Critical Points of (x^2 + y^2) e^(-x)

Solve the easy equation first

Only the $x$ derivative needs the product rule, because $e^{-x}$ is a constant as far as $y$ is concerned.

Start with the second equation, since it is the simpler of the two. The exponential is never zero, so $2y\,e^{-x} = 0$ forces $y = 0$ and nothing else. That is worth doing before touching the first equation: substituting $y = 0$ collapses it from two variables to one.

Exactly two critical points, $(0,0)$ and $(2,0)$, and since every step was an equivalence rather than an approximation, the list is complete.

Now the second partials. Differentiating $f_x$ again in $x$ needs the product rule once more; the other two are quick.

A minimum you can prove twice, and a saddle you can see

At the origin, $f_{xx} = 2$, $f_{yy} = 2$ and $f_{xy} = 0$, so $D = 4 > 0$ with a positive diagonal entry: a local minimum with $f(0,0) = 0$. A second argument confirms it without any derivatives: $x^{2}+y^{2} \ge 0$ and $e^{-x} > 0$, so $f \ge 0$ everywhere and equals 0 only at the origin. The local minimum is in fact the global minimum.

At $(2,0)$, $f_{xy} = 0$ because $y = 0$, while $f_{xx} = e^{-2}(4 - 8 + 2) = -\frac{2}{e^{2}}$ and $f_{yy} = \frac{2}{e^{2}}$. The two diagonal entries have opposite signs, so their product is negative and $D = -\frac{4}{e^{4}} \approx -0.0733$: a saddle.

Opposite signs on the diagonal is the clearest saddle there is, because you can see both directions separately. Along the $x$-axis, $f(x, 0) = x^{2}e^{-x}$, which rises from 0, peaks at $x = 2$ with value $\frac{4}{e^{2}} \approx 0.541$, then decays. So $(2,0)$ is a maximum along that line.

Along the vertical line $x = 2$, the same point is the bottom of an upward parabola. Highest point in one direction, lowest in another: that is a saddle, and the algebra and the picture agree.

The mistake students make

The minus sign in the exponent is where this problem is won or lost. Differentiating $e^{-x}$ gives $-e^{-x}$, so the product rule produces a subtraction, not an addition.

Miss it and you get $2x + x^{2} + y^{2} = 0$, whose solutions with $y = 0$ are $x = 0$ and $x = -2$. The reported saddle then lands at $(-2, 0)$, which is not a critical point at all: the true derivative there is $e^{2}(-4 - 4) = -8e^{2}$, nowhere near zero. A sign slip does not merely shift the answer, it invents a point.

• Differentiate the exponential factor on its own and note the sign it carries.
• Assemble the product rule, then factor $e^{-x}$ back out.
• Substitute your candidate point into the original partials and confirm both come out to zero.

The other habit worth building is choosing which equation to solve first. Attacking $2x - x^{2} - y^{2} = 0$ before using $f_y = 0$ leaves a circle of candidate points and a much messier substitution. The equation with fewer terms usually carries more information, and here $f_y = 0$ pins $y$ down completely in one line.