Analyzing and Sketching Function: xy^2-x^2y+x+y=2 | Help with Homework Problem

In summary: Forbidden Regions 2. $$x=-1, y=-1, x=1, y=1$$[/B]If you think of ##x## as a constant what you have is a quadratic equation in ##y##, which you can solve, and then plot the ##(x,y)## pairs that are the solutions of that equation. By doing that for a number of different values of ##x## you can identify enough points to sketch to form an idea of the shape of the function.This is the equation, that I got this morning, $$y=\frac{\left(x^2-1\right)\pm\sqrt{\left(1
  • #1
Taylor_1989
402
14

Homework Statement


Hi guys I have been stuck on this for the better part of a day today and was wondering if anyone could help. I really can't seem to know even where to start.

I have to Analyse and sketch the function: $$xy^2-x^2y+x+y=2$$

I have not got a lot in the attempt solution because I am very lost. Can someone please give me an idea on how to tackle these types of problems.

Homework Equations

The Attempt at a Solution



1. $$x=0, y=2, y=0, x=2$$#

[/B]
 
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  • #2
If you think of ##x## as a constant what you have is a quadratic equation in ##y##, which you can solve, and then plot the ##(x,y)## pairs that are the solutions of that equation. By doing that for a number of different values of ##x## you can identify enough points to sketch to form an idea of the shape of the function.
 
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  • #3
This is the equation, that I got this morning, $$y=\frac{\left(x^2-1\right)\pm\sqrt{\left(1-x^2\right)^2-4x\left(x-2\right)}}{2x}$$. But this is where I was getting stuck because I could not find values for the quartic >0. This is why I thought this method I was doing was wrong. The values I got for x^4-6x+8x+1=0 was x=-1 and x=-0.544.
 
  • #4
Taylor_1989 said:
This is the equation, that I got this morning, $$y=\frac{\left(x^2-1\right)\pm\sqrt{\left(1-x^2\right)^2-4x\left(x-2\right)}}{2x}$$. But this is where I was getting stuck because I could not find values for the quartic >0. This is why I thought this method I was doing was wrong. The values I got for x^4-6x+8x+1=0 was x=-1 and x=-0.544.
Your solution for y looks fine, as far as you took it, but when you simplify the radical you should get ##x^4 - 6x^2 + 8x + 1##, which is different from what you show. Also, -1 is not a root of the equation ####x^4 - 6x^2 + 8x + 1 = 0##.

Use andrewkirk's suggestion of putting in values of x to get y values. Do this to get a number of points to get an idea of what the graph looks like.
 
  • #5
On a whim, I tried the rotation x+y=u, x-y=v. This gave uv = -2±√(4+8v+v4). Marginally simpler? It makes an asymptote easy to find.
 
  • #6
Thanks @Mark44 and @haruspex for the help. @haruspex at this moment in time I have not come across rotation only in matrices so ur method has sparked interest. Could you either expand on this on maybe point me to a website to read up of this method myself?
 
  • #7
Taylor_1989 said:
Thanks @Mark44 and @haruspex for the help. @haruspex at this moment in time I have not come across rotation only in matrices so ur method has sparked interest. Could you either expand on this on maybe point me to a website to read up of this method myself?
It's just a change of coordinates. The u and v axes are at 45 degrees to the x and y axes. Actually it is not a pure rotation, there is also an expansion by a factor of √2. To make it a pure rotation you would have to use u=(x+y)/√2 etc. but for the purposes of sketching the graph you can fix that up later.
Here is another trick to try...
Figure out some forbidden regions. For what range of x does y have no solutions, and vice versa? Likewise u and v?
 

1. What is the purpose of analyzing a function?

Analyzing a function allows us to understand its behavior and properties, such as its domain and range, extrema, and symmetry. This information is crucial for solving problems and making predictions based on the function.

2. How do you find the domain and range of a function?

To find the domain, we must identify all the values of the independent variable (usually denoted as x) that the function can take. The range, on the other hand, is all the possible output values (usually denoted as y) of the function. We can determine the domain and range by looking at the graph or using algebraic techniques.

3. What is the difference between a local and global maximum/minimum?

A local maximum (or minimum) occurs at a specific point on the graph where the function has the highest (or lowest) value in a small neighborhood, but it may not be the highest (or lowest) value of the entire function. A global maximum (or minimum), on the other hand, is the highest (or lowest) value of the entire function.

4. How can we determine if a function is even, odd, or neither?

A function is even if it remains unchanged when its input is replaced by its opposite (e.g., f(-x) = f(x)). A function is odd if it changes sign when its input is replaced by its opposite (e.g., f(-x) = -f(x)). If a function does not exhibit either of these properties, it is considered neither even nor odd.

5. Can a function have more than one inverse?

No, a function can only have one inverse. The inverse of a function is a reflection of the original function over the line y = x. If a function has multiple inverses, it means that it does not pass the horizontal line test and is not a one-to-one function.

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