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Abipipoune06
@Abipipoune06
May 2019
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Bonjour , vraiment besoin d aide en maths merci d avance je suis bloquée depuis 2 semaines.
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scoladan
Verified answer
Bonjour,
Exo 2
f(x) = ln(x) et g(x) = x
1) voir ci-joint
on peut donc conjecturer que (C) est toujours en-dessous de (Δ).
2) d(x) = f(x) - g(x) = ln(x) - x
d'(x) = 1/x - 1 = (1 - x)/x
x 0 1 +∞
(1 - x) - 0 +
d'(x) || + 0 -
d(x) || crois. décrois.
lim d(x) quand x → 0 = -∞
d(1) = -1
et lim d(x) quand x → +∞ = lim x(ln(x)/x - 1) = lim (-x) = -∞
On en déduit que pour tout x ∈ ]0;+∞[, d(x) < 0
et donc que ln(x) < x
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Bonjour,Exo 2
f(x) = ln(x) et g(x) = x
1) voir ci-joint
on peut donc conjecturer que (C) est toujours en-dessous de (Δ).
2) d(x) = f(x) - g(x) = ln(x) - x
d'(x) = 1/x - 1 = (1 - x)/x
x 0 1 +∞
(1 - x) - 0 +
d'(x) || + 0 -
d(x) || crois. décrois.
lim d(x) quand x → 0 = -∞
d(1) = -1
et lim d(x) quand x → +∞ = lim x(ln(x)/x - 1) = lim (-x) = -∞
On en déduit que pour tout x ∈ ]0;+∞[, d(x) < 0
et donc que ln(x) < x