Critical point of a multivariable function

The reason is the same as in one variable: if \(f\) has a local extremum at \((a,b)\) and its partial derivatives exist, then \(t\mapsto f(a+t,b)\) and \(t\mapsto f(a,b+t)\) have an extremum at \(t=0\), so \(f_x(a,b)=0\) and \(f_y(a,b)=0\). Hence interior local extrema can only occur where \(\nabla f=\mathbf 0\) or where the gradient does not exist.

To find them, solve the system \(f_x=0\), \(f_y=0\). For \(f(x,y)=x^2+y^2-4x+6y\), the equations \(2x-4=0\) and \(2y+6=0\) give the single critical point \((2,-3)\). When the equations are nonlinear there may be several points or none, and it helps to solve for one variable and substitute into the other.

A critical point is not necessarily an extremum. The function \(f(x,y)=x^2-y^2\) has \(\nabla f=\mathbf 0\) at the origin, yet it increases along the \(x\)-axis and decreases along the \(y\)-axis: it is a saddle point. Being critical is a necessary condition, not a sufficient one.

There are also critical points where the gradient does not exist, such as the vertex of the cone \(f=\sqrt{x^2+y^2}\), which is a minimum. To decide what kind of point a critical point is, we will use second derivatives, collected in the Hessian matrix and the second derivative test.