Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the functions domain and range. y = log₃(x + 1)

Explanation:

Step1: Calculate $x + 1$ for $x=-\frac{8}{9}$

$x + 1=-\frac{8}{9}+1=\frac{-8 + 9}{9}=\frac{1}{9}$

Step2: Calculate $y=\log_3(x + 1)$ for $x=-\frac{8}{9}$

$y=\log_3\frac{1}{9}=\log_33^{-2}=-2$

Step3: Calculate $x + 1$ for $x=-\frac{2}{3}$

$x + 1=-\frac{2}{3}+1=\frac{-2 + 3}{3}=\frac{1}{3}$

Step4: Calculate $y=\log_3(x + 1)$ for $x=-\frac{2}{3}$

$y=\log_3\frac{1}{3}=\log_33^{-1}=-1$

Step5: Calculate $x + 1$ for $x = 0$

$x + 1=0 + 1=1$

Step6: Calculate $y=\log_3(x + 1)$ for $x = 0$

$y=\log_31=0$

Step7: Calculate $x + 1$ for $x = 2$

$x + 1=2+1=3$

Step8: Calculate $y=\log_3(x + 1)$ for $x = 2$

$y=\log_33=1$

Step9: Calculate $x + 1$ for $x = 8$

$x + 1=8 + 1=9$

Step10: Calculate $y=\log_3(x + 1)$ for $x = 8$

$y=\log_39=\log_33^{2}=2$

The completed table:

$x$	$x + 1$	$y=\log_3(x + 1)$
$-\frac{2}{3}$	$\frac{1}{3}$	$-1$
$0$	$1$	$0$
$2$	$3$	$1$
$8$	$9$	$2$

For the domain of $y = \log_3(x + 1)$, the argument of the logarithm must be positive. So $x+1>0$, which gives $x>-1$. The domain is $(-1,\infty)$.
For the range, since the logarithmic function $y = \log_3u$ can take on all real - values as $u=x + 1$ ranges over $(0,\infty)$, the range is $(-\infty,\infty)$.

Answer:

$x$	$x + 1$	$y=\log_3(x + 1)$
$-\frac{2}{3}$	$\frac{1}{3}$	$-1$
$0$	$1$	$0$
$2$	$3$	$1$
$8$	$9$	$2$

Domain: $(-1,\infty)$
Range: $(-\infty,\infty)$