Question

What are the domain and range of f(x) =(1/6)^x + 2?

A.domain: {x | x > - 1/6} ; range: {y | y > 0}

B.domain: {x | x > 1/6} ; range: {y | y > 2}

C.domain: {x | x is a real number}; range: {y | y > 2}

D.domain: {x | x is a real number}; range: {y | y > –2}

Answer 1

Answer:

$\large\boxed{C.\ domain:\ \{x\ |\ x\ \text{is a real number}\};\ range:\ \{y\ |\ y>2\}}$

Step-by-step explanation:

$f(x)=\left(\dfrac{1}{6}\right)^x+2\\\\\text{The domain:}\ x\in\mathbb{R}\to\{x\ |\ x\ \text{is a real number\}}\\\\\lim\limits_{x\to\infty}\bigg[\left(\dfrac{1}{6}\right)^x+2\bigg]=\lim\limits_{x\to\infty}\left(\dfrac{1}{6}\right)^x+\lim\limits_{x\to\infty}2=0+2=2\\\\\lim\limits_{x\to-\infty}\bigg[\left(\dfrac{1}{6}\right)^x+2\bigg]=\lim\limits_{x\to-\infty}\left(\dfrac{1}{6}\right)^x+\lim\limits_{x\to\infty}2=\infty+2=\infty\\\\\text{The range:}\ y\in(2,\ \infty)\to\{y\ |\ y>2\}\\\\\bold{Look\ at\ the\ picture}$

Answer 2

Answer: Option C

domain: {x | x is a real number}; range: {y | y > 2}

Step-by-step explanation:

We have the function $f(x) =(\frac{1}{6})^x + 2$

Note that f(x) is an exponential function.

By definition the exponential functions of the form $a(b)^x +k$  have as domain all real numbers and as range {y | y > k}  if $a>0$, $b>0$

Where a is the main coefficient, b is the base and k is the vertical displacement.

In this case $k = 2$, $b=\frac{1}{6}$, $a=1$

Therefore the domain of f(x) is all real numbers and the range of f(x) is

{y | y > 2}