Question 4 of 8 Type the correct answer in each box. Use numerals instead of words. The function \( f(x)=x^{\frac{1}{2}} \) is transformed to get function \( m \). \( m(x)=(x+1)^{\frac{1}{2}}+2 \) What are the domain and the range of function \( m \) ? domain: \( [\square, \infty) \) range: \( [\square, \infty) \)

The function \( m(x) = \sqrt{x + 1} + 2 \) involves the square root transformation of \( f(x) = \sqrt{x} \) along with horizontal and vertical shifts. **Domain:** - The expression inside the square root must be non-negative: \[ x + 1 \geq 0 \implies x \geq -1 \] So, the domain of \( m(x) \) is: \[ [-1, \infty) \] **Range:** - The smallest value of \( \sqrt{x + 1} \) is 0 (when \( x = -1 \)): \[ m(-1) = \sqrt{0} + 2 = 2 \] - As \( x \) increases, \( \sqrt{x + 1} \) increases without bound, so \( m(x) \) can take on any value from 2 upwards. Therefore, the range of \( m(x) \) is: \[ [2, \infty) \] **Final Answer:** - domain: \([-1, \infty)\) - range: \([2, \infty)\)