Practice 1 Solutions

Write the inverse of the given function.

$f (x) = 13^{x}$

$f^{- 1} (x) = \log_{13} x$

$v (x) = 2^{- x}$

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

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

$p (x) = 3^{x} - 4$

$y = 3^{x} - 4 x = 3^{y} - 4 x + 4 = 3^{y} \log_{3} (x + 4) = \log_{3} 3^{y} \log_{3} (x + 4) = y$

$p^{- 1} (x) = \log_{3} (x + 4)$

Note

Recall that $\log_{b} b = 1$ .

$f (x) = e^{x - 2}$

$y = e^{x - 2} x = e^{y - 2} \ln x = \ln e^{y - 2} \ln x = y - 2 y = \ln (x) + 2$

$f^{- 1} (x) = \ln (x) + 2$

Note

Q: Why is it important to use parentheses in your final answer?
A: Because $\ln (x) + 2$ is not the same function as $\ln (x + 2)$

$g (x) = 7^{x} - 6$

$y = 7^{x} - 6 x = 7^{y} - 6 x + 6 = 7^{y} \log_{7} (x + 6) = \log_{7} 7^{y} y = \log_{7} (x + 6)$

$g^{- 1} (x) = \log_{7} (x + 6)$

$h (x) = 5^{- x} + 2$

$y = 5^{- x} + 2 y = {(\frac{1}{5})}^{x} + 2 x = {(\frac{1}{5})}^{y} + 2 x - 2 = {(\frac{1}{5})}^{y} \log_{\frac{1}{5}} (x - 2) = \log_{\frac{1}{5}} {(\frac{1}{5})}^{y} y = \log_{\frac{1}{5}} (x - 2)$

$h^{- 1} (x) = \log_{\frac{1}{5}} (x - 2)$

Graph the inverse without technology. Describe the end behavior.

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

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

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

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

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

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

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

Graph without technology.

$h (x) = \log_{4} x$

$h^{- 1} (x) = 4^{x}$

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

$f (x) = \log x$

$f^{- 1} (x) = 10^{x}$

x	$f^{- 1} (x)$	$f (x) = \log x (f^{- 1} (x), x)$
–1	$\frac{1}{10}$	$(\frac{1}{10}, - 1)$
0	1	(1, 0)
1	10	(10, 1)

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

$f (x) = 5^{x} g (x) = \log_{5} (x - 3) + 1$

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

From f(x), g(x) was reflected over the line $y = x$ , shifted right 3 units, and up 1 unit.

$f (x) = \log_{2} x g (x) = 4 \log_{2} x + 3$

$f (x) : a = 1, h = 0, k = 0 g (x) : a = 4, h = 0, k = 3$

From f(x), g(x) was stretched by a factor of 4 and shifted up 3 units.

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

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

From f(x), g(x) was reflected over the x-axis and shifted left 2 units.

Graph using technology. Name the domain and range.

$y = \ln (x + 4) - 2$

$domain : \{x | x \in ℝ, x > - 4\} range : \{y | y \in ℝ\}$

Note

Q: What line represents the vertical asymptote?
A: x = –4 since the graph was shifted to the left 4.

$y = 3 + \ln x$

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

Note

Q: What is the value of $x = 0$ ?
A: Undefined. $x = 0$ is the vertical asymptote.

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

$domain : \{x | x \in ℝ, x > - 5\} range : \{y | y \in ℝ\}$