How do you find the domain and range of $y = 3 \sqrt{x - 2}$ $y = 3 sqrt (x-2)$ ?

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Answer 1 · 2015-07-17T19:45:04

For $x - 2$ $x-2$ to have a real square root, we require $x - 2 \geq 0$ $x-2 >= 0$ , hence $x \geq 2$ $x >= 2$ .

Given $x \geq 2$ $x >= 2$ we find $y$ $y$ can have any positive value.

So the domain is $[2, \infty)$ $[2, oo)$ and range is $[0, \infty)$ $[0, oo)$

For $\sqrt{x - 2}$ $sqrt(x-2)$ to have a value in $RR$ , we require $x-2 >= 0$

Add $2$ to both sides of this inequality to get:

$x >= 2$ .

If $x >= 2$ , then $sqrt(x-2) >= 0$ is well defined and hence $y = 3sqrt(x-2)$ is well defined.

So the domain is $[2, oo)$

$sqrt(x - 2) >= 0$ so $y = 3sqrt(x-2) >= 0$

In fact, for any $y >= 0$ , let $x = (y/3)^2+2$ .

Then $3sqrt(x-2) = 3sqrt((y/3)^2) = 3(y/3) = y$

So the range is the whole of $[0, oo)$

How do you find the domain and range of y = 3 sqrt (x-2) y=3√x−2y = 3 sqrt (x-2) ?