Question

Sa se determine multimea valorilor lui f:R→R, f(x)=$\frac{x^{2}+3x+1}{x^{2}-x+1}$

Answer 1

Răspuns:

Im f = [-1/3, 5]

Explicație pas cu pas:

$f(x) = y\\\\ \frac{x^2 + 3x+1}{x^2 - x + 1} = y\\\\ x^2 + 3x + 1 = y(x^2 - x + 1)\\\\ x^2 + 3x + 1 = yx^2 - yx + y\\\\ x^2 - yx^2 + 3x + yx + 1 - y = 0\\\\ (1-y)x^2 + (3+y)x + (1-y) = 0\\\\ \Delta = (3+y)^2 - 4\cdot (1-y)\cdot (1-y) = 9 + 6y + y^2 - 4(1 -2y + y^2) = 9 + 6y + y^2 - 4 + 8y - 4y^2 = -3y^2 + 14y + 5\\\\ \Delta \geq 0\\\\ -3y^2 + 14y + 5 \geq 0\\\\ 3y^2 - 14y - 5 \leq 0\\\\ \Delta ' = 196 - 4\cdot 3 \cdot (-5) = 196 + 60 = 256\\\\ \sqrt{\Delta'} = \sqrt{2^8} = 2^4 = 16\\\\ y_{1,2} = \frac{14\pm16}{6} = \frac{7\pm8}{3}\\\\ y_1 = \frac{-1}{3}, y_2 = \frac{15}{3} = 5\\\\ \implies y \in [\frac{-1}{3}, 5] \implies Im\: f = [-\frac{1}{3},5]$

Answer 2

$f:\mathbb{R}\to \mathbb{R},\quad f(x) = \dfrac{x^2+3x+1}{x^2-x+1}$

$f(x) = \dfrac{x^2-x+1}{x^2-x+1}+\dfrac{4x}{x^2-x+1} =\\ \\= 1 +\dfrac{4x}{x^2-x+1} \\ \\ \\ f'(x) = \dfrac{4(x^2-x+1)-4x(2x-1)}{(x^2-x+1)^2} = \dfrac{4x^2-4x+4-8x^2+4x}{(x^2-x+1)^2} = \\ \\ = \dfrac{-4x^2+4}{(x^2-x+1)^2}\\ \\ f'(x) = 0 \Rightarrow -4x^2+4 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1 \\ \\ \lim\limits_{x\to -\infty} f(x) = 1,\quad \lim\limits_{x\to +\infty}f(x) = 1 \\ f(-1) = -\dfrac{1}{3},\quad f(1) = 5$

$\Rightarrow Imf = \Big[-\dfrac{1}{3},\,5\Big]$