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Find the inverse of the function.

[tex]\[ y = 2x^2 - 4 \][/tex]

A. [tex]\( y = \pm \sqrt{x} + 2 \)[/tex]

B. [tex]\( y = \pm \sqrt{\frac{x + 4}{2}} \)[/tex]

C. [tex]\( y = \pm \frac{\sqrt{x + 4}}{2} \)[/tex]

D. [tex]\( y = \pm \sqrt{x} - 2 \)[/tex]


Sagot :

To find the inverse of the function [tex]\( y = 2x^2 - 4 \)[/tex], let's follow a detailed step-by-step solution:

1. Start with the Original Function:
[tex]\[ y = 2x^2 - 4 \][/tex]

2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
To find the inverse, we swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. This means we rewrite the equation as:
[tex]\[ x = 2y^2 - 4 \][/tex]

3. Solve for [tex]\( y \)[/tex]:
Now, our goal is to solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex].

a. Start by isolating [tex]\( y^2 \)[/tex] on one side of the equation:
[tex]\[ x + 4 = 2y^2 \][/tex]

b. Divide both sides of the equation by 2 to further isolate [tex]\( y^2 \)[/tex]:
[tex]\[ \frac{x + 4}{2} = y^2 \][/tex]

c. Take the square root of both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = \pm \sqrt{\frac{x + 4}{2}} \][/tex]

4. Simplify the Expression:
We can simplify the expression further to get two solutions:
[tex]\[ y = \pm \frac{\sqrt{2x + 8}}{2} \][/tex]

Therefore, the inverse of the function [tex]\( y = 2x^2 - 4 \)[/tex] is:
[tex]\[ y = \pm \frac{\sqrt{2x + 8}}{2} \][/tex]

To be more precise, the inverse functions are:
[tex]\[ y = -\frac{\sqrt{2x + 8}}{2} \quad \text{and} \quad y = \frac{\sqrt{2x + 8}}{2} \][/tex]

So, the inverse of [tex]\( y = 2x^2 - 4 \)[/tex] can be written as:
[tex]\[ y = -\frac{\sqrt{2x + 8}}{2} \quad \text{or} \quad y = \frac{\sqrt{2x + 8}}{2}. \][/tex]

Complete and clean steps:
1. Start with the original function [tex]\( y = 2x^2 - 4 \)[/tex].
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to get [tex]\( x = 2y^2 - 4 \)[/tex].
3. Solve for [tex]\( y \)[/tex]: [tex]\( y = \pm \frac{\sqrt{2x + 8}}{2}. \)[/tex]

This confirms that the solutions are [tex]\( y = -\frac{\sqrt{2x + 8}}{2} \)[/tex] and [tex]\( y = \frac{\sqrt{2x + 8}}{2} \)[/tex].