Question

find the inverse of the function $y = x^2 - 12$.
$\circ$ $y = \pm \sqrt{x - 12}$
$\circ$ $y = \pm \sqrt{x + 12}$
$\circ$ $y = \pm \sqrt{x} - 12$
$\circ$ $y = \pm \sqrt{x} + 12$
done

find the inverse of the function $y = 2x^2 + 2$.
$\circ$ $y = \pm \sqrt{x - 2}$
$\circ$ $y = \pm \sqrt{\frac{1}{2}x - 1}$
$\circ$ $y = \pm \sqrt{2x^2 - 4}$
$\circ$ $y = \pm \sqrt{\frac{1}{2}x - 1}$
done

Explanation:

First Problem: Find the inverse of \( y = x^2 - 12 \)

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

To find the inverse, we start by interchanging \( x \) and \( y \) in the equation. So we get \( x = y^2 - 12 \).

Step2: Solve for \( y \)

First, we add 12 to both sides of the equation to isolate the \( y^2 \) term. This gives us \( x + 12 = y^2 \). Then, to solve for \( y \), we take the square root of both sides. Remembering that when we take the square root, we have both a positive and negative solution. So \( y=\pm\sqrt{x + 12} \).

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

We begin by swapping \( x \) and \( y \) in the function. So the equation becomes \( x = 2y^2+2 \).

Step2: Isolate the \( y^2 \) term

Subtract 2 from both sides of the equation: \( x - 2=2y^2 \). Then divide both sides by 2 to get \( \frac{x - 2}{2}=y^2 \), which can be rewritten as \( \frac{1}{2}x - 1=y^2 \).

Step3: Solve for \( y \)

Take the square root of both sides. Since we are solving for \( y \), we have two solutions (positive and negative square roots). So \( y = \pm\sqrt{\frac{1}{2}x - 1} \).

Answer:

\( y=\pm\sqrt{x + 12} \) (the second option in the first problem's choices)

Second Problem: Find the inverse of \( y = 2x^2+2 \)