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Question Number 79883 by MJS last updated on 29/Jan/20

to Sir Jagoll (and of course everybody else) (1) y=((x^2 −x−6)/(x^2 −3x−4))= =(((x−3)(x+2))/((x−4)(x+1))) ⇒ ⇒ { ((zeros at x=−2; x=3)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x−6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y−3))/(y−1))=0 D=((25y^2 −46y+25)/(4(y−1)^2 ))>0∀x∈R ⇒ ⇒ { ((range=R)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=1 ⇒ y=1 y′=−((2(x^2 −2x+7))/((x−4)^2 (x+1)^2 )) no real zeros ⇒ ⇒ no local extremes y′′=((4(x^3 −3x^2 +21x−25))/((x−4)^3 (x+1)^3 ))=0 at x≈1.33131 ⇒ ⇒ turning point (2) y=((x^2 −x−6)/(x^2 −3x+4))= =(((x−3)(x+2))/(x^2 −3x+4)); x^2 −3x+4=0 no real zeros ⇒ ⇒ { ((zeros at x=−2; x=3)),((no vertical asymptote)) :} defined for x∈R range: transforming y=((x^2 −x−6)/(x^2 −3x+4)) to x^2 −((3y−1)/(y−1))x+((2(2y+3))/(y−1))=0 D=−((7y^2 +14y−25)/(4(y−1)^2 ))≥0 for −1−((4(√(14)))/7)≤y≤−1+((4(√(14)))/7) ⇒ ⇒ { ((range=[−1−((4(√(14)))/7); −1+((4(√(14)))/7)])),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=5 ⇒ y=1 y′=−((2(x^2 −10x+11))/((x^2 −3x+4)^2 ))=0 at x=5±(√(14)) y^(′′) =((4(x^3 −15x^2 +33x−13))/((x^2 −3x+4)^3 )) y′′ { ((>0 at x=5−(√(14)) ⇒ local minimum)),((<0 at x=5+(√(14)) ⇒ local maximum)),((=0 at { ((x≈.506699)),((x≈2.06421)),((x≈12.4291)) :} ⇒ 3 turning points )) :} (3) y=((x^2 −x+6)/(x^2 −3x−4))= =((x^2 −x+6)/((x−4)(x+1))) ⇒ ⇒ { ((no real zeros)),((vertical asymptotes at x=−1; x=4)) :} defined for x∈R\{−1; 4} range: transforming y=((x^2 −x+6)/(x^2 −3x−4)) to x^2 −((3y−1)/(y−1))x−((2(2y+3))/(y−1))=0 D=((25y^2 +2y−23)/(4(y−1)^2 ))<0 for −1<y<((23)/(25)) ⇒ ⇒ { ((range=R\]−1; ((23)/(25))[)),((horizontal asymptote at y=1 (∗))) :} (∗) doesn′t mean y=1 is not within range!!! x=−5 ⇒ y=1 y′=−((2(x^2 +10x−11))/((x−4)^2 (x+1)^2 ))=0 at x=−11; x=1 y′′=((4(x^3 +15x^2 −33x+53))/((x−4)^3 (x+1)^3 )) y′′ { ((>0 at x=−11 ⇒ local minimum)),((<0 at x=1 ⇒ local maximum)),((=0 at x≈−17.1098 ⇒ turning point)) :}

$$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}−\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} −\mathrm{46}{y}+\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }>\mathrm{0}\forall{x}\in\mathbb{R}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\mathbb{R}}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{1}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{7}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{no}\:\mathrm{local}\:\mathrm{extremes} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{21}{x}−\mathrm{25}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} }=\mathrm{0}\:\mathrm{at}\:{x}\approx\mathrm{1}.\mathrm{33131}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{turning}\:\mathrm{point} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}};\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0}\:\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{no}\:\mathrm{vertical}\:\mathrm{asymptote}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}+\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=−\frac{\mathrm{7}{y}^{\mathrm{2}} +\mathrm{14}{y}−\mathrm{25}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }\geqslant\mathrm{0}\:\mathrm{for}\:−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\leqslant{y}\leqslant−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{range}=\left[−\mathrm{1}−\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}};\:−\mathrm{1}+\frac{\mathrm{4}\sqrt{\mathrm{14}}}{\mathrm{7}}\right]}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{11}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}\pm\sqrt{\mathrm{14}} \\ $$$${y}^{''} =\frac{\mathrm{4}\left({x}^{\mathrm{3}} −\mathrm{15}{x}^{\mathrm{2}} +\mathrm{33}{x}−\mathrm{13}\right)}{\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}−\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{5}+\sqrt{\mathrm{14}}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:\begin{cases}{{x}\approx.\mathrm{506699}}\\{{x}\approx\mathrm{2}.\mathrm{06421}}\\{{x}\approx\mathrm{12}.\mathrm{4291}}\end{cases}\:\Rightarrow\:\mathrm{3}\:\mathrm{turning}\:\mathrm{points}\:}\end{cases} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{no}\:\mathrm{real}\:\mathrm{zeros}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\ $$$$\mathrm{range}:\:\mathrm{transforming}\:{y}=\frac{{x}^{\mathrm{2}} −{x}+\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}\:\mathrm{to} \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{3}{y}−\mathrm{1}}{{y}−\mathrm{1}}{x}−\frac{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{3}\right)}{{y}−\mathrm{1}}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{25}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{23}}{\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} }<\mathrm{0}\:\mathrm{for}\:−\mathrm{1}<{y}<\frac{\mathrm{23}}{\mathrm{25}}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\left.\mathrm{range}=\mathbb{R}\backslash\right]−\mathrm{1};\:\frac{\mathrm{23}}{\mathrm{25}}\left[\right.}\\{\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\mathrm{1}\:\left(\ast\right)}\end{cases} \\ $$$$\left(\ast\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{mean}\:{y}=\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{within}\:\mathrm{range}!!! \\ $$$$\:\:\:\:\:\:\:\:{x}=−\mathrm{5}\:\Rightarrow\:{y}=\mathrm{1} \\ $$$${y}'=−\frac{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{11}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11};\:{x}=\mathrm{1} \\ $$$${y}''=\frac{\mathrm{4}\left({x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} −\mathrm{33}{x}+\mathrm{53}\right)}{\left({x}−\mathrm{4}\right)^{\mathrm{3}} \left({x}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${y}''\begin{cases}{>\mathrm{0}\:\mathrm{at}\:{x}=−\mathrm{11}\:\Rightarrow\:\mathrm{local}\:\mathrm{minimum}}\\{<\mathrm{0}\:\mathrm{at}\:{x}=\mathrm{1}\:\Rightarrow\:\mathrm{local}\:\mathrm{maximum}}\\{=\mathrm{0}\:\mathrm{at}\:{x}\approx−\mathrm{17}.\mathrm{1098}\:\Rightarrow\:\mathrm{turning}\:\mathrm{point}}\end{cases} \\ $$

Commented by TawaTawa last updated on 29/Jan/20

Thanks for your time sir, God bless you.

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Commented by jagoll last updated on 29/Jan/20

waw...thank you mister. i was confused, my teacher said that horizontal asymptotes were not include in the range.

$$\mathrm{waw}...\mathrm{thank}\:\mathrm{you}\:\mathrm{mister}.\:\mathrm{i}\:\mathrm{was}\: \\ $$$$\mathrm{confused},\:\mathrm{my}\:\mathrm{teacher}\:\mathrm{said}\:\mathrm{that}\:\mathrm{horizontal} \\ $$$$\mathrm{asymptotes}\:\mathrm{were}\:\mathrm{not}\:\mathrm{include}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{range}. \\ $$

Commented by jagoll last updated on 29/Jan/20

mr. if given function y= ((ax+b)/(cx+d )) , is it true that the range is not the same as the horizontal asymptote? R_f : y ≠ (a/c) ?

$$\mathrm{mr}.\:\mathrm{if}\:\mathrm{given}\:\mathrm{function}\: \\ $$$$\mathrm{y}=\:\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}\:}\:,\:\mathrm{is}\:\mathrm{it}\:\mathrm{true}\:\mathrm{that}\:\mathrm{the}\:\mathrm{range} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{the}\:\mathrm{horizontal} \\ $$$$\mathrm{asymptote}?\: \\ $$$$\mathrm{R}_{\mathrm{f}} \::\:\mathrm{y}\:\neq\:\frac{\mathrm{a}}{\mathrm{c}}\:? \\ $$

Commented by john santu last updated on 29/Jan/20

for example if f(x) = ((2x−1)/(x+3)) domain x ≠ −3 if y = 2 ⇒ 2=((2x−1)/(x+3)) 2x+6 = 2x−1 ⇒ 6=−1 ?? it true for function y = ((ax+b)/(cx+d)) range y ≠ (a/c)

$$\mathrm{for}\:\mathrm{example}\: \\ $$$$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}+\mathrm{3}}\: \\ $$$$\mathrm{domain}\:\mathrm{x}\:\neq\:−\mathrm{3}\: \\ $$$$\mathrm{if}\:\mathrm{y}\:=\:\mathrm{2}\:\Rightarrow\:\mathrm{2}=\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}+\mathrm{3}} \\ $$$$\mathrm{2x}+\mathrm{6}\:=\:\mathrm{2x}−\mathrm{1}\:\Rightarrow\:\mathrm{6}=−\mathrm{1}\:?? \\ $$$$\mathrm{it}\:\mathrm{true}\:\mathrm{for}\:\mathrm{function}\:\mathrm{y}\:=\:\frac{\mathrm{ax}+\mathrm{b}}{\mathrm{cx}+\mathrm{d}} \\ $$$$\mathrm{range}\:\mathrm{y}\:\neq\:\frac{\mathrm{a}}{\mathrm{c}} \\ $$

Commented by jagoll last updated on 29/Jan/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 29/Jan/20

y=((ax+b)/(cx+d)) ⇒ { ((zero at x=−(b/a))),((vertical asymptote at x=−(d/c) )) :} defined for x∈R\{−(d/c)} y=((ax+b)/(cx+d)) ⇔ x=−((dy−b)/(cy−a)) ⇒ range=R\{(a/c)} horizontal asymptote at y=(a/c) so you are right

$${y}=\frac{{ax}+{b}}{{cx}+{d}}\:\Rightarrow\:\begin{cases}{\mathrm{zero}\:\mathrm{at}\:{x}=−\frac{{b}}{{a}}}\\{\mathrm{vertical}\:\mathrm{asymptote}\:\mathrm{at}\:{x}=−\frac{{d}}{{c}}\:}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\frac{{d}}{{c}}\right\} \\ $$$${y}=\frac{{ax}+{b}}{{cx}+{d}}\:\Leftrightarrow\:{x}=−\frac{{dy}−{b}}{{cy}−{a}}\:\Rightarrow\:\mathrm{range}=\mathbb{R}\backslash\left\{\frac{{a}}{{c}}\right\} \\ $$$$\mathrm{horizontal}\:\mathrm{asymptote}\:\mathrm{at}\:{y}=\frac{{a}}{{c}} \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$

Commented by jagoll last updated on 29/Jan/20

sir your notation is ]14−4(√(6 )) ; 14+4(√6) [ it same as y ≤ 14−4(√6) ∨y ≥ 14+4(√6) ?

$$\mathrm{sir}\:\mathrm{your}\:\mathrm{notation}\:\mathrm{is} \\ $$$$\left.\right]\mathrm{14}−\mathrm{4}\sqrt{\mathrm{6}\:}\:;\:\mathrm{14}+\mathrm{4}\sqrt{\mathrm{6}}\:\left[\:\right. \\ $$$$\mathrm{it}\:\mathrm{same}\:\mathrm{as}\:\mathrm{y}\:\leqslant\:\mathrm{14}−\mathrm{4}\sqrt{\mathrm{6}}\:\vee\mathrm{y}\:\geqslant\:\mathrm{14}+\mathrm{4}\sqrt{\mathrm{6}} \\ $$$$? \\ $$

Commented by MJS last updated on 29/Jan/20

depends... y∈[a; b] is the same as a≤y≤b ⇒ y∈R\[a; b] means y<a∨y>b y∈]a; b[ is the same as a<y<b ⇒ y∈R\]a; b[ means y≤a∨y≥b

$$\mathrm{depends}... \\ $$$${y}\in\left[{a};\:{b}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:{a}\leqslant{y}\leqslant{b} \\ $$$$\Rightarrow\:{y}\in\mathbb{R}\backslash\left[{a};\:{b}\right]\:\mathrm{means}\:{y}<{a}\vee{y}>{b} \\ $$$$\left.{y}\in\right]{a};\:{b}\left[\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:{a}<{y}<{b}\right. \\ $$$$\left.\Rightarrow\:{y}\in\mathbb{R}\backslash\right]{a};\:{b}\left[\:\mathrm{means}\:{y}\leqslant{a}\vee{y}\geqslant{b}\right. \\ $$

Commented by jagoll last updated on 29/Jan/20

thanks you mister

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{mister} \\ $$

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