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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−−−−−−√

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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:

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