Consider the function d: R² × R² → R defined by {} d(x, y) = min{|y2x₂, 1} if x₁ = y₁ otherwise, 1 for x = (x₁, x₂), Y = (Y₁, Y2) (That is, for example for x = (1,2) and y = (2,3), d(x, y) = y = (1, 1.5), d(x, y) = 0.5, and for x = (1,2), y = (1,3), d(x, y) = 1.) = 1, for x = (1, 2) and Is this a well-define metric on R2? Argue for each condition of a metric if it is satisfied and provide a justification or counterexample.

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A: Thanks for the question :)And your upvote will be really appreciable ;)

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A: Thanks for the question :)And your upvote will be really appreciable ;)

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A: We can solve this problem using partial derivatives.

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A: The required answer is in below step.

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A: Solution

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A: The given function is  f(x, y, z) = y5 + xz

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