71I'm getting 6/√13 . Not sure though!
\hspace{-16}$If $\bf{x_{1},x_{2}\in \mathbb{R}}$ and $\bf{f\big(x_{1},x_{2}\big)=\big(x_{1}-x_{2}\big)^2+\big(3\sqrt{8}+\sqrt{(4-x^2_{1})}-\sqrt{(x^2_{2}-16)}\big)^2}$\\\\ Then $\bf{f\big(x_{1},x_{2}\big)_{Min.}}$ is
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4 Answers
262vivek , your answer seems wrong .
|x1| ≤2 and |x2|≥4
thus (x1-x2)2 ≥ 4
f(x1,x2) > 4
71Yes, I got my error later. Was reluctant a bit. :)
But I don;t get your answer either.
262the above post is not the answer ,
i was trying to show that your answer is wrong .
i tried partial derivative but the equation seems to be quite complex.
anyways , solution ?
