The stationary point for $ y=(2+x)e^{-x} $ is:

Question

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Restating the problem: determine the stationary point(s) of the function y = (2 + x) e^{-x}.

First, I compute the derivative using the product rule on (2 + x) and e^{-x}:
- The derivative of (2 + x) is 1.
- The derivative of e^{-x} is -e^{-x}.
Thus, y' = 1 * e^{-x} + (2 + x) * (-e^{-x}) = e^{-x} - (2 + x) e^{-x}.

Factor out e^{-x} (which is never zero for any real x): y' = e^{-x} [1 - (2 + x)] = e^{-x} (-1 - x).

To find stationary points, set y' = 0. Since e^{-x} ≠ 0 for all x, we require -1 - x = 0, which yields x = -1.

Next, compute the corresponding y-coordinate by substituting x = -1 into the original function:
y = (2 + (-1)) e^{-(-1)} = 1 * e^{1} = e.

Therefore, the calculations indicate that the coordinate where the slope is zero is at x = -1, with the corresponding y-value equal to e. The result is the ordered pair consisting of these computed values.

If you compare this with any options given, you would look for the point where the x-coordinate is -1 and the y-coordinate is e. Note that the derivative analysis hinges on the fact that e^{-x} never vanishes, which is why the zero of y' comes solely from -1 - x = 0.

In this solution, the key steps were: differentiate via product rule, factor out the nonzero exponential term, set the remaining linear factor to zero, and substitute back to find the y-coordinate.

MUF0091 Mathematics Unit 1 - Semester 1, 2025 12.7 Study Area 3 revision quiz

The stationary point for \( y=(2+x)e^{-x} \) is:

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