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Question Number 79816 by jagoll last updated on 28/Jan/20
hello mister. i need help explaining how determine the range of function of rational functions like (i) f(x)=((ax^2 +bx+c)/(px^2 +qx+r)) (ii) f(x)=((ax^2 +bx+c)/(px+q))
$$\mathrm{hello}\:\mathrm{mister}. \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{help}\:\mathrm{explaining}\:\mathrm{how}\:\mathrm{determine} \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function} \\ $$$$\mathrm{of}\:\mathrm{rational}\:\mathrm{functions} \\ $$$$\mathrm{like}\:\left(\mathrm{i}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{px}^{\mathrm{2}} +\mathrm{qx}+\mathrm{r}} \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}}{\mathrm{px}+\mathrm{q}} \\ $$
Answered by MJS last updated on 28/Jan/20
y=((ax^2 +bx+c)/(px^2 +qx+r)) leads to x^2 +((qy−b)/(py−a))+((ry−c)/(py−a))=0 D=(((4pr−q^2 )y^2 −2(2ar−bq+2cp)y+4ac−b^2 )/(4(py−a)^2 )) D<0 ⇒ no real solution for x ⇒ no y exists the range then is R\[interval for y where D<0] same when p=0 which is your (ii)
$${y}=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{px}^{\mathrm{2}} +{qx}+{r}} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} +\frac{{qy}−{b}}{{py}−{a}}+\frac{{ry}−{c}}{{py}−{a}}=\mathrm{0} \\ $$$${D}=\frac{\left(\mathrm{4}{pr}−{q}^{\mathrm{2}} \right){y}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}{ar}−{bq}+\mathrm{2}{cp}\right){y}+\mathrm{4}{ac}−{b}^{\mathrm{2}} }{\mathrm{4}\left({py}−{a}\right)^{\mathrm{2}} } \\ $$$${D}<\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{x}\:\Rightarrow\:\mathrm{no}\:{y}\:\mathrm{exists} \\ $$$$\mathrm{the}\:\mathrm{range}\:\mathrm{then}\:\mathrm{is}\:\mathbb{R}\backslash\left[\mathrm{interval}\:\mathrm{for}\:{y}\:\mathrm{where}\:{D}<\mathrm{0}\right] \\ $$$$\mathrm{same}\:\mathrm{when}\:{p}=\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{your}\:\left(\mathrm{ii}\right) \\ $$
Commented by jagoll last updated on 28/Jan/20
if the function y=((2x^2 −2)/(3x^2 +3x−2)) ⇒3x^2 y+3xy−2y=2x^2 −2 (3y−2).x^2 +3y.x+2−2y=0 D= 9y^2 −4.(3y−2)(2−2y) D= 9y^2 −8(3y−2)(1−y) D=9y^2 −8(−3y^2 +5y−2) D=33y^2 +40y+16 >0 for ∀y∈R R_f = {y∣ y∈R } that right sir?
$$\mathrm{if}\:\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{y}=\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}}\:\Rightarrow\mathrm{3x}^{\mathrm{2}} \mathrm{y}+\mathrm{3xy}−\mathrm{2y}=\mathrm{2x}^{\mathrm{2}} −\mathrm{2} \\ $$$$\left(\mathrm{3y}−\mathrm{2}\right).\mathrm{x}^{\mathrm{2}} +\mathrm{3y}.\mathrm{x}+\mathrm{2}−\mathrm{2y}=\mathrm{0} \\ $$$$\mathrm{D}=\:\mathrm{9y}^{\mathrm{2}} −\mathrm{4}.\left(\mathrm{3y}−\mathrm{2}\right)\left(\mathrm{2}−\mathrm{2y}\right) \\ $$$$\mathrm{D}=\:\mathrm{9y}^{\mathrm{2}} −\mathrm{8}\left(\mathrm{3y}−\mathrm{2}\right)\left(\mathrm{1}−\mathrm{y}\right) \\ $$$$\mathrm{D}=\mathrm{9y}^{\mathrm{2}} −\mathrm{8}\left(−\mathrm{3y}^{\mathrm{2}} +\mathrm{5y}−\mathrm{2}\right) \\ $$$$\mathrm{D}=\mathrm{33y}^{\mathrm{2}} +\mathrm{40y}+\mathrm{16}\:>\mathrm{0}\:\mathrm{for}\:\forall\mathrm{y}\in\mathbb{R} \\ $$$$\mathrm{R}_{\mathrm{f}} \:=\:\left\{\mathrm{y}\mid\:\mathrm{y}\in\mathbb{R}\:\right\} \\ $$$$\mathrm{that}\:\mathrm{right}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 28/Jan/20
range function not related to flat asymtot sir?
$$\mathrm{range}\:\mathrm{function}\:\mathrm{not}\:\mathrm{related}\:\mathrm{to} \\ $$$$\mathrm{flat}\:\mathrm{asymtot}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 28/Jan/20
for example y= ((2x^2 +3x−2)/(3x^2 −5x+2)) R_f : y ≠ lim_(x→∞) ((2x^2 +3x−2)/(3x^2 −5x+2)) R_f : y≠ (2/3) ?
$$\mathrm{for}\:\mathrm{example}\: \\ $$$$\mathrm{y}=\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}}{\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{2}} \\ $$$$\mathrm{R}_{\mathrm{f}} \::\:\mathrm{y}\:\neq\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{2}}{\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{2}} \\ $$$$\mathrm{R}_{\mathrm{f}} :\:\mathrm{y}\neq\:\frac{\mathrm{2}}{\mathrm{3}}\:? \\ $$
Commented by MJS last updated on 28/Jan/20
y=((2x^2 −2)/(3x^2 +3x−2)) x^2 +((3y)/(3y−2))x−((2(y−1))/(3y−2))=0 [x^2 +Px+Q=0 ⇒ D=(P^2 /4)−Q] D=((33y^2 −40y+16)/(4(3y−2)^2 ))>0∀y∈R ⇒ range is R
$${y}=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}{y}}{\mathrm{3}{y}−\mathrm{2}}{x}−\frac{\mathrm{2}\left({y}−\mathrm{1}\right)}{\mathrm{3}{y}−\mathrm{2}}=\mathrm{0} \\ $$$$\:\:\:\:\:\left[{x}^{\mathrm{2}} +{Px}+{Q}=\mathrm{0}\:\Rightarrow\:{D}=\frac{{P}^{\mathrm{2}} }{\mathrm{4}}−{Q}\right] \\ $$$${D}=\frac{\mathrm{33}{y}^{\mathrm{2}} −\mathrm{40}{y}+\mathrm{16}}{\mathrm{4}\left(\mathrm{3}{y}−\mathrm{2}\right)^{\mathrm{2}} }>\mathrm{0}\forall{y}\in\mathbb{R} \\ $$$$\Rightarrow\:\mathrm{range}\:\mathrm{is}\:\mathbb{R} \\ $$
Commented by MJS last updated on 28/Jan/20
y=((2x^2 +3x−2)/(3x^2 −5x+2)) x^2 −((5y+3)/(3y−2))x+((2(y+1))/(3y−2))=0 D=((y^2 +22y+25)/(4(3y−2)^2 )) D<0 for −11−4(√6)<y<−11+4(√6) ⇒ range is R\]−11−4(√6); −11+4(√6)[
$${y}=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{5}{y}+\mathrm{3}}{\mathrm{3}{y}−\mathrm{2}}{x}+\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{\mathrm{3}{y}−\mathrm{2}}=\mathrm{0} \\ $$$${D}=\frac{{y}^{\mathrm{2}} +\mathrm{22}{y}+\mathrm{25}}{\mathrm{4}\left(\mathrm{3}{y}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${D}<\mathrm{0}\:\mathrm{for}\:−\mathrm{11}−\mathrm{4}\sqrt{\mathrm{6}}<{y}<−\mathrm{11}+\mathrm{4}\sqrt{\mathrm{6}} \\ $$$$\left.\Rightarrow\:\mathrm{range}\:\mathrm{is}\:\mathbb{R}\backslash\right]−\mathrm{11}−\mathrm{4}\sqrt{\mathrm{6}};\:−\mathrm{11}+\mathrm{4}\sqrt{\mathrm{6}}\left[\right. \\ $$
Commented by jagoll last updated on 28/Jan/20
sir how to get in line 2? still understand
$$\mathrm{sir}\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{in}\:\mathrm{line}\:\mathrm{2}?\:\mathrm{still}\:\mathrm{understand} \\ $$
Commented by jagoll last updated on 28/Jan/20
3x^2 y+3xy−2y−2x^2 +2=0 (3y−2)x^2 +(3y)x+2−2y=0 x^2 +((3y)/(3y−2)) x+((2(1−y))/(3y−2))=0 that right sir?
$$\mathrm{3x}^{\mathrm{2}} \mathrm{y}+\mathrm{3xy}−\mathrm{2y}−\mathrm{2x}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{3y}−\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{3y}\right)\mathrm{x}+\mathrm{2}−\mathrm{2y}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{3y}}{\mathrm{3y}−\mathrm{2}}\:\mathrm{x}+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{y}\right)}{\mathrm{3y}−\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{that}\:\mathrm{right}\:\mathrm{sir}? \\ $$
Commented by MJS last updated on 28/Jan/20
y=((2x^2 +3x−2)/(3x^2 −5x+2)) (3x^2 −5x+2)y=2x^2 +3x−2 (3y−2)x^2 −(5y+3)x+2(y+1)=0 x^2 −((5y+3)/(3y−2))x+((2(y+1))/(3y−2))=0 you′ve made mistakes in line 1 p=−((5y+3)/(3y−2)); q=((2(y+1))/(3y−2)) D=(p^2 /4)−q=((y^2 +22y+25)/(4(3y−2)^2 ))
$${y}=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}} \\ $$$$\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2}\right){y}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{2} \\ $$$$\left(\mathrm{3}{y}−\mathrm{2}\right){x}^{\mathrm{2}} −\left(\mathrm{5}{y}+\mathrm{3}\right){x}+\mathrm{2}\left({y}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{5}{y}+\mathrm{3}}{\mathrm{3}{y}−\mathrm{2}}{x}+\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{\mathrm{3}{y}−\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{you}'\mathrm{ve}\:\mathrm{made}\:\mathrm{mistakes}\:\mathrm{in}\:\mathrm{line}\:\mathrm{1} \\ $$$$ \\ $$$${p}=−\frac{\mathrm{5}{y}+\mathrm{3}}{\mathrm{3}{y}−\mathrm{2}};\:{q}=\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{\mathrm{3}{y}−\mathrm{2}} \\ $$$${D}=\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}=\frac{{y}^{\mathrm{2}} +\mathrm{22}{y}+\mathrm{25}}{\mathrm{4}\left(\mathrm{3}{y}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$
Commented by MJS last updated on 29/Jan/20
look at qu. 79883
$$\mathrm{look}\:\mathrm{at}\:\mathrm{qu}.\:\mathrm{79883} \\ $$

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