Determine the stationary points of y = (x²+ lambda)e^-x-1
Determine the stationary points of $y = \bigl(x^{2}+\lambda\bigr)e^{-x}-1$ for the various values of the parameter $\lambda\in\mathbb{R}$.
Explanation of Method
A discriminant decides how many there are.
Step 1: The derivative
\[ y' = 2x\,e^{-x}-\bigl(x^{2}+\lambda\bigr)e^{-x} = e^{-x}\left(-x^{2}+2x-\lambda\right). \tag{1} \]
Since $e^{-x}\gt 0$, the stationary points are the roots of
\[ x^{2}-2x+\lambda = 0, \qquad \Delta = 4\bigl(1-\lambda\bigr). \tag{2} \]
Step 2: The three cases
\[ \lambda\gt 1:\ \Delta\lt 0,\ \text{no stationary points}; \qquad \lambda = 1:\ x = 1 \ \text{(double)}; \qquad \lambda\lt 1:\ x_{1,2} = 1\mp\sqrt{1-\lambda}. \]
Step 3: The height of a stationary point
At a root, $\lambda = 2x_{0}-x_{0}^{2}$, so $x_{0}^{2}+\lambda = 2x_{0}^{2}-2x_{0}$ and
\[ y_{0} = 2x_{0}\bigl(x_{0}-1\bigr)e^{-x_{0}}-1. \tag{3} \]
Step 4: Classification
Differentiating (1), $y'' = e^{-x}\left(x^{2}-4x+\lambda+2\right)$. and substituting $\lambda = 2x_{0}-x_{0}^{2}$ collapses the bracket: \[y''\bigl(x_{0}\bigr) = 2e^{-x_{0}}\bigl(1-x_{0}\bigr). \tag{4}\]
Hence $x_{0}\lt 1$ gives a minimum, $x_{0}\gt 1$ a maximum. For $\lambda\lt 1$: \[x_{1} = 1-\sqrt{1-\lambda} \ \text{(minimum)}, \qquad x_{2} = 1+\sqrt{1-\lambda} \ \text{(maximum)}.\]
Step 5: The borderline case $\lambda = 1$
Here $y''(1) = 0$, so the second-derivative test fails. Differentiating once more, \[y''' = e^{-x}\left(-x^{2}+6x-\lambda-6\right), \qquad y'''(1) = e^{-1}\bigl(-1+6-1-6\bigr) = -\frac{2}{e}\neq0,\] so with $y' = y'' = 0$ and $y'''\neq0$ the point $x = 1$ is a stationary inflection, not an extremum, and $y(1) = 2\cdot1\cdot0\cdot e^{-1}-1 = -1$.
Step 6: Verification
\[\lambda = 0:\ x_{1} = 0,\ y_{1} = -1 \ (\text{min}); \qquad x_{2} = 2,\ y_{2} = 4e^{-2}-1 = -0.4586589 \ (\text{max}) \;\;\checkmark\]
\[\lambda = -3:\ x_{1} = -1,\ y_{1} = 4e-1 = 9.8731273 \ (\text{min}); \qquad x_{2} = 3,\ y_{2} = 12e^{-3}-1 = -0.4025552 \ (\text{max}) \;\;\checkmark\]
\[\text{The minimum value can exceed the maximum value (as at } \lambda = -3),\ \text{the two lying far apart} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lambda>1:\ \text{no stationary points}\,}\]
Final answer:
\[\boxed{\,\lambda = 1:\ \text{one, at } \bigl(1,-1\bigr),\ \text{a stationary inflection}\,}\]
Final answer:
\[\boxed{\,\lambda<1:\ \text{two, } x_{1,2} = 1\mp\sqrt{1-\lambda}:\ \text{a minimum at } x_{1},\ \text{a maximum at } x_{2},\ \text{heights from (3)}\,}\]
Remark
Remark. The parameter enters only through the constant term of the quadratic $x^{2}-2x+\lambda$, so the two stationary abscissae are always symmetric about $x = 1$ — and $x = 1$ is precisely where (4) changes sign. That is why the left root is always the minimum and the right one always the maximum, and why at $\lambda = 1$, when they merge, an inflection is all that remains. The last verification line is another instance of Problem 430.