We are given the function: \( f(x) = \frac{2}{\pi} \tan^{-1}(x), \quad x \geq 0 \).
The inverse tangent function, \( \tan^{-1}(x) \), has the following properties:
- Domain: \( (-\infty, \infty) \)
- Range: \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
For \( x \geq 0 \):
As \( x \) increases from 0 to \( \infty \), \( \tan^{-1}(x) \) increases from 0 towards \( \frac{\pi}{2}. \)
The given function is scaled by \( \frac{2}{\pi} \), therefore:
At \( x = 0: f(0) = \frac{2}{\pi} \cdot \tan^{-1}(0) = 0 \)
As \( x \to \infty: f(x) \to \frac{2}{\pi} \cdot \frac{\pi}{2} = 1 \)
Since \( \tan^{-1}(x) \) is continuous and strictly increasing for \(x \geq 0 \), and \( f(x) \) is from the minimum value at \(x=0 \) to the supremum as \(x \to \infty \).
Therefore, the range of the function \( f(x) = \frac{2}{\pi} \tan^{-1}(x), \quad x \geq 0 \) is \( \boxed{[0, 1)} \).
Please let me know in the comments if you find any errors in this solution.