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11:01
This is perhaps known: e^{-x} and \exp(-x) are treated differently by Approach0.
playing with Approach0 I found another duplicate: math.stackexchange.com/questions/933462/…. However, the search engine is quite sensitive : searching with $exp(-x)$ doesn't pick proper results while $e^{-x}$ does! — Pierre H. 1 hour ago
I can confirm that $e^{-x} \geq \frac{2-x}{2+x} $ and $\exp(-x) \geq \frac{2-x}{2+x} $ return quite different results.

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