Question

Find the inverse of the translated quadratic function y=x2+11
with domain restricted to x≥0
.(1 point)
Responses

f−1(x)=x−11−−−−−√
f inverse left parenthesis x right parenthesis equals Start Root x minus 11 End Root

f−1(x)=x+11−−−−−√
f inverse left parenthesis x right parenthesis equals Start Root x plus 11 End Root

f−1(x)=x2−11−−−−−−√
f inverse left parenthesis x right parenthesis equals Start Root x squared minus 11 End Root

f−1(x)=x2−−√+11
f inverse left parenthesis x right parenthesis equals Start Root x squared End Root plus 11

Answer 1

To find the inverse of the function \( y = x^2 + 11 \) with the domain restricted to \( x \geq 0 \), we will follow these steps:

1. **Set the function equal to \( y \)**:
\[
y = x^2 + 11
\]

2. **Solve for \( x \)** in terms of \( y \)**:
\[
y - 11 = x^2
\]
\[
x = \sqrt{y - 11}
\]

Since the domain of the original function is restricted to \( x \geq 0 \), we take the positive root.

3. **Express the inverse function**:
Now, replace \( y \) with \( x \) to find \( f^{-1}(x) \):
\[
f^{-1}(x) = \sqrt{x - 11}
\]

Thus, the correct answer for the inverse function is:
\[
f^{-1}(x) = \sqrt{x - 11}
\]

So the response is:
**f^{-1}(x) = \sqrt{x - 11}**.