Question
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: \( [\square, \infty) \)
Ask by Martin Clarke. in the United States
Jan 24,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
domain: \([-1, \infty)\)
range: \([2, \infty)\)
Solution
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: \([2, \infty)\)
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Mind Expander
The original function \( f(x) = x^{\frac{1}{2}} \) has a domain of \( [0, \infty) \). However, with the transformation applied in \( m(x) = (x+1)^{\frac{1}{2}} + 2 \), we determine that the smallest value of \( x \) in this function shifts to \( -1 \) because it becomes valid for \( x + 1 \geq 0 \). Thus, the domain of \( m \) is \( [-1, \infty) \). As for the range, the minimum output occurs when \( x = -1 \), giving \( m(-1) = (0)^{\frac{1}{2}} + 2 = 2 \). Since the function can increase indefinitely as \( x \) grows, the range of \( m \) is \( [2, \infty) \). So, for the final answer: domain: \( [-1, \infty) \) range: \( [2, \infty) \)
