Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the function’s domain and range.

$y = 2^x + 1$

$x$	$2^x$	$y = 2^x + 1$
$-1$
$0$
$1$
$2$
$3$

Explanation:

Step1: Calculate $2^x$ for $x=-2$

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

Step2: Calculate $y$ for $x=-2$

$y = \frac{1}{4} + 1 = \frac{5}{4}$

Step3: Calculate $2^x$ for $x=-1$

$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$

Step4: Calculate $y$ for $x=-1$

$y = \frac{1}{2} + 1 = \frac{3}{2}$

Step5: Calculate $2^x$ for $x=0$

$2^{0} = 1$

Step6: Calculate $y$ for $x=0$

$y = 1 + 1 = 2$

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

$2^{1} = 2$

Step8: Calculate $y$ for $x=1$

$y = 2 + 1 = 3$

Step9: Calculate $2^x$ for $x=2$

$2^{2} = 4$

Step10: Calculate $y$ for $x=2$

$y = 4 + 1 = 5$

Step11: Calculate $2^x$ for $x=3$

$2^{3} = 8$

Step12: Calculate $y$ for $x=3$

$y = 8 + 1 = 9$

Step13: Identify domain of $y=2^x+1$

Exponential functions accept all real $x$, so domain is all real numbers.

Step14: Identify range of $y=2^x+1$

$2^x > 0$ for all real $x$, so $2^x + 1 > 1$.

Answer:

Completed Table:

$x$	$2^x$	$y=2^x+1$
$-1$	$\frac{1}{2}$	$\frac{3}{2}$
$0$	$1$	$2$
$1$	$2$	$3$
$2$	$4$	$5$
$3$	$8$	$9$

Domain: All real numbers ($(-\infty, \infty)$)

Range: All real numbers greater than 1 ($(1, \infty)$)