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Question Number 94366 by i jagooll last updated on 18/May/20
what is the value of x if f(x+1) = x^2 −1 g(x)= 2x+7 and f(g^(−1) (x))= 3
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{if}\:\mathrm{f}\left({x}+\mathrm{1}\right)\:=\:{x}^{\mathrm{2}} −\mathrm{1} \\ $$$${g}\left({x}\right)=\:\mathrm{2}{x}+\mathrm{7}\:\mathrm{and}\:{f}\left({g}^{−\mathrm{1}} \left({x}\right)\right)=\:\mathrm{3}\: \\ $$
Commented by mr W last updated on 18/May/20
g^(−1) (x)=((x−7)/2) f(x)=(x−1)^2 −1 f(g^(−1) (x))=(((x−7)/2)−1)^2 −1=3 ((x−7)/2)−1=±2 ⇒x=13 or 5
$${g}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−\mathrm{7}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$${f}\left({g}^{−\mathrm{1}} \left({x}\right)\right)=\left(\frac{{x}−\mathrm{7}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}=\mathrm{3} \\ $$$$\frac{{x}−\mathrm{7}}{\mathrm{2}}−\mathrm{1}=\pm\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{13}\:{or}\:\mathrm{5} \\ $$
Commented by i jagooll last updated on 18/May/20
thank you both
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Commented by Abdulrahman last updated on 18/May/20
please solve this 3^x +x^3 =17 with steps
$$\mathrm{please} \\ $$$$\mathrm{solve} \\ $$$$\mathrm{this} \\ $$$$\mathrm{3}^{\mathrm{x}} +\mathrm{x}^{\mathrm{3}} =\mathrm{17} \\ $$$$\mathrm{with} \\ $$$$\mathrm{steps} \\ $$
Commented by mr W last updated on 18/May/20
1. we see (if you don′t see, then i can′t help you), that x=2 is a root, since 3^2 +2^3 =9+8=17. 2. both 3^x and x^3 are strictly increasing functions, so 3^x +x^3 is also strictly increasing. when a strictly increasing function has a zero, then this zero is the only one zero. that means x=2 is the only one root of eqn. x^2 +3^x =17.
$$\mathrm{1}.\:{we}\:{see}\:\left({if}\:{you}\:{don}'{t}\:{see},\:{then}\:{i}\:{can}'{t}\right. \\ $$$$\left.{help}\:{you}\right),\:{that}\:{x}=\mathrm{2}\:{is}\:{a}\:{root},\:{since} \\ $$$$\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{3}} =\mathrm{9}+\mathrm{8}=\mathrm{17}. \\ $$$$\mathrm{2}.\:{both}\:\mathrm{3}^{{x}} \:{and}\:{x}^{\mathrm{3}} \:{are}\:{strictly}\:{increasing} \\ $$$${functions},\:{so}\:\mathrm{3}^{{x}} +{x}^{\mathrm{3}} \:{is}\:{also}\:{strictly} \\ $$$${increasing}.\:{when}\:{a}\:{strictly}\:{increasing} \\ $$$${function}\:{has}\:{a}\:{zero},\:{then}\:{this}\:{zero} \\ $$$${is}\:{the}\:{only}\:{one}\:{zero}.\:{that}\:{means} \\ $$$${x}=\mathrm{2}\:{is}\:{the}\:{only}\:{one}\:{root}\:{of}\:{eqn}. \\ $$$${x}^{\mathrm{2}} +\mathrm{3}^{{x}} =\mathrm{17}. \\ $$
Commented by Abdulrahman last updated on 18/May/20
bundle of thanks
$$\mathrm{bundle}\:\mathrm{of}\:\mathrm{thanks} \\ $$
Commented by Abdulrahman last updated on 18/May/20
∫(((x−2)^3 )/(2−x^2 ))dx=?
$$\int\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=? \\ $$
Commented by abdomathmax last updated on 18/May/20
I =∫ (((x−2)^3 )/(2−x^2 ))dx ⇒I =∫ ((x^3 −3x^2 ×2 +3x×2^2 −2^3 )/(2−x^2 ))dx =∫ ((x^3 −6x^2 +12x+8)/(2−x^2 )) dx =−∫((x^3 −6x^2 +12x+8)/(x^2 −2))dx =−∫ ((x(x^2 −2)−6x^2 +14x +8)/(x^2 −2))dx =−∫ xdx +∫ ((6x^2 −14x+8)/(x^2 −2))dx =−(x^2 /2) +∫ ((6(x^2 −2)−14x +20)/(x^2 −2))dx =−(x^2 /2) +6x −2∫ ((7x −10)/(x^2 −2))dx F(x)=((7x−10)/(x^2 −2)) =((7x−10)/((x−(√2))(x+(√2)))) =(a/(x−(√2))) +(b/(x+(√2))) a =((7(√2)−10)/(2(√2))) and b =((−7(√2)−10)/(−2(√2))) =((7(√2)+10)/(2(√2))) ∫((7x−10)/(x^2 −2))dx =((7(√2)−10)/(2(√2)))ln∣x−(√2)∣+((7(√2)+10)/(2(√2)))ln∣x+(√2)∣ +c I =−(x^2 /2) +6x−(7(√2)−10)ln∣x−(√2)∣−(7(√2)+10)ln∣x+(√2)∣ +C
$$\mathrm{I}\:=\int\:\:\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} }{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} ×\mathrm{2}\:+\mathrm{3x}×\mathrm{2}^{\mathrm{2}} −\mathrm{2}^{\mathrm{3}} }{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int\:\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} \:+\mathrm{12x}+\mathrm{8}}{\mathrm{2}−\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=−\int\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} \:+\mathrm{12x}+\mathrm{8}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$=−\int\:\:\frac{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{6x}^{\mathrm{2}} \:+\mathrm{14x}\:+\mathrm{8}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$=−\int\:\mathrm{xdx}\:+\int\:\frac{\mathrm{6x}^{\mathrm{2}} −\mathrm{14x}+\mathrm{8}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\int\:\:\frac{\mathrm{6}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{14x}\:+\mathrm{20}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{6x}\:−\mathrm{2}\int\:\:\frac{\mathrm{7x}\:−\mathrm{10}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{7x}−\mathrm{10}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\:=\frac{\mathrm{7x}−\mathrm{10}}{\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)\left(\mathrm{x}+\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{a}}{\mathrm{x}−\sqrt{\mathrm{2}}}\:+\frac{\mathrm{b}}{\mathrm{x}+\sqrt{\mathrm{2}}} \\ $$$$\mathrm{a}\:=\frac{\mathrm{7}\sqrt{\mathrm{2}}−\mathrm{10}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\mathrm{and}\:\mathrm{b}\:=\frac{−\mathrm{7}\sqrt{\mathrm{2}}−\mathrm{10}}{−\mathrm{2}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{7}\sqrt{\mathrm{2}}+\mathrm{10}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int\frac{\mathrm{7x}−\mathrm{10}}{\mathrm{x}^{\mathrm{2}} −\mathrm{2}}\mathrm{dx}\:=\frac{\mathrm{7}\sqrt{\mathrm{2}}−\mathrm{10}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\mathrm{x}−\sqrt{\mathrm{2}}\mid+\frac{\mathrm{7}\sqrt{\mathrm{2}}+\mathrm{10}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{2}}\mid\:+\mathrm{c} \\ $$$$\mathrm{I}\:=−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{6x}−\left(\mathrm{7}\sqrt{\mathrm{2}}−\mathrm{10}\right)\mathrm{ln}\mid\mathrm{x}−\sqrt{\mathrm{2}}\mid−\left(\mathrm{7}\sqrt{\mathrm{2}}+\mathrm{10}\right)\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{2}}\mid\:+\mathrm{C} \\ $$
Commented by mr W last updated on 18/May/20
abdulraman sir: please open a new thread if you want to ask a new question!
$${abdulraman}\:{sir}: \\ $$$${please}\:{open}\:{a}\:{new}\:{thread}\:{if}\:{you}\:{want} \\ $$$${to}\:{ask}\:{a}\:{new}\:{question}! \\ $$
Commented by Abdulrahman last updated on 18/May/20
ok thanks alot
$$\mathrm{ok}\:\mathrm{thanks}\:\mathrm{alot} \\ $$
Answered by john santu last updated on 18/May/20
g^(−1) (x) = f^(−1) (3) ⇒f^(−1) (x^2 −1) = x+1 ⇒ { ((x^2 −1= 3)),((⇒ { ((x = 2)),((x = −2)) :})) :} case(1) f^(−1) (3)= 3 & g^(−1) (x)=((x−7)/2) ⇒ ((x−7)/2) = 3 ⇒ x = 13 case(2) f^(−1) (3)=−1 & g^(−1) (x)=((x−7)/2) ⇒ ((x−7)/2) = −1 ⇒ x = 5 solution x = 5 or 13
$$\mathrm{g}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{3}\right)\: \\ $$$$\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\:=\:\mathrm{x}+\mathrm{1}\:\Rightarrow\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}=\:\mathrm{3}}\\{\Rightarrow\begin{cases}{\mathrm{x}\:=\:\mathrm{2}}\\{\mathrm{x}\:=\:−\mathrm{2}}\end{cases}}\end{cases} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=\:\mathrm{3}\:\&\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{x}−\mathrm{7}}{\mathrm{2}}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{x}\:=\:\mathrm{13}\: \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{3}\right)=−\mathrm{1}\:\&\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{x}\right)=\frac{\mathrm{x}−\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{x}−\mathrm{7}}{\mathrm{2}}\:=\:−\mathrm{1}\:\Rightarrow\:\mathrm{x}\:=\:\mathrm{5}\: \\ $$$$\mathrm{solution}\:\mathrm{x}\:=\:\mathrm{5}\:\mathrm{or}\:\mathrm{13}\: \\ $$

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