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Question 513411: Find the domain of g(x)=sqrt(1-x)
its a homework and i cant figure it out
my options are
a. -infinity, 1
b. 1, infinity
c. and all real numbers
Found 2 solutions by tinbar, drcole:
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! So the domain is pretty much a set of numbers that 'works' for the given function. In this case, you'll notice the only thing that might cause problems is the sqrt part. Clearly you CANNOT take sqrt of negative numbers. Like the sqrt(-5) does not exist (Why? If you cannot figure it out, assume it does exist, then it's either positive or negative, but then what?...)
Anyway, you cannot let the argument of the sqrt be negative. In this case, the part in the sqrt, 1-x, cannot be less than 0.
We want 1-x>=0, rearranging gives x>=1. So we want all numbers in the set of 1 to +infinity, which is choice b for you.
Answer by drcole(72) (Show Source):
You can put this solution on YOUR website! Work from the outside in. The function g(x) is a composition of functions: first you take x and find 1 - x, then you take the square root of 1 - x. Each layer of this function has a natural domain (possible set of inputs) and range (possible set of outputs). The trick here is to look at the outermost layer first and see what sorts of inputs can go inside of it.
In this case, the outermost layer is sqrt(u) (where u represents the inside, i.e., u = 1 - x). What sorts of u can we take the square root of and get back a real number? The answer is that u must be a non-negative real number. Written as a formula, we get:
u >= 0 (where >= means "greater than or equal")
Now we recall that u was the inside of this composition of functions: u = 1 - x. Thus our inequality becomes:
1 - x >= 0
1 >= x (adding x to both sides)
x <= 1 (flipping both the left and right hand sides AND the inequality)
So for u to be a valid input into sqrt(u), x must be less than or equal to 1. This corresponds to all numbers in the interval (-infinity, 1], or choice (a).
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