Question Number 12552 by @ANTARES_VY last updated on 25/Apr/17
$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{expression}}. \\ $$$$\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{8}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{3}.} \\ $$
Answered by ajfour last updated on 25/Apr/17
![if you mean the range then y=((x^2 +2x+8)/(x^2 +2x+3)) = 1+(5/((x+1)^2 +2)) y_(max) =1+(5/2)=(7/2) when (x+1)=0 y_(min) → 1 when x→∞ so the range of the expression is y∈(1,(7/2)] .](Q12559.png)
$${if}\:{you}\:{mean}\:{the}\:{range}\:{then} \\ $$$${y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{8}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\:=\:\mathrm{1}+\frac{\mathrm{5}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$${y}_{{max}} =\mathrm{1}+\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{2}}\:\:{when}\:\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${y}_{{min}} \rightarrow\:\mathrm{1}\:{when}\:{x}\rightarrow\infty \\ $$$${so}\:{the}\:{range}\:{of}\:{the}\:{expression}\:{is} \\ $$$${y}\in\left(\mathrm{1},\frac{\mathrm{7}}{\mathrm{2}}\right]\:. \\ $$