If (-1, -1) is an extremum of [tex]f[/tex], then both partial derivatives vanish at this point.
Compute the gradients and evaluate them at the given point.
- [tex]f(x,y)=xy+\frac1x +\frac1y[/tex]
[tex]\nabla f = \left\langle y - \dfrac1{x^2}, x - \dfrac1{y^2}\right\rangle \implies \nabla f (-1,-1) = \langle-2,-2\rangle \neq \langle0,0,\rangle[/tex]
- [tex]f(x,y) = x^2+2x[/tex]
[tex]\nabla f = \langle 2x+2,0\rangle \implies \nabla f(-1,-1) = \langle0,0\rangle[/tex]
[tex]\nabla f = \langle y, x-2y\rangle \implies \nabla f(-1,1) = \langle-1,1\rangle \neq\langle0,0\rangle[/tex]
- [tex]f(x,y) = xy-\frac1x-\frac1y[/tex]
[tex]\nabla f = \left\langle y + \frac1{x^2}, x + \frac1{y^2}\right\rangle \implies \nabla f(-1,1) = \langle0,0\rangle[/tex]
The first and third functions drop out.
The second function depends only on [tex]x[/tex]. Compute the second derivative and evaluate it at the critical point [tex]x=-1[/tex].
[tex]f(x,y) = x^2+2x \implies f'(x) = 2x + 2 \implies f''(x) = 2 > 0[/tex]
This indicates a minimum when [tex]x=-1[/tex]. In fact, since this function is independent of [tex]y[/tex], every point with this [tex]x[/tex] coordinate is a minimum. However,
[tex]x^2 + 2x = (x + 1)^2 - 1 \ge -1[/tex]
for all [tex]x[/tex], so (-1, 1) and all the other points [tex](-1,y)[/tex] are actually global minima.
For the fourth function, check the sign of the Hessian determinant at (-1, 1).
[tex]H(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -2/x^3 & 1 \\ 1 & -2/y^3 \end{bmatrix} \implies \det H(-1,-1) = 3 > 0[/tex]
The second derivative with respect to [tex]x[/tex] is -2/(-1) = 2 > 0, so (-1, -1) is indeed a local minimum.
The correct choice is the fourth function.
