Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
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].
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].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.