how do we find the range for root (9-x)

Question

how do we find the range for  root (9-x)

mr.math · Answer

$$\sqrt{9-x}\ge 0.$$

mr.math · Answer

For a any polynomial P(x), $\sqrt{P(x)}\ge 0$.

mr.math · Answer

For any*

anonymous · Answer

You can't take the square root of a negative number. So the domain is $$\Large x \in ( - \infty ,9]$$This means that the lowest possible output is 0, when x = 9. So the range is all numbers from 0 to infinity.

ash2326 · Answer

we havy y= root(9-x)
let's see the domain of y
9-x>=0 ( expression inside root should be greater than 0)
x<=9
so domain is (-infinity, 9]
range can be found by puttin the values of x
if x=9
y=0
y can't be less than 0
if x=-infinity
then y is infinity
so range is [0,infinity) or range is>=0

anonymous · Answer

Hmm OpenStudy is giving me grief... It wasn't showing me your answer before I posted Mr.Math :'(

anonymous · Answer

I think answer is all non-negative real numbers cause here we have to see that its sqrt operator and range of it is the above said and so the answer is [0,infnty)