(x), g(x) was reflected over the line $y = x$ , shifted up 3 units.

$f (x) = \ln x g (x) = - \ln (x + 1) - 5$

$f (x) : a = 1, h = 0, k = 0 g (x) : a = - 1, h = - 1, k = 5$

From f(x), g(x) was reflected over the x-axis, shifted left 1 unit, and down 5 units.

Graph with technology.

$y = - \log_{5} (x - 3) + 1$

Note

Q: What is the value of $x = 3$ ?
A: Undefined. $x = 3$ is the vertical asymptote.

$y = 3 \ln x - 6$

$y = - \ln (x + 2)$