Question

Can someone please describe the end behavior of the function?

Answer 1

As `x` increases `f(x)` approaches `-2`

Explanation:

`lim_(x->oo) k^x ->0` when `k<1`

Therefore

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

Answer 2

As x increases, `f(x)` approaches `-2`.

Explanation:

From a simple algebraic perspective, just note that any number larger than one (1) raised to any positive exponent will increase to infinity. Any number LESS than one will rapidly decrease - approaching zero (0), if not quite ever actually reaching it.

Thus, we can see that the first term, `(2/3)^x` , is going to approach `0` as x increases. Thus the final result of f(x) will be approaching `-2`.

A couple of quick calculations will show that trend very simply.

For x = 0
`f(x) = (2/3)^0 - 2 = 1 - 2 = -1`
For x = 1
`f(x) = (2/3)^1 - 2 = (2/3) - 2 = -1.33`
For x = 10
`f(x) = (2/3)^10 - 2 = (0.017) - 2 = -1.983`
For x = 100
`f(x) = (2/3)^100 - 2 = (2.435 xx10^(-18)) - 2 = -2.0000...`