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Sagot :

ECTO220

Réponse :

Bonsoir

[tex]e^{-x^{2}-16 } \leq \frac{1}{e^{10x} }[/tex]

⇔ [tex]e^{-x^{2}-16 } \leq e^{-10x}[/tex]

⇔ -x² - 16 ≤ -10x

⇔ -x² + 10x - 16 ≤ 0

Δ = 36

x1 = (-10-6)/(-2) = 8   et x2 = (-10+6)/(-2) = 2

S = ]-∞ ; 2] ∪ [8 ; +∞[

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