Question

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

a) domain: {xx>- 1/6); range: {yly>0}
b) domain: {xx> 1/6); range: {yly> 2}
c) domain: x ; range: {yly > 2}
d) domain: x is a real number; range:{yly>-2}

Answer 1

The domain of f(x) = (1/6)^x + 2 is all real numbers, and the range is all real numbers greater than 2. The correct answer is c) domain: x is a real number; range: y .

Step-by-step explanation:

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

The function f(x) = (1/6)^x + 2 is an exponential function plus a constant. For exponential functions, the domain (the set of all possible x-values) is all real numbers, because you can raise a positive number to any real power. Therefore, the domain is x is a real number.

As for the range (the set of all possible y-values), since (1/6)^x is always positive and diminishes as x grows, and since the smallest value it can take is 0 (as x approaches infinity), the smallest y-value the function can take is 2 (when (1/6)^x is zero). In other words, y can never be less than 2, but it can take any value larger than 2. So, the range is y > 2.

Therefore, the correct answer is c) domain: x ; range: y > 2.