Question

Find the inverse of the translated quadratic function y=x2+11
𝑦
=
𝑥
2
+
11
with domain restricted to x≥0
𝑥
≥
0
.(1 point)
Responses

f−1(x)=x+11−−−−−√
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
−
1
(
𝑥
)
=
𝑥
2
+
11
f inverse left parenthesis x right parenthesis equals Start Root x squared End Root plus 11

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

f−1(x)=x2−11−−−−−−√

Answer 1

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

1. **Replace \( y \) with \( f(x) \)**:
\[
f(x) = x^2 + 11
\]

2. **Switch \( x \) and \( y \)**:
\[
x = y^2 + 11
\]

3. **Solve for \( y \)**:
\[
y^2 = x - 11
\]
\[
y = \sqrt{x - 11} \quad (\text{since } y \geq 0 \text{ due to the restriction } x \geq 0)
\]

4. **Express as the inverse function**:
\[
f^{-1}(x) = \sqrt{x - 11}
\]

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