Find the domain, range, and inverse of the function f(x)=arcsin(sqrt(x))

given:

y = arcsin √x
or
√x = sin y

domain: clearly 0 ≤ x ≤ 1
and the domain is : y any real number

for inverse:
interchange the x and y in
√x = sin y
to get
√y = sinx
y = sin^2 x

so f^-1 (x) = sin^2 x or (sinx)^2

verification:
http://www.wolframalpha.com/input/?i=plot+√x+%3D+sin%28y%29

To find the domain of the function f(x) = arcsin(sqrt(x)), we need to consider the restrictions on the values of x.

The arcsin function, also known as the inverse sine function, has a domain of [-1, 1]. Since we are taking the square root of x in the function, x must be greater than or equal to 0 to ensure a real output. Therefore, the domain of f(x) is the set of all non-negative real numbers: [0, +∞).

To find the range of the function, we need to determine the possible values of the output f(x). The range of the arcsin function is [-π/2, π/2]. By taking the square root of x, the output will always be non-negative. Therefore, the range of f(x) is [0, π/2].

To find the inverse of the function, we swap the roles of x and f(x) and solve for x. Let y = arcsin(sqrt(x)). To find the inverse, we need to express x in terms of the inverse function y.

Step 1: Start with the equation y = arcsin(sqrt(x)).
Step 2: Rewrite the equation as x = sin^2(y).
Step 3: Simplify by using the trigonometric identity sin^2(y) = 1 - cos^2(y).
Step 4: Substitute x = sin^2(y) = 1 - cos^2(y).
Step 5: Solve for cos^2(y): cos^2(y) = 1 - x.
Step 6: Take the square root of both sides: cos(y) = sqrt(1 - x).
Step 7: Solve for y by taking the inverse cosine: y = arccos(sqrt(1 - x)).

Therefore, the inverse of the function f(x) = arcsin(sqrt(x)) is given by f^(-1)(x) = arccos(sqrt(1 - x)).

Note: Keep in mind that the domain and range of the inverse function are the reverse of the original function. So, the domain of the inverse function is [0, π/2], and the range is [0, +∞).