Question

f(x)=x^2-x+2 supra x^2+1.Calculati Imf​

Answer 1

$y = \dfrac{x^2-x+2}{x^2+1} \\ \\ y(x^2+1) = x^2-x+2 \\ \\yx^2+y-x^2+x-2 = 0\\ \\(y-1)x^2+x+y-2 = 0$

Condiția de existență pentru x este:

$\Delta_x \geq 0 \Rightarrow 1 - 4(y-1)(y-2) \geq 0 \Rightarrow 4(y-1)(y-2) -1 \leq 0 \Rightarrow \\ \\ \Rightarrow 4(y^2-3y+2)-1 \leq 0 \Rightarrow 4y^2-12y+7\leq 0\\ \\ \Delta_y = 144 - 16\cdot 7 = 32 \Rightarrow y_{1,2} = \dfrac{12\pm 4\sqrt{2}}{8}= \dfrac{3\pm \sqrt{2}}{2}\\ \\ \\ \Rightarrow y \in \Bigg[\dfrac{3-\sqrt 2}{2},\, \dfrac{3+\sqrt 2}{2}\Bigg] \Rightarrow Imf = \Bigg[\dfrac{3-\sqrt 2}{2},\, \dfrac{3+\sqrt 2}{2}\Bigg]$