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Partial Dirivatives conceptual question

TheNerdyGinger

New member
Joined
Oct 16, 2017
Messages
6
Hello. I have a calculus problem that goes as follows:

Is it possible to find a function for which it is true that, for all x > 0 and y > 0, f[SUB]x [/SUB]> 0 and f[SUB]y [/SUB]< 0, and f (x,y) > 0? If so, give an example. If not, why not?

I'm not sure how to go about solving this. I can't think of an example to make it true, but I don't know how to prove it false. Can someone please help me?
 

gregk

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Joined
Sep 18, 2017
Messages
17
Is it possible to find a function for which it is true that, for all x > 0 and y > 0, f[SUB]x [/SUB]> 0 and f[SUB]y [/SUB]< 0, and f (x,y) > 0?
Let's consider a similar problem in a single-variable: is there a function \(\displaystyle f\colon\mathbb{R}\to\mathbb{R}\) such that for all \(\displaystyle x\), we have \(\displaystyle f(x) > 0\) and \(\displaystyle f'(x) < 0\)? Think about what that sort of function would look like. Then see if you can find a very similar two-variable function satisfying your requirements.
 

TheNerdyGinger

New member
Joined
Oct 16, 2017
Messages
6
Let's consider a similar problem in a single-variable: is there a function \(\displaystyle f\colon\mathbb{R}\to\mathbb{R}\) such that for all \(\displaystyle x\), we have \(\displaystyle f(x) > 0\) and \(\displaystyle f'(x) < 0\)? Think about what that sort of function would look like. Then see if you can find a very similar two-variable function satisfying your requirements.
By that I think x +1/y would work
 

ksdhart2

Full Member
Joined
Mar 25, 2016
Messages
929
By that I think x +1/y would work
Well, if you're ever unsure of an answer, you can always check it yourself. Let's see if it fulfills all of the necessary criteria. For any x > 0 and y > 0, we have f(x, y) = x + 1/y. Obviously, x is positive, and we can easily see that 1/y can only be negative if y < 0, which we've temporarily disallowed. Hence, we have f(x, y) = (positive) + (positive) = (positive). So it satisfies the first criterion. Great!

Next, let's check the partial derivative with respect to x. f[SUB]x[/SUB] = 1. One is always positive, no matter what, so we're good here too.

Finally, let's check the partial derivative with respect to y. f[SUB]y[/SUB] = -1/y[SUP]2[/SUP]. We know that y[SUP]2[/SUP] is always going to be positive, so we have f[SUB]y[/SUB] = (negative)/(positive) = (negative). Hooray! The function passes all three criteria!

The question only asks you to find one example of a function which satisfies the criteria, and you've done just that. So, problem solved. Good job.
 
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