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

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Answer 1 · 2018-03-05T19:48:51

#D_f=(-\infty, 2]#
Range #= [0,infty)#

Since we have a square root, the value under it cannot be negative:

#2-x >=0 \implies x<= 2#

Therefore, the Domain is:

#D_f=(-\infty, 2]#

We now construct the equation from the domain, finding the Range:

#y(x\to-\infty) \to sqrt(\infty) \to \infty#

#y(x=2) = sqrt(2-2) = 0#

Range #= [0,infty)#