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

1 Answer

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

Explanation:

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

2-x >=0 \implies x<= 22x0x2

Therefore, the Domain is:

D_f=(-\infty, 2]Df=(,2]

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

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

y(x=2) = sqrt(2-2) = 0y(x=2)=22=0

Range = [0,infty)=[0,)