Match the function with its graph (labeled I-VI). F(x, y) = 1/1 + x^2 + y^2 f(x, y) = 1/1 + x^2 y^2 f(x, y) = ln(x^2 + y^2) f(x, y) = cos (Squareroot x^2 + y^2) f(x, y) = (xy) f(x, y) = cos(xy)

a)
1/(1 + x^2 + y^2)
Clearly as x and y approaches very small positive or negative values,
we approx get 1 / (1 + 0 + 0), which is z = 1 almost
Also, as x or y approach very large values, z is almost 0
All this happens only in graph III

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b) 1/(1 + x^2y^2) :
This is graph I

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c) z = ln(x^2 + y^2)
Clearly as x and y approach 0, ln(0) is almost negative infinity
And as x and y are large, z reaches +infinity
This is true only in graph IV

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d) cos(sqrt(x^2 + y^2)) :
This is graph V
When x and y approach close to 0, cos(sqrt0) --> cos(0) = 1 gets its max value and this is observed by the small hill near wheer x and y are zero
and where z approx equals 1
Graph V

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e) z = |xy|
We know that this must lie only in positive z-region
and it must look almost like a plane
This is true for option VI

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f) cos(xy) :
This is graph II
When x = y = 0, z = cos(0) = 1
So, there is a point (0,0,1) which is true for this graph
Also, keeping y constant, we get z = cos(cx) in the xz plane
and looking at the xz plane, we have a cosine curve trace
So, graph II