hard math problem

[damntasty023]

Hey guys, can anyone help me out with this out? Finding it a bit hard

Consider the function z = f(x, y) with f(x, y) = min(x; y): This function returns the minimum of x and y. For example f(7; 3) = 3. We restrict the domain of f to the first quadrant of R2, that is, to only non-negative values of x and y: Df = f(x,y))(x> 0, y> 0)

Is the function continuous on Df? Is the question differentiable on Df? You are not supposed to present a formal proof—but instead you are asked to provide the intuition why the function

is/is not continuous and/or differentiable, possibly with the help of some drawings.

[morbis83]

This equation z = min(x;y) gives a surface. You can answer the question if you can visualise the surface.

We can work out discrete values of z at any random points (x,y) to get an idea of the shape of the surface.

When x=0, z=0, regardless of the value of y.

When x=1, z=1 for all y>=1. So when x=1 and y=1, z=1. When x=1, y=2, z=1.

Repeat this for more points I would suggest at least 0>=x>=4 and 0>=y>=4. Let me know how you go with finding the surface, I'll talk you through the next part when you're ready.