May someone help me with a functions questions?
$begingroup$
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
$endgroup$
|
show 1 more comment
$begingroup$
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
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
$endgroup$
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
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51
add a comment |
$begingroup$
$g^{-1}(x)=x+6$. 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.
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
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.
$endgroup$
$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Jan 14 at 16:50
lulululu
43.9k25182
43.9k25182
$begingroup$
Thanks my friend.
$endgroup$
– Jie Gao
Jan 14 at 16:51
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$g^{-1}(x)=x+6$. 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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$g^{-1}(x)=x+6$. 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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
$g^{-1}(x)=x+6$. 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.
$endgroup$
$g^{-1}(x)=x+6$. 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.
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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