May someone help me with a functions questions?












0












$begingroup$


I was wondering if someone could help me with this question. It is genuinely not homework, it is extra work set by my angry asian mum, and I am desperately in need of help for just this one question. I hope someone can save my life or will get a beating from my angry Asian mum. I $100%$ guarantee that you will say that this is school homework, for that matter I am only $11$ years old already doing secondary school/ high school (that is what you Americans call it, I presume). So please just not disregard this question and say "it is homework" or my my mum will genuinely hit me for not completing all my extracurricular maths work (btw I get a crap ton of work worth about $12$ hours every day).



Anyways that is besides the point, here is the only one maths question that I am desperately in need for help. Please save me by just helping me with one question☺



$$f(x)=x^2, g(x)=x-6$$



Solve the equation $fg(x)=g^{-1}(x)$



x=



Thanks in advance, fellow mathematicians.






functions real-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    How can we solve an equation if the variables $f,g$ are given?
    $endgroup$
    – Yanko
    Jan 14 at 16:41






  • 3




    $begingroup$
    Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
    $endgroup$
    – lulu
    Jan 14 at 16:41












  • $begingroup$
    @lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
    $endgroup$
    – Yanko
    Jan 14 at 16:42








  • 1




    $begingroup$
    @Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
    $endgroup$
    – lulu
    Jan 14 at 16:43






  • 1




    $begingroup$
    @Yanko I think the OP is trying to solve for $x$
    $endgroup$
    – saulspatz
    Jan 14 at 16:44
















0












$begingroup$


I was wondering if someone could help me with this question. It is genuinely not homework, it is extra work set by my angry asian mum, and I am desperately in need of help for just this one question. I hope someone can save my life or will get a beating from my angry Asian mum. I $100%$ guarantee that you will say that this is school homework, for that matter I am only $11$ years old already doing secondary school/ high school (that is what you Americans call it, I presume). So please just not disregard this question and say "it is homework" or my my mum will genuinely hit me for not completing all my extracurricular maths work (btw I get a crap ton of work worth about $12$ hours every day).



Anyways that is besides the point, here is the only one maths question that I am desperately in need for help. Please save me by just helping me with one question☺



$$f(x)=x^2, g(x)=x-6$$



Solve the equation $fg(x)=g^{-1}(x)$



x=



Thanks in advance, fellow mathematicians.






functions real-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    How can we solve an equation if the variables $f,g$ are given?
    $endgroup$
    – Yanko
    Jan 14 at 16:41






  • 3




    $begingroup$
    Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
    $endgroup$
    – lulu
    Jan 14 at 16:41












  • $begingroup$
    @lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
    $endgroup$
    – Yanko
    Jan 14 at 16:42








  • 1




    $begingroup$
    @Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
    $endgroup$
    – lulu
    Jan 14 at 16:43






  • 1




    $begingroup$
    @Yanko I think the OP is trying to solve for $x$
    $endgroup$
    – saulspatz
    Jan 14 at 16:44














0












0








0





$begingroup$


I was wondering if someone could help me with this question. It is genuinely not homework, it is extra work set by my angry asian mum, and I am desperately in need of help for just this one question. I hope someone can save my life or will get a beating from my angry Asian mum. I $100%$ guarantee that you will say that this is school homework, for that matter I am only $11$ years old already doing secondary school/ high school (that is what you Americans call it, I presume). So please just not disregard this question and say "it is homework" or my my mum will genuinely hit me for not completing all my extracurricular maths work (btw I get a crap ton of work worth about $12$ hours every day).



Anyways that is besides the point, here is the only one maths question that I am desperately in need for help. Please save me by just helping me with one question☺



$$f(x)=x^2, g(x)=x-6$$



Solve the equation $fg(x)=g^{-1}(x)$



x=



Thanks in advance, fellow mathematicians.






functions real-numbers







share|cite|improve this question











$endgroup$




I was wondering if someone could help me with this question. It is genuinely not homework, it is extra work set by my angry asian mum, and I am desperately in need of help for just this one question. I hope someone can save my life or will get a beating from my angry Asian mum. I $100%$ guarantee that you will say that this is school homework, for that matter I am only $11$ years old already doing secondary school/ high school (that is what you Americans call it, I presume). So please just not disregard this question and say "it is homework" or my my mum will genuinely hit me for not completing all my extracurricular maths work (btw I get a crap ton of work worth about $12$ hours every day).



Anyways that is besides the point, here is the only one maths question that I am desperately in need for help. Please save me by just helping me with one question☺



$$f(x)=x^2, g(x)=x-6$$



Solve the equation $fg(x)=g^{-1}(x)$



x=



Thanks in advance, fellow mathematicians.






functions real-numbers




functions real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 16:52







Jie Gao

















asked Jan 14 at 16:39









Jie GaoJie Gao

32




32












  • $begingroup$
    How can we solve an equation if the variables $f,g$ are given?
    $endgroup$
    – Yanko
    Jan 14 at 16:41






  • 3




    $begingroup$
    Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
    $endgroup$
    – lulu
    Jan 14 at 16:41












  • $begingroup$
    @lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
    $endgroup$
    – Yanko
    Jan 14 at 16:42








  • 1




    $begingroup$
    @Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
    $endgroup$
    – lulu
    Jan 14 at 16:43






  • 1




    $begingroup$
    @Yanko I think the OP is trying to solve for $x$
    $endgroup$
    – saulspatz
    Jan 14 at 16:44


















  • $begingroup$
    How can we solve an equation if the variables $f,g$ are given?
    $endgroup$
    – Yanko
    Jan 14 at 16:41






  • 3




    $begingroup$
    Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
    $endgroup$
    – lulu
    Jan 14 at 16:41












  • $begingroup$
    @lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
    $endgroup$
    – Yanko
    Jan 14 at 16:42








  • 1




    $begingroup$
    @Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
    $endgroup$
    – lulu
    Jan 14 at 16:43






  • 1




    $begingroup$
    @Yanko I think the OP is trying to solve for $x$
    $endgroup$
    – saulspatz
    Jan 14 at 16:44
















$begingroup$
How can we solve an equation if the variables $f,g$ are given?
$endgroup$
– Yanko
Jan 14 at 16:41




$begingroup$
How can we solve an equation if the variables $f,g$ are given?
$endgroup$
– Yanko
Jan 14 at 16:41




3




3




$begingroup$
Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
$endgroup$
– lulu
Jan 14 at 16:41






$begingroup$
Well, just go step by step. First of all, does $fg$ mean the product $f(x)times g(x)$ or does it mean the composition $fcirc g(x)$? Either way, I assume you are looking for a value $x_0$ such that the two sides are equal for that particular value, yes?
$endgroup$
– lulu
Jan 14 at 16:41














$begingroup$
@lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
$endgroup$
– Yanko
Jan 14 at 16:42






$begingroup$
@lulu $g^{-1}(x)=x+6$ so whether it is product or composition we will not get an equality. (This might be a troll question, though I'm not sure)
$endgroup$
– Yanko
Jan 14 at 16:42






1




1




$begingroup$
@Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
$endgroup$
– lulu
Jan 14 at 16:43




$begingroup$
@Yanko I think (but am not sure) that the OP is looking for a particular value of $x$ that solves that equation.
$endgroup$
– lulu
Jan 14 at 16:43




1




1




$begingroup$
@Yanko I think the OP is trying to solve for $x$
$endgroup$
– saulspatz
Jan 14 at 16:44




$begingroup$
@Yanko I think the OP is trying to solve for $x$
$endgroup$
– saulspatz
Jan 14 at 16:44










2 Answers
2






active

oldest

votes


















2












$begingroup$

I am going to assume that $fg$ denotes the composition $fcirc g$ because, in this case, using the product makes for a cubic equation with no convenient roots.



So, assuming I have it right, you are asked to solve the equation $$fcirc g(x)=g^{-1}(x)$$



Now, for the left hand: $$fcirc g(x)=f(g(x))=g(x)^2=(x-6)^2$$



While, for the right hand, $$g^{-1}(x)=x+6$$



So you want to solve the quadratic equation $$(x-6)^2=x+6$$



I assume you know the Quadratic Formula? If so, you can just expand this out and solve in the usual way. Hint: The roots are pleasant, so you can also just factor the quadratic by inspection.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks my friend.
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51



















0












$begingroup$

$g^{-1}(x)=x+6$. Then, $$(fg)(x)=x^2(x-6)=x+6=g^{-1}(x)$$ which implies that $$x^3-6x^2-x-6=0.$$ But the roots aren't very nice, just see here by using Wolfram:Alpha.



If by $fg(x)$ you mean $(fcirc g)(x)$, you obtain $$(fcirc g)(x)=f(x-6)=(x-6)^2=x+6=g^{-1}(x),$$ thus $$x^2-13x+30=0$$ which you can solve using pq-formula and obtain $x=3$ and $x=10$ as solutions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks also, truly a wonderful person
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51










  • $begingroup$
    @JieGao If your question was successfully answered, please consider accepting one of the answers.
    $endgroup$
    – James
    Jan 16 at 13:33












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

I am going to assume that $fg$ denotes the composition $fcirc g$ because, in this case, using the product makes for a cubic equation with no convenient roots.



So, assuming I have it right, you are asked to solve the equation $$fcirc g(x)=g^{-1}(x)$$



Now, for the left hand: $$fcirc g(x)=f(g(x))=g(x)^2=(x-6)^2$$



While, for the right hand, $$g^{-1}(x)=x+6$$



So you want to solve the quadratic equation $$(x-6)^2=x+6$$



I assume you know the Quadratic Formula? If so, you can just expand this out and solve in the usual way. Hint: The roots are pleasant, so you can also just factor the quadratic by inspection.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks my friend.
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51
















2












$begingroup$

I am going to assume that $fg$ denotes the composition $fcirc g$ because, in this case, using the product makes for a cubic equation with no convenient roots.



So, assuming I have it right, you are asked to solve the equation $$fcirc g(x)=g^{-1}(x)$$



Now, for the left hand: $$fcirc g(x)=f(g(x))=g(x)^2=(x-6)^2$$



While, for the right hand, $$g^{-1}(x)=x+6$$



So you want to solve the quadratic equation $$(x-6)^2=x+6$$



I assume you know the Quadratic Formula? If so, you can just expand this out and solve in the usual way. Hint: The roots are pleasant, so you can also just factor the quadratic by inspection.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks my friend.
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51














2












2








2





$begingroup$

I am going to assume that $fg$ denotes the composition $fcirc g$ because, in this case, using the product makes for a cubic equation with no convenient roots.



So, assuming I have it right, you are asked to solve the equation $$fcirc g(x)=g^{-1}(x)$$



Now, for the left hand: $$fcirc g(x)=f(g(x))=g(x)^2=(x-6)^2$$



While, for the right hand, $$g^{-1}(x)=x+6$$



So you want to solve the quadratic equation $$(x-6)^2=x+6$$



I assume you know the Quadratic Formula? If so, you can just expand this out and solve in the usual way. Hint: The roots are pleasant, so you can also just factor the quadratic by inspection.






share|cite|improve this answer









$endgroup$



I am going to assume that $fg$ denotes the composition $fcirc g$ because, in this case, using the product makes for a cubic equation with no convenient roots.



So, assuming I have it right, you are asked to solve the equation $$fcirc g(x)=g^{-1}(x)$$



Now, for the left hand: $$fcirc g(x)=f(g(x))=g(x)^2=(x-6)^2$$



While, for the right hand, $$g^{-1}(x)=x+6$$



So you want to solve the quadratic equation $$(x-6)^2=x+6$$



I assume you know the Quadratic Formula? If so, you can just expand this out and solve in the usual way. Hint: The roots are pleasant, so you can also just factor the quadratic by inspection.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 16:50









lulululu

43.9k25182




43.9k25182












  • $begingroup$
    Thanks my friend.
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51


















  • $begingroup$
    Thanks my friend.
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51
















$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51




$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51











0












$begingroup$

$g^{-1}(x)=x+6$. Then, $$(fg)(x)=x^2(x-6)=x+6=g^{-1}(x)$$ which implies that $$x^3-6x^2-x-6=0.$$ But the roots aren't very nice, just see here by using Wolfram:Alpha.



If by $fg(x)$ you mean $(fcirc g)(x)$, you obtain $$(fcirc g)(x)=f(x-6)=(x-6)^2=x+6=g^{-1}(x),$$ thus $$x^2-13x+30=0$$ which you can solve using pq-formula and obtain $x=3$ and $x=10$ as solutions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks also, truly a wonderful person
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51










  • $begingroup$
    @JieGao If your question was successfully answered, please consider accepting one of the answers.
    $endgroup$
    – James
    Jan 16 at 13:33
















0












$begingroup$

$g^{-1}(x)=x+6$. Then, $$(fg)(x)=x^2(x-6)=x+6=g^{-1}(x)$$ which implies that $$x^3-6x^2-x-6=0.$$ But the roots aren't very nice, just see here by using Wolfram:Alpha.



If by $fg(x)$ you mean $(fcirc g)(x)$, you obtain $$(fcirc g)(x)=f(x-6)=(x-6)^2=x+6=g^{-1}(x),$$ thus $$x^2-13x+30=0$$ which you can solve using pq-formula and obtain $x=3$ and $x=10$ as solutions.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks also, truly a wonderful person
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51










  • $begingroup$
    @JieGao If your question was successfully answered, please consider accepting one of the answers.
    $endgroup$
    – James
    Jan 16 at 13:33














0












0








0





$begingroup$

$g^{-1}(x)=x+6$. Then, $$(fg)(x)=x^2(x-6)=x+6=g^{-1}(x)$$ which implies that $$x^3-6x^2-x-6=0.$$ But the roots aren't very nice, just see here by using Wolfram:Alpha.



If by $fg(x)$ you mean $(fcirc g)(x)$, you obtain $$(fcirc g)(x)=f(x-6)=(x-6)^2=x+6=g^{-1}(x),$$ thus $$x^2-13x+30=0$$ which you can solve using pq-formula and obtain $x=3$ and $x=10$ as solutions.






share|cite|improve this answer









$endgroup$



$g^{-1}(x)=x+6$. Then, $$(fg)(x)=x^2(x-6)=x+6=g^{-1}(x)$$ which implies that $$x^3-6x^2-x-6=0.$$ But the roots aren't very nice, just see here by using Wolfram:Alpha.



If by $fg(x)$ you mean $(fcirc g)(x)$, you obtain $$(fcirc g)(x)=f(x-6)=(x-6)^2=x+6=g^{-1}(x),$$ thus $$x^2-13x+30=0$$ which you can solve using pq-formula and obtain $x=3$ and $x=10$ as solutions.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 14 at 16:48









JamesJames

2,122422




2,122422












  • $begingroup$
    Thanks also, truly a wonderful person
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51










  • $begingroup$
    @JieGao If your question was successfully answered, please consider accepting one of the answers.
    $endgroup$
    – James
    Jan 16 at 13:33


















  • $begingroup$
    Thanks also, truly a wonderful person
    $endgroup$
    – Jie Gao
    Jan 14 at 16:51










  • $begingroup$
    @JieGao If your question was successfully answered, please consider accepting one of the answers.
    $endgroup$
    – James
    Jan 16 at 13:33
















$begingroup$
Thanks also, truly a wonderful person
$endgroup$
– Jie Gao
Jan 14 at 16:51




$begingroup$
Thanks also, truly a wonderful person
$endgroup$
– Jie Gao
Jan 14 at 16:51












$begingroup$
@JieGao If your question was successfully answered, please consider accepting one of the answers.
$endgroup$
– James
Jan 16 at 13:33




$begingroup$
@JieGao If your question was successfully answered, please consider accepting one of the answers.
$endgroup$
– James
Jan 16 at 13:33


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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