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Question Number 171032 by Mastermind last updated on 06/Jun/22

Find the domain and range of the  function, f(x)=((x^2 +2)/(2x+1))    Mastermind

$${Find}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{the} \\ $$$${function},\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by cortano1 last updated on 06/Jun/22

 D_f = x≠−(1/2) ; x∈R   2xy+y=x^2 +2  ⇒x^2 −2xy+2−y=0  ⇒Δ = 4y^2 −4.1.(2−y)≥0  ⇒y^2 +y−2≥0  ⇒(y+2)(y−1)≥0  ⇒R_f  : y≤−2 ∨ y≥1

$$\:{D}_{{f}} =\:{x}\neq−\frac{\mathrm{1}}{\mathrm{2}}\:;\:{x}\in\mathbb{R} \\ $$$$\:\mathrm{2}{xy}+{y}={x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{2}−{y}=\mathrm{0} \\ $$$$\Rightarrow\Delta\:=\:\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\left(\mathrm{2}−{y}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{y}^{\mathrm{2}} +{y}−\mathrm{2}\geqslant\mathrm{0} \\ $$$$\Rightarrow\left({y}+\mathrm{2}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow{R}_{{f}} \::\:{y}\leqslant−\mathrm{2}\:\vee\:{y}\geqslant\mathrm{1} \\ $$$$ \\ $$

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