Answer:
1. (-4,6) there is no a solution to the equation through this point
2. (2,−6) there is no a solution to the equation through this point
3. (−5,39) there is a solution to the equation through this point
4. (−1,45) there is a solution to the equation through this point
Step-by-step explanation:
Using the existence and uniqueness theorem:
Let:
[tex]F(x,y)=\sqrt{y^2-36} \\\\and\\\\\frac{\partial F}{\partial y} =\frac{y}{\sqrt{y^2-36} }[/tex]
Now, let's find the domain of [tex]F(x,y)[/tex], due to the square root:
[tex]y^2-36 \geq 0\\\\y^2\geq 36\\\\y \geq 6\hspace{12}or\hspace{12}y\leq-6[/tex]
So the domain of the function is:
[tex]y \in R\hspace{12}y\geq6\hspace{12}or\hspace{12}y\leq-6[/tex]
Now, due to the fraction [tex]\frac{\partial F}{\partial y}[/tex] the denominator must be also different from 0, so:
[tex]y^2-36\neq0\\\\y \neq \pm6[/tex]
So, the theorem tells us that for each [tex]y_0\in R:\hspace{12}y_0>6\hspace{12}or\hspace{12}y_0<-6[/tex] there exists a unique solution defined in an open interval around [tex]x_0[/tex].
1. (-4,6) there is no a solution to the equation through this point because [tex]y_0=6[/tex]
2. (2,−6) there is no a solution to the equation through this point because
[tex]y_0=-6[/tex]
3. (−5,39) there is a solution to the equation through this point because
[tex]y_0>6[/tex]
4. (−1,45) there is a solution to the equation through this point because
[tex]y_0>6[/tex]
