We have a function: $f(x,y)= xy-x$ defined on the set $$D = \{(x,y)\in\mathbb{R}^2 : 0 ≤ y ≤ 2-x^2\}.$$
I have found the critical point to be $(0,1)$, which I'm not sure is true. But I also have to find the biggest and smallest value that the function has in the set $D$.
I have defined the following, but can't come further:
$$D_1 = \{(x,y)\in\mathbb{R}^2: x = 0, 0 ≤ y ≤ 2\}$$ $$D_2 = \{(x,y)\in\mathbb{R}^2: y = 0, -\sqrt{2}≤ x ≤ \sqrt{2}\}$$ $$D_3 = \{(x,y)\in\mathbb{R}^2: y = 2-x^2, -\sqrt{2}≤ x ≤ \sqrt{2}\}$$
And then I'm supposed to find the values from those definitions, but I'm kind of stuck.
The curve $D_1$ is not part of the boundary of $D$. If you change the definition of $D_3$ to $$ D_3 = \{(x,y) \in \mathbb{R}^2 : \color{blue}{-\sqrt{2} \leq x \leq \sqrt{2}},\ y=2-x^2\} $$ Then $D_2$ and $D_3$ form the boundary of $D$.
Since $D$ is closed and bounded, the global max and min exist. Either they occur in the interior of $D$, or on the boundary of $D$. You have already found the one critical point in the interior of $D$. To find critical points on the boundary, look at the functions \begin{align*} g_1(x) &= f(x,0) & - \sqrt{2} \leq x \leq \sqrt{2} \\ g_2(x) &= f(x,2-x^2) & - \sqrt{2} \leq x \leq \sqrt{2} \\ \end{align*} These you can analyze with single-variable calculus techniques.