Function of 2 variables. I know the theoretical steps, but cant put them in practise

[belgiumrob]

Hi, So i have this function of 2 variables. The theoretical steps are partly clear, and i basically know what to do, but with the following problem, I cant put it into practise. I hope some one can shine some light on this question?

Consider the function f : R^2 −→ R : (x, y) → ln (1 + x^2y)
- Determine the domain of the function f
- Construct a sign diagram for the function f
- Compute all stationary points for the function f.
- Prove that the second-order conditions for the characterization of stationary pointsfail in this case
- Use the solution of the sign diagram of this question to classify the stationary points found. Does the function have a global minimum, respectively a global maximum ?If so, at which point(s) ?

So, how far did I go:
dom f={ (x,y) R²| conditions}

There is only 1 condition, the argument of the ln should be strictly positive, hence:
dom f={ (x,y) R²| 1+x²y>0}

So, i should investigate g(x,y)= 1+x²y
* The zeros satisfy 1+x²y=0 which is also written as: y= -1/x²

And here I am already stuck. As far as I know, I have to take the following steps, but for some reason i dont see it with this function??!?!?
*compute the zeros
* look where it is not continuous
* exclude the zeros from the domain
* and then....

Hope someone can shine some light on the question?

Thanks

 

[Deleted member 4993]

Hi, So i have this function of 2 variables. The theoretical steps are partly clear, and i basically know what to do, but with the following problem, I cant put it into practise. I hope some one can shine some light on this question?

Consider the function f : R^2 −→ R : (x, y) → ln (1 + x^2y)
- Determine the domain of the function f
- Construct a sign diagram for the function f
- Compute all stationary points for the function f.
- Prove that the second-order conditions for the characterization of stationary pointsfail in this case
- Use the solution of the sign diagram of this question to classify the stationary points found. Does the function have a global minimum, respectively a global maximum ?If so, at which point(s) ?

So, how far did I go:
dom f={ (x,y) R²| conditions}

There is only 1 condition, the argument of the ln should be strictly positive, hence:
dom f={ (x,y) R²| 1+x²y>0} → x²y > -1 → y > - (1/x)²

So, i should investigate g(x,y)= 1+x²y
* The zeros satisfy 1+x²y=0 which is also written as: y= -1/x²

And here I am already stuck. As far as I know, I have to take the following steps, but for some reason i dont see it with this function??!?!?
*compute the zeros
* look where it is not continuous
* exclude the zeros from the domain
* and then....

Hope someone can shine some light on the question?

Thanks

.

 

[belgiumrob]

Thanks Subhotosh Khan.

But what you wrote in red, is also what i wrote 2 lines below that, right? or am i missing something?

But, for some reason, I still dont see what manipulations I can do now, in order to effectively calculate the stationary points or zeros...?

What am i not seeing?

Thanks

 

[Ishuda]

Hi, So i have this function of 2 variables. The theoretical steps are partly clear, and i basically know what to do, but with the following problem, I cant put it into practise. I hope some one can shine some light on this question?

Consider the function f : R^2 −→ R : (x, y) → ln (1 + x^2y)
- Determine the domain of the function f
- Construct a sign diagram for the function f
- Compute all stationary points for the function f.
- Prove that the second-order conditions for the characterization of stationary pointsfail in this case
- Use the solution of the sign diagram of this question to classify the stationary points found. Does the function have a global minimum, respectively a global maximum ?If so, at which point(s) ?

So, how far did I go:
dom f={ (x,y) R²| conditions}

There is only 1 condition, the argument of the ln should be strictly positive, hence:
dom f={ (x,y) R²| 1+x²y>0}

So, i should investigate g(x,y)= 1+x²y
* The zeros satisfy 1+x²y=0 which is also written as: y= -1/x²

And here I am already stuck. As far as I know, I have to take the following steps, but for some reason i dont see it with this function??!?!?
*compute the zeros
* look where it is not continuous
* exclude the zeros from the domain
* and then....

Hope someone can shine some light on the question?

Thanks

Some hints
- Determine the domain of the function f
You have said you want g(x,y)>0. Unless you want more information, such as a sketch (or graph) of g(x,y) with a shaded area showing the domain, this is the best you can do [unkless you carry it a step further like Subhotosh Khan did]. If you do want that sketch then when looking for a function greater (less than) than some value, it is sometimes good to graph where the function is equal to that value. Given that, if we want g(x,y)>0, look at g(x,y) = 0 or, as you have it, the function
y = \(\displaystyle -\, \dfrac{1}{x^2}\)
and sketch this curve. It will basically divide the plane into two regions. So in which of those regions is g(x,y) positive.

- Construct a sign diagram for the function f
This is what you started to do with g(x,y) above. Where is ln(1 + x² y) = 0? It is at g(x,y)=1. When does that happen? The answer to that question will divide the plane into four regions. In which of those four regions is f(x,y) positive [g(x,y)>1] and in which is it negative [0<g(x,y)<1]?

Can you continue from there? If not, please show your work and let us know where you are stuck.

 

[belgiumrob]

Thanks Ishuda !

I never used graphing a function before, but as it shows, it is actually very handy! Don't know if im allowed to do that on the exam though...

Okay, but now i am even more confused maybe...

So I drawed first the graph g(x,y), which shows a convex parabola, which will never be positive right?

So, thats started me wondering how the graph of f(x,y) looks like, which is just a horizontal line at 0, so for every x, y is zero.

To be sure, would this mean that every x is a stationary point?

if that is the case, i know how to continue, if that is not the case, then im still stuck? (hoping im right)

Regards

 

[Ishuda]

Thanks Ishuda !

I never used graphing a function before, but as it shows, it is actually very handy! Don't know if im allowed to do that on the exam though...

Okay, but now i am even more confused maybe...

So I drawed first the graph g(x,y), which shows a convex parabola, which will never be positive right? Actually a hyperbola in t=x²

So, thats started me wondering how the graph of f(x,y) looks like, which is just a horizontal line at 0, so for every x, y is zero.

To be sure, would this mean that every x is a stationary point?

if that is the case, i know how to continue, if that is not the case, then im still stuck? (hoping im right)

Regards

O.K. We have two functions:
f(x,y) = ln(g(x,y))
g(x,y) = 1 + x² y
Let's consider the first question
- Determine the domain of the function f
This, as has been indicated above by several is where g(x,y) > 0. Now if we want to go further with a shaded graph [which we wouldn't on a timed test because of time], we could let t=x², then g(x,y)=0 could be written as
y(t) = -\(\displaystyle \dfrac{1}{t}\)
which is a hyperbola. But we are restricted to one sheet since t=x² \(\displaystyle \ge\) 0. So we have the sheet in the 4th quadrant (t>0) as our solution. Going back to x [=\(\displaystyle \pm\)t], we have a sort of single sheet of a hyperbola in the fourth quadrant with asymptotes of x=0 and y=0 plus a reflection of this in the third quadrant, i.e. y(-x) = y(x) [and g(-x,y) = g(x,y)].

Now we two regions to consider: (1) The two regions 'below' the 'hyperbola sheets' such as points like (10,-10) and (-10,-10) and (2) those points 'above' the 'hyperbola sheets' like (0,y₊) for any y=y₊ and (x₊,0) for any x=x₊. So if we shade the region 'above the hyperbola sheets', this would indicate the domain of the function f.

Now, the second question
- Construct a sign diagram for the function f
Simply stated, a sign diagram of f is just a partition of the (in this case) plane into regions where f is positive and f is negative. You should have had a more formal definition in your course. So, f(x,y) is positive when g(x,y) > 1 and is negative when 0 < g(x,y) < 1.

Since we already know from above where g(x,y)>0, we just follow the same kind of process we did above to find out where g(x,y)=1. Well g(x,y)=1 implies yx²=0 which means either y is zero (a horizontal line along the x-axis) or x is zero (a vertical line along the y-axis). Well these two line divide the plane into four areas (the four quadrants). Keeping in mind the domain of f(x,y), is g(x,y) greater than one (put a + sign for f positive) or less than one (put a - sign for f is negative) in quadrant 1 (x>0, y>0). What about the other three quadrants?

Continue on to the other questions showing your work so we may help you further if needed.

EDIT