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I'm struggling with something that I think should be pretty basic. I am taking a multi-variable calculus course at university where we are finding relative max/mins and saddle points of multi-variable functions. On some of these problems, I can find a couple of critical points, but can't seem to find all of them, for instance: $$f(x, y) = 2xy - \frac12(x^4 + y^4) + 1 $$

I find the partial derivatives of x and y and get: $$f_x(x, y) = 2y - 2x^3$$ $$f_y(x, y) = 2x - 2y^3$$

I set both of these to 0 and get a system of equations. Eventually I end up with a $y=x^3$ and subbing that into my other equation gives me: $$2x(1-x^5)=0$$

From this, I can come up with the answers $x=0$, and $x=1$. However, I also know from graphing this function that the remaining critical point is $x=-1$. I am sure that I just need to factor out the $(1-x^5)$ further, but I do not know a convenient way to do that. I'll have to be doing these fairly quickly for an upcoming exam. Any advice?

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  • $\begingroup$ Powers of powers multiply, not add. $\endgroup$ – Michael Burr Oct 29 '17 at 20:51
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your System is given by $$y=x^3$$ $$x=y^3$$ thus $x=x^9$ and you will get $$0=x(x^8-1)$$ note that $$x^8-1=\left( x-1 \right) \left( x+1 \right) \left( {x}^{2}+1 \right) \left( {x}^{4}+1 \right) $$

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