Find all critical values of $f(x,y,z)=(2x+4y-3z)^2$ And find all $c$ such that $f(x,y,z)=c$ is regular

Question

Ok, so I know to find the gradient and find all values where the gradient equals zero.

In this case, I believe we have critical points when ($2x+4y-3z)=0$ and the critical value is $f(x,y,z)=0$, ie the image under f of all critical points.

To find values of c i think i look at the case where our original function has values $(0,0,0)$ and we want this point not to be in our inverse image of our critical value

So, $(0,0,0)$ does not belong to $f$$^-1$($0$)

So, $(2x+4y-3z)^2-c=0$

And at $(0,0,0)$ we have that $ c=0$.

Assuming I’m on the right track, at this point do i consider when $(2x+4y-3z)^2$ has a value other than zero and conclude that $c\geq0$

Therefore, $f(x,y,z)=c$ is regular for all $c\geq0$?

Thanks

Answer 1

You are on the right track and your intuition is correct, however you can be sure by solving for x and noticing that y and z are independent, $$x=\frac{-4y+3z}{2}$$ $$\therefore\text{critical points}\Rightarrow\{(x,y,z)\in \mathbb{R}^3 | 2x+4y-3z=0\}$$ thus you see $\forall c\geq 0,\ f(x,y,z)=c$ is regular.