Compute the domain of the real-valued function f(x)=sqrt(1-sqrt(2-x)). Thank you

To compute the domain of the function f(x) = √(1 - √(2 - x)), we need to consider the restrictions imposed by the square root function.

The domain of a function is the set of all possible values that the input variable (in this case, x) can take.

In the given function, we have two square root functions, √(1 - √(2 - x)). Let's analyze them separately:

1. The inner square root function, √(2 - x), is defined when the expression inside the square root is greater than or equal to zero:
2 - x ≥ 0
x ≤ 2

2. The outer square root function, √(1 - √(2 - x)), is defined when the expression inside the square root is greater than or equal to zero:
1 - √(2 - x) ≥ 0

Solving this inequality step-by-step:

1 - √(2 - x) ≥ 0
√(2 - x) ≥ 1
(2 - x) ≥ 1^2
2 - x ≥ 1
-x ≥ -1
x ≤ 1

So, combining the results from both inequalities, x ≤ 2 (from the inner square root) and x ≤ 1 (from the outer square root), we find that the domain of the function is:

Domain: x ≤ 1

To find the domain of the given function f(x) = √(1 - √(2 - x)), we need to consider the restrictions imposed by the square root operations.

1. The expression inside the inner square root (√(2 - x)) must be non-negative or zero, as the square root of a negative number is undefined in the real number system. Therefore, we set 2 - x ≥ 0 and solve for x:
2 - x ≥ 0
-x ≥ -2
x ≤ 2

This gives us the first part of the domain: x ≤ 2.

2. Next, we consider the expression inside the outer square root (√(1 - √(2 - x))). This square root is only defined for values where its argument (1 - √(2 - x)) is non-negative or zero. So we set 1 - √(2 - x) ≥ 0 and solve for x:
1 - √(2 - x) ≥ 0

To remove the square root, we square both sides of the inequality:
(1 - √(2 - x))^2 ≥ 0
(1 - √(2 - x)) × (1 - √(2 - x)) ≥ 0
1 - 2√(2 - x) + (2 - x) ≥ 0
2 - x - 2√(2 - x) ≥ 0
2 - x ≥ 2√(2 - x)
(2 - x) / 2 ≥ √(2 - x)
(2 - x)^2 / 4 ≥ 2 - x

Expanding and simplifying the inequality:
(x^2 - 4x + 4) / 4 ≥ 2 - x
x^2 - 4x + 4 ≥ 8 - 4x

Moving everything to one side of the inequality:
x^2 - 4x + 4 - 8 + 4x ≥ 0
x^2 - 4x - 4 ≥ 0

To solve this quadratic inequality, we can factor it:
(x - 2)(x - 2) ≥ 0

From this factorization, we see that the inequality is true when x ≤ 2.

Combining the two parts of the domain we found earlier:
Domain: x ≤ 2

first of all,

2-x > 0
-x > -2
x < 2

furthermore,
1-√(2-x) > 0
1 > √(2-x)
√(2-x) < 1
so 0 < 2-x < 1
-2 < -x < -1
2 > x > 1
or 1 < x < 2

so if x < 2 AND 1 < x < 2
then

1 < x < 2

test:
take a value outside that domain:
let x = 5
f(5) = √( 1 - √-3)) , not defined for the reals
let x = .6
f(.6) = √(1 - √1.4)
= √(-.1832...) which is undefined as well

but x = 1.3
f(1.3) = √(1 - √.7)
= √.16333997...) = a real number