How do you find the domain and range of $y = 2^{- x}$ $y=2^(-x)$ ?

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Answer 1 · 2017-05-09T06:31:54

The domain is $(- \infty, \infty)$ $(-oo, oo)$ and the range is $(0, \infty)$ $(0, oo)$ .

Given:

$y = 2^{- x}$ $y = 2^(-x)$

First note that this is well defined for any real value of $x$ $x$ .

So the domain is the whole of $RR$ , or in interval notation $(-oo, oo)$ .

Next note that $2^(-x) > 0$ for any real value of $x$ . So the range is $(0, oo)$ or a subset of it.

Let $y > 0$ . Then $log(y)$ is well defined. So we can take logs of both sides to find:

$log(y) = log(2^(-x)) = (-x) log(2)$

Dividing both sides by $-log(2)$ we find:

$x = -log(y)/log(2)$

So that tells us that for any $y > 0$ there is an $x$ such that:

$y = 2^(-x)$

So the range is the whole of $(0, oo)$ .

How do you find the domain and range of y=2^(-x)y=2−xy=2^(-x)?