Final answer:
The maximum value of the function f(x,y)=e^{xy} on the curve x^2+y^2=2 occurs at points (1,1) and (-1,-1), with a value of e. The minimum value occurs at points (1,-1) and (-1,1), with a value of 1/e.
Step-by-step explanation:
To find the points on the intersection of the function f(x,y)=e^{xy} and the curve x^2+y^2=2 where the function f takes its maximum and minimum values, we need to consider the behavior of the exponential function. Since e^{xy} is always positive, it reaches its maximum value when the exponent xy is as large as possible given the constraint that x^2+y^2=2.
To maximize xy under the given constraint, we can use calculus methods such as Lagrange multipliers or simply inspect the symmetry of the circle represented by x^2+y^2=2. This yields the points (1,1) and (-1,-1) where the maximum function value occurs. The maximum value at these points is e.
Similarly, to minimize the function f we look for the minimum value of xy, which happens at points (1,-1) and (-1,1) on the circle, yielding the minimum value of 1/e.