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Thread: Partial Dirivatives conceptual question

  1. #1

    Partial Dirivatives conceptual question

    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, fx > 0 and fy < 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?

  2. #2
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    Quote Originally Posted by TheNerdyGinger View Post
    Is it possible to find a function for which it is true that, for all x > 0 and y > 0, fx > 0 and fy < 0, and f (x,y) > 0?
    Let's consider a similar problem in a single-variable: is there a function [tex]f\colon\mathbb{R}\to\mathbb{R}[/tex] such that for all [tex]x[/tex], we have [tex]f(x) > 0[/tex] and [tex]f'(x) < 0[/tex]? 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.

  3. #3
    Quote Originally Posted by gregk View Post
    Let's consider a similar problem in a single-variable: is there a function [tex]f\colon\mathbb{R}\to\mathbb{R}[/tex] such that for all [tex]x[/tex], we have [tex]f(x) > 0[/tex] and [tex]f'(x) < 0[/tex]? 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

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    Quote Originally Posted by TheNerdyGinger View Post
    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. fx = 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. fy = -1/y2. We know that y2 is always going to be positive, so we have fy = (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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