To find the range of the function \( f(x) = \sin^{-1}(x - \sqrt{x}) \), we will follow these steps:
### Step 1: Define the function
We start with the function:
\[
f(x) = \sin^{-1}(x - \sqrt{x})
\]
### Step 2: Simplify the expression inside the inverse sine
We need to analyze the expression \( x - \sqrt{x} \). To do this, we can rewrite it in a more manageable form. We can express \( x \) in terms of \( \sqrt{x} \):
Let \( y = \sqrt{x} \). Then \( x = y^2 \), and we can rewrite the expression as:
\[
x - \sqrt{x} = y^2 - y
\]
### Step 3: Find the minimum and maximum values of \( y^2 - y \)
Now, we need to find the range of \( y^2 - y \). This is a quadratic function in \( y \):
\[
g(y) = y^2 - y
\]
To find the vertex of this parabola, we can use the vertex formula \( y = -\frac{b}{2a} \):
\[
y = -\frac{-1}{2 \cdot 1} = \frac{1}{2}
\]
Now, we can find the value of \( g(y) \) at this point:
\[
g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}
\]
### Step 4: Determine the range of \( g(y) \)
The function \( g(y) = y^2 - y \) opens upwards (since the coefficient of \( y^2 \) is positive). Therefore, the minimum value occurs at \( y = \frac{1}{2} \), which we found to be \( -\frac{1}{4} \). As \( y \) approaches infinity, \( g(y) \) approaches infinity as well.
Thus, the range of \( g(y) \) is:
\[
[-\frac{1}{4}, \infty)
\]
### Step 5: Find the range of the inverse sine function
Now, we need to consider the function \( f(x) = \sin^{-1}(g(y)) \). The output of the inverse sine function is defined for inputs in the range \([-1, 1]\). Therefore, we need to find the intersection of the range of \( g(y) \) with \([-1, 1]\).
The minimum value of \( g(y) \) is \( -\frac{1}{4} \), which is within the range of the inverse sine function. The maximum value of \( g(y) \) can go to infinity, but we only consider values up to 1 for the inverse sine function.
### Step 6: Determine the final range of \( f(x) \)
Thus, the range of \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is:
\[
\left[\sin^{-1}(-\frac{1}{4}), \sin^{-1}(1)\right]
\]
Since \( \sin^{-1}(1) = \frac{\pi}{2} \), we need to compute \( \sin^{-1}(-\frac{1}{4}) \):
\[
f(x) \text{ has a range of } \left[-\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2}\right]
\]
### Final Answer
The range of the function \( f(x) = \sin^{-1}(x - \sqrt{x}) \) is:
\[
\left[-\sin^{-1}\left(\frac{1}{4}\right), \frac{\pi}{2}\right]
\]
