# Finding the minima for a function of a vector

A potentially naïve question about finding the minimum of a function:

If I have a scalar function of a scalar variable $y=f(x)$, which is continuously differentiable, I can find minima but finding the values of $x$ for which $f'(x)=0$.

On the other hand, if I have a scalar function of a vector variable, $y=f(\mathbf{x})$, can I do the same? Can I always find the value of $\mathbf{x}$ where $f'(\mathbf{x})=0$?

I am aware I can take partial derivatives for each value of $\mathbf{x}$, i.e.

$\nabla y = \left[\frac{\partial y}{\partial x_1}, \frac{\partial y}{\partial x_2}, \dots \frac{\partial y}{\partial x_N} \right]$

But how do I go from this vector of derivatives to a final value for $\mathcal{x}$? Do I just set each derivative to zero?

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Yes, that is a necessary condition for extremum for $C^1$ functions on $\mathbb R^n$: $\frac{\partial y}{\partial x_i}(\hat{\mathbf x}) = 0$ for all $i=1,\dots,n$ whenever $\hat{\mathbf x}$ is an extremum argument of the function $y$. – Ilya Aug 1 '12 at 16:17

The theorem goes as follows: Let $U\subseteq \mathbb{R}^n$ be an open subset, and let $x_0\in U$. Let $f:U\to \mathbb{R}$ be any function such that $x_0$ is a local extremum of $f$. If all partial derivatives of $f$ exist at $x_0$ then $\nabla f(x_0) = 0$. In particular, if $f$ is differentiable at $x_0$ then $\nabla f(x_0) = 0$.
This means that if you want to find extrema of a differentiable multivariable function defined on an open set you may search for points at which all the partial derivatives vanish simultaneously. Such points are called critical points. Any local extremum of $f$ must be a critical point, however not every critical point is a local extremum. Classifying the critical points requires some more work.