Using Lagrange Multipliers, find all extremum points and the extremum values of the function $$f(x,y,z) = x^2-y^2-z^2$$ subject to the constraints: $$x^2+y^2=4$$ $$y=\sqrt{3+z^2}$$
So far I've done: $$\text{Lagrange multipliers:}\quad \alpha, \beta \in \Bbb R$$ $$g_1\overset{\large\text{set}}{=}x^2+y^2-4$$ $$g_2\overset{\large\text{set}}{=}\sqrt{3+z^2}-y$$
$$\vec \nabla f = \alpha\vec\nabla g_1 + \beta\vec\nabla g_2$$ $$(2x,-2y,-2z) = \alpha(2x,2y,0) + \beta\left(0,-1,\frac{z}{\sqrt{3+z^2}}\right)$$ So we have $$\cases{2x=2\alpha x & (1) \\ -2y=2\alpha y - \beta & (2) \\ -2z=\large \frac{\beta z}{\sqrt{3+z^2}} & (3)}$$
From $(1)$ we get $\alpha = 1$.
Putting $\alpha =1$ into $(2)$ we get $$-2y=2y-\beta$$ $$\beta = 4y$$
Putting $\beta = 4y$ into $(3)$, we get $$-2z=\frac{4yz}{\sqrt{3+z^2}}$$ $$-2z\sqrt{3+z^2}=4yz$$ $$4yz+2z\sqrt{3+z^2}=0$$ $$z\left(4y+2\sqrt{3+z^2}\right)=0$$ So $$z=0 \quad \text{or} \quad 4y+2\sqrt{3+z^2}=0$$
Putting $z=0$ into $y=\sqrt{3+z^2}$, we get $y=\sqrt{3}$. Putting $y=\sqrt{3}$ into $x^2+y^2=4$, we get $x=\pm 1$. So two critical points are $$(1,\sqrt{3},0) \quad \text{and}\quad (-1,\sqrt{3},0)$$
But I encounter a problem with the next two critical points:
We have $4y=-2\sqrt{3+z^2}$. If I put this into $y=\sqrt{3+z^2}$, I get a false result.
From here I don't know how to proceed. Any help is appreciated, many thanks in advance!
It is possible that you arrive at contradictory equations. You just found that in the case $4y + 2\sqrt{3+z^2}=0$, there are no solutions.
I would like to point out that in equation (1) you should consider two different cases. Case 1: $\alpha = 1$ and case 2: $x=0$. The latter case $x=0$ will give you two extra solutions. It follows then from $x^2+y^2=4$ that $y^2=4$, so $y= \pm 2$. But $y=-2$ will give no solution: substituting $y=-2$ in $y-\sqrt{3+z^2}=0$ gives a contradiction. So $y=2$. You can then find that $z=1$ or $-1$.