<s>[INST] Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I do to plot this function by myself not by WolframAlpha. Thank you in advance.
 [/INST] Since the equation implies $\sqrt{x}+\sqrt{y}\geq 0$, therefore it is equivalent to $\sqrt{x}+\sqrt{y}=\sqrt{5}$. Therefore because $y\geq 0$, $y=(\sqrt{5}-\sqrt{x})^2=x+5-2 \sqrt{5}\cdot \sqrt{x}$. And we know from the condition of $\sqrt{x}, \sqrt{y}\geq 0$ that $0\leq x, y\leq 5$. Therefore, finally, the original statement is the same as $y=x+5-2\sqrt{5} \sqrt{x} (x\in[0, 5])$.
To do a simple graphical analysis (though may not be necessary), We consider the $1/4$ arc of $(x-5)^2+(y-5)^2=25 (0\leq x, y\leq 5)$, which we can rewrite as
\begin{align*}
y&=5-\sqrt{10x-x^2}\\
&=x+5-\sqrt{10x+2x\sqrt{10x-x^2}}\\
&\geq x+5-\sqrt{10x+2x\cdot \sqrt{25}}\\
&=x+5-2\sqrt{5}\sqrt{x}
\end{align*}
, thus indicating that the whole graph lies 'below' the circle.

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