Question

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

Answer 1

Answer: The domain of f(x) is all real numbers, $(-\infty,\infty)$ and range of f(x) is $(2,\infty)$.

Explanation:

The given function is,

$f(x)=(\frac{1}{6})^x+2$

The domain is the set of all possible values for which the value of function exist.

The given function is defined for any real value of x, therefore the domain of the function is set of all real number. it can be written in interval notation $(-\infty,\infty)$.

The range is the set of possible value of f(x) at different value of x. Range is the set of outputs.

In the given function,

$(\frac{1}{6})^x>0$

for any value of x.

So,

$(\frac{1}{6})^x+2>0+2$

$f(x)>2$

Therefore the value of f(x) is always greater than 2 and the range of the function is  $(2,\infty)$.

Answer 2

Answer:

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

Step-by-step explanation:

C on edge