The stationary point for [math: y=(2+x)e−x] y=(2+x)e^{-x} is:

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We start by restating the problem to ensure clarity: given y = (2 + x) e^{-x}, find the stationary point.
Option a says the point is (-3, e^3). This would imply x = -3, so y should be (2 + (-3)) e^{3} = (-1) e^{3} = -e^{3}, which is neither e^3 nor any positive value. The coordinates in a do not satisfy the function, so this is incorrect.
Option b claims (1, 3 e^{-1}). If x = 1, then y = (2 + 1) e^{-1} = 3 e^{-1}. So the coordinate pair (1, 3 e^{-1}) does lie on the curve y = (x + 2) e^{-x}. However, a stationary point requires the derivative to be zero, not just lying on the curve.
Option c provides (-2, 0). Substituting x = -2 gives y = (0) e^{2} = 0, so the point is on the curve, but again this does not guarantee a stationary point without checking the derivative. We must compute y' to verify stationarity.
Option d presents (-1, e). If x = -1, then y = (2 + (-1)) e^{-(-1)} = 1 \cdot e^{1} = e, so this point lies on the curve. To test for a stationary point, differentiate: y = (x + 2) e^{-x}. Use the product rule: y' = (1) e^{-x} + (x + 2)(-e^{-x}) = e^{-x} - (x + 2) e^{-x} = [1 - (x + 2)] e^{-x} = (-x - 1) e^{-x}. Setting y' = 0 requires -x - 1 = 0, hence x = -1 (since e^{-x} ≠ 0 for all x). Therefore, the stationary point occurs at x = -1, and the corresponding y-value is e, giving the point (-1, e). Thus, option d correctly identifies the stationary point.
In summary, option a is incompatible with the derivative and the algebra of the function; option b, while on the curve, does not check the critical condition for a stationary point; option c lies on the curve but again does not satisfy y' = 0; option d satisfies both being on the curve and having y' = 0, making it the stationary point. The logical flow requires first confirming the derivative equals zero to locate stationary points, then verifying the coordinates on the original function, which confirms (-1, e) as the stationary point.

The stationary point for [math: y=(2+x)e−x] y=(2+x)e^{-x} is:

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