Question

What is the inverse of f(x)=x+9−−−−√ for x≥−9?
Responses

f−1(x)=(x−9)2, x≥9
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals left parenthesis x minus 9 right parenthesis squared , x is greater than or equal to 9

f−1(x)=−(x−9)2, x≤9
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals negative left parenthesis x minus 9 right parenthesis squared , x is less than or equal to 9

f−1(x)=x2−9, x≥0
f to the power of negative 1 end exponent left parenthesis x right parenthesis equals x squared minus 9 , x is greater than or equal to 0

f−1(x)=x2−9, x≤0

Answer 1

To find the inverse of the function \( f(x) = \sqrt{x + 9} \) for \( x \geq -9 \), we follow these steps:

1. **Rewrite the function**:
\[
y = \sqrt{x + 9}
\]

2. **Swap \( x \) and \( y \)**:
\[
x = \sqrt{y + 9}
\]

3. **Square both sides to eliminate the square root**:
\[
x^2 = y + 9
\]

4. **Solve for \( y \)**:
\[
y = x^2 - 9
\]

5. **Identify the domain of the inverse function**:
Since \( f(x) = \sqrt{x + 9} \) is defined for \( x \geq -9 \), the output \( f(x) \) will be in the range \( [0, \infty) \). Therefore, the inverse function \( f^{-1}(x) \) will have a domain of \( x \geq 0 \).

Thus, the inverse function is:
\[
f^{-1}(x) = x^2 - 9, \quad x \geq 0
\]

Therefore, the correct response is:
\[
\text{f}^{-1}(x) = x^2 - 9, \quad x \geq 0
\]