Polynomial Functions Problem [closed]
$begingroup$
If $f(x)=x^2+1$ and $g(x)=x-1$, for all real numbers $x$, for what real number $a$ does $f(g(-a))= g(f(-a))$?
polynomials
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
$endgroup$
closed as off-topic by Did, Gibbs, Brahadeesh, José Carlos Santos, drhab Feb 7 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Gibbs, Brahadeesh, José Carlos Santos, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $f(x)=x^2+1$ and $g(x)=x-1$, for all real numbers $x$, for what real number $a$ does $f(g(-a))= g(f(-a))$?
polynomials
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
$endgroup$
closed as off-topic by Did, Gibbs, Brahadeesh, José Carlos Santos, drhab Feb 7 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Gibbs, Brahadeesh, José Carlos Santos, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $f(x)=x^2+1$ and $g(x)=x-1$, for all real numbers $x$, for what real number $a$ does $f(g(-a))= g(f(-a))$?
polynomials
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
$endgroup$
If $f(x)=x^2+1$ and $g(x)=x-1$, for all real numbers $x$, for what real number $a$ does $f(g(-a))= g(f(-a))$?
polynomials
polynomials
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
edited Dec 29 '18 at 14:53
Key Flex
8,59761233
8,59761233
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
asked Dec 29 '18 at 2:31
ObsculyzeObsculyze
43
43
closed as off-topic by Did, Gibbs, Brahadeesh, José Carlos Santos, drhab Feb 7 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Gibbs, Brahadeesh, José Carlos Santos, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, Gibbs, Brahadeesh, José Carlos Santos, drhab Feb 7 at 13:47
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, Gibbs, Brahadeesh, José Carlos Santos, drhab
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Given $f(x)=x^2+1$ and $g(x)=x-1$
$$f(g(-a))=g(f(-a))$$
$$f(-a-1)=g(a^2+1)$$
$$a^2+2+2a=a^2+1-1$$
$$a=-1$$
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
$endgroup$
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
add a comment |
$begingroup$
Write
$$f(g(-x)) = (-(x-1))^2+1$$
so, expand it. Do the same for $g(f(-x))$.
Thus, you get a equality of the form
$$ax^2+bx+c=px^2+qx+r.$$
Just solve it.
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given $f(x)=x^2+1$ and $g(x)=x-1$
$$f(g(-a))=g(f(-a))$$
$$f(-a-1)=g(a^2+1)$$
$$a^2+2+2a=a^2+1-1$$
$$a=-1$$
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
$endgroup$
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
add a comment |
$begingroup$
Given $f(x)=x^2+1$ and $g(x)=x-1$
$$f(g(-a))=g(f(-a))$$
$$f(-a-1)=g(a^2+1)$$
$$a^2+2+2a=a^2+1-1$$
$$a=-1$$
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
$endgroup$
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
add a comment |
$begingroup$
Given $f(x)=x^2+1$ and $g(x)=x-1$
$$f(g(-a))=g(f(-a))$$
$$f(-a-1)=g(a^2+1)$$
$$a^2+2+2a=a^2+1-1$$
$$a=-1$$
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
$endgroup$
Given $f(x)=x^2+1$ and $g(x)=x-1$
$$f(g(-a))=g(f(-a))$$
$$f(-a-1)=g(a^2+1)$$
$$a^2+2+2a=a^2+1-1$$
$$a=-1$$
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
edited Dec 29 '18 at 3:11
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
answered Dec 29 '18 at 2:51
Key FlexKey Flex
8,59761233
8,59761233
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
add a comment |
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
You made a small mistake in going to the $3$rd line. Note that $fleft(-a-1right) = left(-a-1right)^2 + 1 = a^2 + 2a + 2$. As such, the final answer should be $a = -1$ instead. You can easily verify this by seeing that $fleft(gleft(1right)right) = fleft(0right) = 1$ and $gleft(fleft(1right)right) = gleft(2right) = 1$.
$endgroup$
– John Omielan
Dec 29 '18 at 3:06
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
$begingroup$
@JohnOmielan Thanks, for pointing out my error!
$endgroup$
– Key Flex
Dec 29 '18 at 3:11
add a comment |
$begingroup$
Write
$$f(g(-x)) = (-(x-1))^2+1$$
so, expand it. Do the same for $g(f(-x))$.
Thus, you get a equality of the form
$$ax^2+bx+c=px^2+qx+r.$$
Just solve it.
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
add a comment |
$begingroup$
Write
$$f(g(-x)) = (-(x-1))^2+1$$
so, expand it. Do the same for $g(f(-x))$.
Thus, you get a equality of the form
$$ax^2+bx+c=px^2+qx+r.$$
Just solve it.
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
add a comment |
$begingroup$
Write
$$f(g(-x)) = (-(x-1))^2+1$$
so, expand it. Do the same for $g(f(-x))$.
Thus, you get a equality of the form
$$ax^2+bx+c=px^2+qx+r.$$
Just solve it.
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
$endgroup$
Write
$$f(g(-x)) = (-(x-1))^2+1$$
so, expand it. Do the same for $g(f(-x))$.
Thus, you get a equality of the form
$$ax^2+bx+c=px^2+qx+r.$$
Just solve it.
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
answered Dec 29 '18 at 2:48
Lucas CorrêaLucas Corrêa
1,5351421
1,5351421
add a comment |
add a comment |
