Question

find the inverse of the function $y = x^2 - 12$.
$y = \pm\sqrt{x - 12}$
$y = \pm\sqrt{x + 12}$ (checked)
$y = \pm\sqrt{x} - 12$
$y = \pm\sqrt{x} + 12$
complete

find the inverse of the function $y = 2x^2 + 2$.
$\circ$ $y = \pm\sqrt{x - 2}$
$\circ$ $y = \pm\sqrt{\frac{1}{2}x - 1}$
$\circ$ $y = \pm\sqrt{2x^2 - 4}$
$\circ$ $y = \pm\sqrt{\frac{1}{2}x - 1}$ (note: likely a typo, maybe $y = \pm\sqrt{\frac{1}{2}(x - 2)}$ or similar)
done

Explanation:

For the first function \( y = x^2 - 12 \) (already solved, but let's recheck):

Step1: Swap \( x \) and \( y \)

\( x = y^2 - 12 \)

Step2: Solve for \( y \)

Add 12 to both sides: \( x + 12 = y^2 \)
Take square roots: \( y = \pm\sqrt{x + 12} \)

Step1: Swap \( x \) and \( y \)

\( x = 2y^2 + 2 \)

Step2: Subtract 2 from both sides

\( x - 2 = 2y^2 \)

Step3: Divide by 2

\( \frac{x - 2}{2} = y^2 \) or \( \frac{1}{2}x - 1 = y^2 \)

Step4: Take square roots

\( y = \pm\sqrt{\frac{1}{2}x - 1} \) (note: there was a typo in the original options, the correct one should have the square root over the entire \( \frac{1}{2}x - 1 \), so the option \( y = \pm\sqrt{\frac{1}{2}x - 1} \) (the second option with the correct square root coverage) is correct)

Answer:

\( y = \pm\sqrt{x + 12} \) (matches the checked option)

For the function \( y = 2x^2 + 2 \):