Find the inverse of the function [tex] y = 2x^2 + 2 [/tex]

A. [tex] y = \pm \sqrt{x-2} [/tex]
B. [tex] y = \pm \sqrt{\frac{1}{2} x - 1} [/tex]
C. [tex] y = \pm \sqrt{2x^2 - 4} [/tex]
D. [tex] y = \pm \sqrt{\frac{1}{2} x - 1} [/tex]



Answer :

Certainly! Let me guide you through the process of finding the inverse of the function \( y = 2x^2 + 2 \) step-by-step.

### Step-by-Step Solution:

1. Express the function: Start by writing down the original function:
[tex]\[ y = 2x^2 + 2 \][/tex]

2. Solve for \( x \) in terms of \( y \): To find the inverse, we need to solve this equation for \( x \):

a. Subtract 2 from both sides to isolate the quadratic term:
[tex]\[ y - 2 = 2x^2 \][/tex]

b. Divide both sides by 2:
[tex]\[ \frac{y - 2}{2} = x^2 \][/tex]

c. Take the square root of both sides. Because squaring can yield both a positive and a negative result, we get two solutions:
[tex]\[ x = \pm \sqrt{\frac{y - 2}{2}} \][/tex]

3. Express the inverse function: Now we have solved for \( x \) in terms of \( y \), representing the inverse function. To write it in the conventional \( y \)-form, we swap \( x \) and \( y \):

[tex]\[ y = \pm \sqrt{\frac{x - 2}{2}} \][/tex]

### Conclusion:

The inverse of the function \( y = 2x^2 + 2 \) is:

[tex]\[ y = \pm \sqrt{\frac{x - 2}{2}} \][/tex]

This means for every [tex]\( x \)[/tex] in the domain of the inverse function, there can be two possible [tex]\( y \)[/tex] values, one positive and one negative, corresponding to the square roots.