Question

f:R->R f(x)= $x^{4} -4x+4$ demonstrati ca f(x) + f(1/x) ≥2

Answer 1

$f:\mathbb{R}\to\mathbb{R},\quad f(x) = x^4-4x+4\\ \\ f'(x) = 4x^3-4\\ f'(x) = 0 \Rightarrow 4x^3 = 4 \Rightarrow x^3 = 1 \Rightarrow x = 1\\ \\ \lim\limits_{x\to -\infty}f(x) = +\infty,\quad \lim\limits_{x\to +\infty}f(x) = +\infty\\ \\ \Rightarrow x = 1 \text{ punct de minim global} \\ \\ \Rightarrow f(x) \geq f(1) \Rightarrow f(x) \geq 1^4-4+4$

$\\\Rightarrow f(x) \geq 1,\quad \forall x\in \mathbb{R} \\ \\ \text{Notez }\,t = \dfrac{1}{x},\quad t\in \mathbb{R}\\\\ \Leftrightarrow f(t)\geq 1,\quad \forall t\in \mathbb{R}\\ \\ \Leftrightarrow f\Big(\dfrac{1}{x}\Big)\geq 1,\quad \forall\dfrac{1}{x}\in \mathbb{R}\,\Leftrightarrow \,x\in \mathbb{R}^*$

Adunăm cele două inegalități:

$\Rightarrow f(x)+f\Big(\dfrac{1}{x}\Big)\geq 1+1 \\ \\ \Rightarrow f(x) + f\Big(\dfrac{1}{x}\Big)\geq 2,\quad \forall x\in \mathbb{R}^*$

Answer 2

Explicație pas cu pas:

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f(x)+f(1/x)≥2 ⇒(x+1)²≥0⇒x∈R*

1/x⇒x≠0