Finding a function from the given equation
up vote
1
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Let f be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x in R^+$. Then which of the following is correct?
A) f is an even function
B) f is an odd function
C) $lim_{xto infty} frac{f(x)}{x^3}=1$
D $lim_{xto infty} bigg(frac{f(x)}{x^2}-x bigg)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
Let f be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x in R^+$. Then which of the following is correct?
A) f is an even function
B) f is an odd function
C) $lim_{xto infty} frac{f(x)}{x^3}=1$
D $lim_{xto infty} bigg(frac{f(x)}{x^2}-x bigg)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
1
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let f be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x in R^+$. Then which of the following is correct?
A) f is an even function
B) f is an odd function
C) $lim_{xto infty} frac{f(x)}{x^3}=1$
D $lim_{xto infty} bigg(frac{f(x)}{x^2}-x bigg)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let f be a polynomial satisfying $f(x^2)-xf(x) = x^4(x^2-1), x in R^+$. Then which of the following is correct?
A) f is an even function
B) f is an odd function
C) $lim_{xto infty} frac{f(x)}{x^3}=1$
D $lim_{xto infty} bigg(frac{f(x)}{x^2}-x bigg)$ exist and is equal to a non-zero quantity.
I have no idea what to do here.
Looking at the options, one thing I could guess is that the question wants us to find $f(x)$. After an analysis of few minutes, I could guess $f(x) = x^3$. But that gives me the answer as B), C). But the answer given is just C).
Any help would be appreciated.
calculus limits functions even-and-odd-functions
calculus limits functions even-and-odd-functions
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Tony
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Tony
112
asked 1 hour ago
Tony
112
112
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tony is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
1
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago
add a comment |
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
1
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
1
1
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Hints: look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
add a comment |
up vote
0
down vote
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Hints: look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
add a comment |
up vote
5
down vote
accepted
Hints: look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Hints: look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
Hints: look at the degree of the polynomial. If $f$ has degree $n$ then LHS has degree $2n$ and RHS has degree $6$. Hence $f$ is a polynomial of degree 3. The constant term is $0$ because $f(0)=0$. Show that the term in $x^{2}$ also must vanish because $xf(x)=f(x^{2})-x^{6}+x^{4}$ is a polynomial with only even powers of $x$. I guess you can take it from here. [All solutions are of the form $ax+x^{3}$]. Final anwser: A) and D) are false (the limit in D) exists but it is $0$); B) and C) are true.
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
edited 5 mins ago
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
answered 50 mins ago
Kavi Rama Murthy
48.2k31854
48.2k31854
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
add a comment |
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
But the answer given is C).
– Tony
44 mins ago
But the answer given is C).
– Tony
44 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
I double checked my solution and it looks right. I think the answer given is wrong.
– Kavi Rama Murthy
41 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
See Sauhard Sharma's comment. Can that be an explanation?
– Tony
32 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
@Kavi One doubt: If a function gas the set of domain restricted to positive reals only, then how can we judge abouts odd character. I mean we can't do $f(-x) =-f(x)$ (taking x as positive) as it is not permissible for $f(x)$ to obtain negative value of x as its domain. Am I right? Or, am I missing something obvious?
– jayant98
30 mins ago
add a comment |
up vote
0
down vote
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
add a comment |
up vote
0
down vote
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
add a comment |
up vote
0
down vote
up vote
0
down vote
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
To expand on @KaviRamaMurthy's answer:
Let $f=sum^n a_kx^k$ where $a_nne 0$ for $xge 0$. Then
$$sum a_kx^{2k}-sum a_kx^{k+1}=x^6-x^4$$
$$sum a_kx^{2k}=sum a_kx^{k+1}+x^6-x^4$$
So $2nin [0,n+1]cup{6}$ which means $nle 1$ or $n=3$. So
$$f=b_0+b_1x+b_2x^2+b_3x^3$$ for $xge 0$. And then
$$b_0+b_1x^2+b_2x^4+b_3x^6=b_0x+b_1x^2+b_2x^3+b_3x^4+x^6-x^4$$
which means $b_3=1$ and then $b_2=0$: $b_1$ is unconstrained and $b_0=0$:
$$f(x)=x^3+b_1x.$$
The only way (B) can hold is if they meant
for $xin R^+$, $f$ is equal to a polynomial such that... but $f$ need not be a polynomial on all of $R$.
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
answered 28 mins ago
Bjørn Kjos-Hanssen
2,066818
2,066818
add a comment |
add a comment |
Tony is a new contributor. Be nice, and check out our Code of Conduct.
Tony is a new contributor. Be nice, and check out our Code of Conduct.
Tony is a new contributor. Be nice, and check out our Code of Conduct.
Tony is a new contributor. Be nice, and check out our Code of Conduct.
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To ay that B) is true it is not enough to give one example. There may be other solutions which are not odd.
– Kavi Rama Murthy
59 mins ago
1
Check the condition that says this polynomial is only satisfied when $x epsilon R^+$. So you don't know if the function is odd or not.
– Sauhard Sharma
55 mins ago