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Question Number 60287 by naka3546 last updated on 19/May/19
f(x) = x^3 + 3x − 7 f^(−1) (x) = ?
$${f}\left({x}\right)\:\:=\:\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}\:−\:\mathrm{7} \\ $$$${f}\:^{−\mathrm{1}} \left({x}\right)\:\:=\:\:? \\ $$
Commented by mr W last updated on 19/May/19
f^(−1) (x)=((((√(x^2 +14x+53))+x+7)/2))^(1/3) −((2/( (√(x^2 +14x+53))+x+7)))^(1/3)
$${f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}+{x}+\mathrm{7}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}+{x}+\mathrm{7}}} \\ $$
Commented by naka3546 last updated on 19/May/19
please show the workings, sir .
$${please}\:\:{show}\:\:{the}\:\:{workings},\:\:{sir}\:\:. \\ $$
Answered by mr W last updated on 19/May/19
x^3 +3x−7=y x^3 +3x−(7+y)=0 let c=7+y ⇒x^3 +3x−c=0 due to (u+v)^3 −3uv(u+v)−(u^3 +v^3 )=0 x=u+v uv=−1 ⇒u^3 v^3 =−1 u^3 +v^3 =c ⇒u^3 and v^3 are roots of eqn. t^2 −ct−1=0 t=((c±(√(c^2 +4)))/2) i.e. u^3 =((c+(√(c^2 +4)))/2), v^3 =((c−(√(c^2 +4)))/2)=−(1/u^3 ) x=u+v=u−(1/u)=(((c+(√(c^2 +4)))/2))^(1/3) −((2/(c+(√(c^2 +4)))))^(1/3) c+(√(c^2 +4))=7+y+(√(y^2 +14y+49+4))=(√(y^2 +14y+53))+y+7 x=((((√(y^2 +14y+53))+y+7)/2))^(1/3) −((2/( (√(y^2 +14y+53))+y+7)))^(1/3) ⇒f^(−1) (x)=((((√(x^2 +14x+53))+x+7)/2))^(1/3) −((2/( (√(x^2 +14x+53))+x+7)))^(1/3)
$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{7}={y} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}−\left(\mathrm{7}+{y}\right)=\mathrm{0} \\ $$$${let}\:{c}=\mathrm{7}+{y} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{3}{x}−{c}=\mathrm{0} \\ $$$${due}\:{to}\:\left({u}+{v}\right)^{\mathrm{3}} −\mathrm{3}{uv}\left({u}+{v}\right)−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${x}={u}+{v} \\ $$$${uv}=−\mathrm{1}\:\Rightarrow{u}^{\mathrm{3}} {v}^{\mathrm{3}} =−\mathrm{1} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} ={c} \\ $$$$\Rightarrow{u}^{\mathrm{3}} \:{and}\:{v}^{\mathrm{3}} \:{are}\:{roots}\:{of}\:{eqn}. \\ $$$${t}^{\mathrm{2}} −{ct}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{{c}\pm\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$${i}.{e}.\:{u}^{\mathrm{3}} =\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}},\:{v}^{\mathrm{3}} =\frac{{c}−\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}=−\frac{\mathrm{1}}{{u}^{\mathrm{3}} } \\ $$$${x}={u}+{v}={u}−\frac{\mathrm{1}}{{u}}=\sqrt[{\mathrm{3}}]{\frac{{c}+\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{{c}+\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}}} \\ $$$${c}+\sqrt{{c}^{\mathrm{2}} +\mathrm{4}}=\mathrm{7}+{y}+\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{49}+\mathrm{4}}=\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}+{y}+\mathrm{7} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}+{y}+\mathrm{7}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}+{y}+\mathrm{7}}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}+{x}+\mathrm{7}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}+{x}+\mathrm{7}}} \\ $$
Commented by naka3546 last updated on 19/May/19
thank you, sir Mr W
$${thank}\:{you},\:{sir}\:\:{Mr}\:{W} \\ $$
Answered by MJS last updated on 19/May/19
x^3 +3x−7−y=0 D=(3^3 /(27))+(((−7−y)^2 )/4)=(1/4)y^2 +(7/2)y+((53)/4)>0 ∀y∈R (df/dx)=3x^2 +3=0 ⇒ x∉R ⇒ f(x) has no “knee” ⇒ we can use Cardano′s formula to solve y=x^3 +3x−7 x=u^(1/3) −v^(1/3) u=((y+7)/2)+((√(y^2 +14y+53))/2) v=−((y+7)/2)+((√(y^2 +14y+53))/2) x=((((y+7)/2)+((√(y^2 +14y+53))/2)))^(1/3) −((−((y+7)/2)+((√(y^2 +14y+53))/2)))^(1/3) ⇒ f^(−1) (x)=((((x+7)/2)+((√(x^2 +14x+53))/2)))^(1/3) −((−((x+7)/2)+((√(x^2 +14x+53))/2)))^(1/3) which is the same as MrW′s solution
$${x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{7}−{y}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{3}^{\mathrm{3}} }{\mathrm{27}}+\frac{\left(−\mathrm{7}−{y}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{2}}{y}+\frac{\mathrm{53}}{\mathrm{4}}>\mathrm{0}\:\forall{y}\in\mathbb{R} \\ $$$$\frac{{df}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}=\mathrm{0}\:\Rightarrow\:{x}\notin\mathbb{R}\:\Rightarrow\:{f}\left({x}\right)\:\mathrm{has}\:\mathrm{no}\:“\mathrm{knee}'' \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{Cardano}'\mathrm{s}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{solve} \\ $$$${y}={x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{7} \\ $$$${x}={u}^{\mathrm{1}/\mathrm{3}} −{v}^{\mathrm{1}/\mathrm{3}} \\ $$$${u}=\frac{{y}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}}{\mathrm{2}} \\ $$$${v}=−\frac{{y}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}}{\mathrm{2}} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\frac{{y}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{−\frac{{y}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{14}{y}+\mathrm{53}}}{\mathrm{2}}} \\ $$$$\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\sqrt[{\mathrm{3}}]{\frac{{x}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{−\frac{{x}+\mathrm{7}}{\mathrm{2}}+\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{53}}}{\mathrm{2}}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{MrW}'\mathrm{s}\:\mathrm{solution} \\ $$
Commented by naka3546 last updated on 19/May/19
thank you , sir MJS
$${thank}\:\:{you}\:\:,\:\:{sir}\:\:{MJS} \\ $$
Commented by MJS last updated on 19/May/19
f(x)=ax^3 +bx^2 +cx+d f^(−1) (x)=? we have to solve ax^3 +bx^2 +cx+(d−y)=0 x^3 +(b/a)x^2 +(c/a)x+((d−y)/a)=0 x=z−(b/(3a)) z^3 +((c/a)−(b^2 /(3a^2 )))z+((d/a)−((bc)/(3a^2 ))+((2b^3 )/(27a^3 ))−(y/a))=0 p=(c/a)−(b^2 /(3a^2 )) q=(d/a)−((bc)/(3a^2 ))+((2b^3 )/(27a^3 )) D=(p^3 /(27))+(((q−(y/a))^2 )/4)=(1/(4a^2 ))y^2 −(q/(2a))y+((4p^3 +27q^2 )/(108)) we have to decide { (((1) D≥0 ⇒ 1 real solution for z ⇒ Cardano)),(((2) D<0 ⇒ 3 real solutions for z ⇒ trigonometric method)) :} ⇒ f^(−1) (x)= { (((1) curve in 1 part)),(((2) curve in 3 parts)) :}
$${f}\left({x}\right)={ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+\left({d}−{y}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\frac{{b}}{{a}}{x}^{\mathrm{2}} +\frac{{c}}{{a}}{x}+\frac{{d}−{y}}{{a}}=\mathrm{0} \\ $$$${x}={z}−\frac{{b}}{\mathrm{3}{a}} \\ $$$${z}^{\mathrm{3}} +\left(\frac{{c}}{{a}}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }\right){z}+\left(\frac{{d}}{{a}}−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} }−\frac{{y}}{{a}}\right)=\mathrm{0} \\ $$$${p}=\frac{{c}}{{a}}−\frac{{b}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} } \\ $$$${q}=\frac{{d}}{{a}}−\frac{{bc}}{\mathrm{3}{a}^{\mathrm{2}} }+\frac{\mathrm{2}{b}^{\mathrm{3}} }{\mathrm{27}{a}^{\mathrm{3}} } \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{\left({q}−\frac{{y}}{{a}}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} }{y}^{\mathrm{2}} −\frac{{q}}{\mathrm{2}{a}}{y}+\frac{\mathrm{4}{p}^{\mathrm{3}} +\mathrm{27}{q}^{\mathrm{2}} }{\mathrm{108}} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{decide} \\ $$$$\begin{cases}{\left(\mathrm{1}\right)\:{D}\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{for}\:{z}\:\Rightarrow\:\mathrm{Cardano}}\\{\left(\mathrm{2}\right)\:{D}<\mathrm{0}\:\Rightarrow\:\mathrm{3}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{for}\:{z}\:\Rightarrow\:\mathrm{trigonometric}\:\mathrm{method}}\end{cases} \\ $$$$\Rightarrow\:{f}^{−\mathrm{1}} \left({x}\right)=\begin{cases}{\left(\mathrm{1}\right)\:\mathrm{curve}\:\mathrm{in}\:\mathrm{1}\:\mathrm{part}}\\{\left(\mathrm{2}\right)\:\mathrm{curve}\:\mathrm{in}\:\mathrm{3}\:\mathrm{parts}}\end{cases} \\ $$

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