What is the Domain of the Function y = sqrt(x^2 - 1) for Finding Critical Point?

[donjt81]

Ok so I am trying to do this problem and I have a question

So based on the definition given in the book "An interior point of the domain of a function f where f' is zero or undefined is a critical point of f"

This is the problem:
y = sqrt(x^2 - 1)
so
y' = x/sqrt(x^2 - 1)

to find a critical point
y' = 0
x/sqrt(x^2 - 1) = 0
x = 0

also to find the critical point we have to see if y' will be undefined at any value of x. as we can see y' will be undefined at x = 0.

so from the first condition when we solved for y' = 0, we got x = 0 and now for the second condition y' is undefined at x = 0.

So both conditions are satisfied at x = 0 so does that mean the critical point is at x = 0. In the definition it says first condition or second condition has to be satisfied. I might be reading too much into the definition.

 

[donjt81]

anyone...?

 

[FrogPad]

if x = -1, or 1 then f' is undefined. 1 and -1 also exist in f(x).

-1 < x < 1 do not exist (real) in f(x) so they are not critical points
when x = 0 (included in the inequality above) f(x) does not exist.

Notice f(x=0) means, sqrt(0^2-1) = sqrt(-1) = i

 

[HallsofIvy]

What IS the domain of that function?