Solution:

Given, $$f(x)=\sqrt{x}-1$$ Since the function is a square root, it can never produce negative results. Hence, the minimum value at $$x=1$$ can only be 0 . As we increase \(x\), the maximum value can reach infinity. Therefore, the function's domain is $$[0, \infty)$$.\[\] function's range is $$[-1, \infty)$$.\[\]

Question 6:Find the Domain and Range of \[f(x) = \frac{1}{x - |x|}\]

Solution:

\[ \text{Let's analyze the behavior of } f(x) = \frac{1}{x - |x|} \] and determine the appropriate domain.

1. **Absolute Value Properties:** \(\text{The absolute value function behaves differently} \) depending on the sign of \(x\) - When \( x \geq 0 \), \( |x| = x \). - When \( x < 0 \), \( |x| = -x \).\[\] 2. **Evaluating \( f(x) \) for Different Cases:** - **Case 1: \( x > 0 \):** \[ f(x) = \frac{1}{x - x} = \frac{1}{0}, \quad \text{which is undefined}\] due to division by zero. - **Case 2: \( x = 0 \):** \[ f(x) = \frac{1}{0 - |0|} = \frac{1}{0}, \quad \text{which is also undefined.} \] - **Case 3: \( x < 0 \):** \[ f(x) = \frac{1}{x - (-x)} = \frac{1}{x + x} = \frac{1}{2x},\] which is defined as long as \(x \neq 0\)

3. **Conclusion:** \[ \text{The function } f(x) = \frac{1}{x - |x|} \text{ is only defined}\] for negative values of\(x\) \[ \text{Therefore, the domain is } (-\infty, 0)\text{.} \] \[ \textbf{Range Analysis:} \] \[ \text{Since \( x \) is negative, the expression \( 2x \)} \]will also be negative. Thus, \(\frac{1}{2x}\) will be negative. \[ \text{As \( x \) approaches zero from the left (negative side),}\] the denominator \( 2x \) becomes increasingly close to zero, making \(\frac{1}{2x}\) approach negative infinity.\[\] \(\text{As \( x \) approaches negative infinity,}\) the denominator \( 2x \) will also be very large and negative, meaning that \(\frac{1}{2x}\) approaches zero from the negative side. \[ \text{Therefore, the range of the function is:} \] \[ \text{Range} = (-\infty, 0). \]