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1D2010
@1D2010
May 2019
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Bonjour svp aider moi svp c'est les fonctions
merci d'avance
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scoladan
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Bonjour,
1) ∀x∈R, x ≥ E(x)
⇒ x/3 ≥ E(x/3)
⇒ x ≥ 3E(x/3)
⇒ x - 3E(x/3) ≥ 0
⇒ Df = R
∀k∈N*,
f(x + 3k) = √[x + 3k - 3E((x + 3k)/3)]
= √[x + 3k - 3E(x/3 + k)]
= √[x + 3k - 3E(x/3) - 3E(k)]
= √[x + 3k - 3E(x/3) - 3k]
= f(x)
donc f périodique de période T = 3
2) ∀x∈R, il existe n∈Z / n ≤ x < n+1 (et E(x) = n)
⇒ ∀ x∈[n;n+1[, u(x) = x - 3E(x/3) = x - 3E(n/3) est une fonction affine croissante et positive
Et v(x) = √x est croissante sur R+.
Donc f est croissante sur [n;n+1[
⇒ f(0) = 0 et ∀ x ∈ [0;3[, f(x) ≤ √3
3) ∀ x ∈ [0;3[, f(x) = √[x - 3E(x/3)] = √[x - 3x0] = √x
4) ci-joint
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Salut tout le monde! !C vraiment trop urgent! !!Merci d'avance.
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Lista de comentários
Verified answer
Bonjour,1) ∀x∈R, x ≥ E(x)
⇒ x/3 ≥ E(x/3)
⇒ x ≥ 3E(x/3)
⇒ x - 3E(x/3) ≥ 0
⇒ Df = R
∀k∈N*,
f(x + 3k) = √[x + 3k - 3E((x + 3k)/3)]
= √[x + 3k - 3E(x/3 + k)]
= √[x + 3k - 3E(x/3) - 3E(k)]
= √[x + 3k - 3E(x/3) - 3k]
= f(x)
donc f périodique de période T = 3
2) ∀x∈R, il existe n∈Z / n ≤ x < n+1 (et E(x) = n)
⇒ ∀ x∈[n;n+1[, u(x) = x - 3E(x/3) = x - 3E(n/3) est une fonction affine croissante et positive
Et v(x) = √x est croissante sur R+.
Donc f est croissante sur [n;n+1[
⇒ f(0) = 0 et ∀ x ∈ [0;3[, f(x) ≤ √3
3) ∀ x ∈ [0;3[, f(x) = √[x - 3E(x/3)] = √[x - 3x0] = √x
4) ci-joint