Question

Which of the following is the inverse of the function f(x)=6x2+12 with a domain restricted to x≥0 ?(1 point) Responses f−1(x)=x6−2−−−−−√ f inverse of x is equal to square root of x over 6 minus 2 end root f−1(x)=x−12−−−−−√ f inverse of x is equal to square root of x minus 12 end root f−1(x)=x6+2−−−−−√ f inverse of x is equal to square root of x over 6 plus 2 end root f−1(x)=x6−12−−−−−−√

Answer 1

To find the inverse of the function \( f(x) = 6x^2 + 12 \) with the domain restricted to \( x \geq 0 \), we can follow these steps:

1. **Replace \( f(x) \) with \( y \)**:
\[
y = 6x^2 + 12
\]

2. **Swap \( x \) and \( y \)**:
\[
x = 6y^2 + 12
\]

3. **Solve for \( y \)**:
\[
x - 12 = 6y^2
\]
\[
y^2 = \frac{x - 12}{6}
\]
\[
y = \sqrt{\frac{x - 12}{6}} \quad (\text{since } y \geq 0 \text{ because the domain of } f \text{ is } x \geq 0)
\]

4. **Write the inverse function**:
\[
f^{-1}(x) = \sqrt{\frac{x - 12}{6}}
\]

Now, let's analyze the answer choices provided:

1. \( f^{-1}(x) = \frac{x}{6} - 2 \)
2. \( f^{-1}(x) = \sqrt{x} - 12 \)
3. \( f^{-1}(x) = \frac{x}{6} + 2 \)
4. \( f^{-1}(x) = \sqrt{x} - 12 \)
5. \( f^{-1}(x) = \sqrt{\frac{x - 12}{6}} \)

The correct choice corresponds to \( f^{-1}(x) = \sqrt{\frac{x - 12}{6}} \).

Thus, the inverse function is:

\[
\boxed{f^{-1}(x) = \sqrt{\frac{x - 12}{6}}}
\]