Question

Can someone explain to me why for y=3^(2-x)+1, the domain is (-infinity, +infinity) and the range is [1,+infinity)..

I found out by using a graphing calculator, but I would like to know the logic.. thanks :)

Answer 1

X can be any number , that's why domain is from -infinity to infinity.
3^(2-x) can never be negative, lowest it can be is 0, so range is from 0 to infinity

Since we are adding one, it would be 1 to infinity

Answer 2

It cannot actually even be 0. It just approaches 0 as x becomes very large. No power of 3 is less than or equal to 0.

Answer 3

Thank you~

Answer 4

Sorry, but I thought of another question.. with 3^(2-x), the range will be (0, +infinity) as it can never reach 0... but with 3^(2-x) + 1, it will be [1, +infinity) as it includes 1.. (proved by graphing calculator).

Why is that..? The +1 is like vertical shifting, but with 3^(2-x), it could not be 0, so if it was shifted up 1, how can it include 1?

Answer 5

it approaches 1 , but never 1
http://www.wolframalpha.com/input/?i=3^%282-x%29%2B1%3D%3D1

Answer 6

It does when x gets bigger... y equals 1 when x = 11.25 and up.

Answer 7

Oh I see now. It never reaches 1 but the calculator can't calculate it precise enough so it rounds to 1 ^^