Question

What is the domain and range of `(2/3)^x – 9`?

Answer 1

Domain: `(-oo, oo)`
Range: `(-9, oo)`

Explanation:

First note that `(2/3)^x-9` is well defined for any Real value of `x`. So the domain is the whole of `RR`, i.e. `(-oo, oo)`

Since `0 < 2/3 < 1`, the function `(2/3)^x` is an exponentially decreasing function which takes large positive values when `x` is large and negative, and is asymptotic to `0` for large positive values of `x`.

In limit notation, we can write:

`lim_(x->-oo) (2/3)^x = -oo`

`lim_(x->oo) (2/3)^x = 0`

`(2/3)^x` is continuous and strictly monotonically decreasing, so its range is `(0, oo)`.

Subtract `9` to find that the range of `(2/3)^x` is `(-9, oo)`.

Let:

`y = (2/3)^x-9`

Then:

`y+9 = (2/3)^x`

If `y > -9` then we can take logs of both sides to find:

`log(y+9) = log((2/3)^x) = x log(2/3)`

and hence:

`x = log(y+9)/log(2/3)`

So for any `y in (-9, oo)` we can find a corresponding `x` such that:

`(2/3)^x-9 = y`

That confirms that the range is the whole of `(-9, oo)`.