How can you be sure that the equation ez(x2 + y2 + z2) + y = 0 has a solution z

Question

How can you be sure that the equation ez(x2 + y2 + z2) + y = 0 has a solution z = f(x, y) which is continuous at x = 1, y =0, with f(1.0) = 0? Using the tangent plane as an approximation to the surface, calculate f(1 + h, k) approximately when h and k are small.

Explanation / Answer

At 1,0 the function becomes e^z(1+z^2) = (1+z^2)^0.5 ==>e^z(1+z^2)^0.5=1 Applying the limits for x-->1 and y---> 0 we have the function defined. But it will not be continous unless the tangent plane at 1,0 is defined For tangent plane we have f = e^z(1+z^2)^0.5-1 Given e^z(1+z^2)^0.5-1 = 0 (given). Hence f is continous with f(1,0) =0

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