Mixing
Identical Function Query
Clash Royale CLAN TAG#URR8PPP
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If $f(x)=fracxln x$ & $g(x)=fracln xx$. Then identify the correct statement.
A) $frac1g(x)$ and $f(x)$ are identical functions
B) $frac1f(x)$ and $g(x)$ are identical functions
C) $f(x)cdot g(x)=1 forall x>0$
D) $frac1f(x)cdot g(x)=1 forall x>0$
I don't have the solution but as per the answer key Only A is the correct statement , B,C,D are incorrect statement .
My Approach for B let $t(x)=frac1f(x)$ , now the question is whether $t(x)$ & $g(x)$ are identical function, my thought would be that they are identical function because for identical function we need to check domain and range on $t(x)$ and not on its reciprocal.But on contrary in the ANSWER Key this is mentioned as INCORRECT.
Regarding C and D I don't know why it is incorrect.
functions
edited 1 hour ago
zipirovich
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
add a comment |Â
up vote
3
down vote
favorite
If $f(x)=fracxln x$ & $g(x)=fracln xx$. Then identify the correct statement.
A) $frac1g(x)$ and $f(x)$ are identical functions
B) $frac1f(x)$ and $g(x)$ are identical functions
C) $f(x)cdot g(x)=1 forall x>0$
D) $frac1f(x)cdot g(x)=1 forall x>0$
I don't have the solution but as per the answer key Only A is the correct statement , B,C,D are incorrect statement .
My Approach for B let $t(x)=frac1f(x)$ , now the question is whether $t(x)$ & $g(x)$ are identical function, my thought would be that they are identical function because for identical function we need to check domain and range on $t(x)$ and not on its reciprocal.But on contrary in the ANSWER Key this is mentioned as INCORRECT.
Regarding C and D I don't know why it is incorrect.
functions
edited 1 hour ago
zipirovich
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If $f(x)=fracxln x$ & $g(x)=fracln xx$. Then identify the correct statement.
A) $frac1g(x)$ and $f(x)$ are identical functions
B) $frac1f(x)$ and $g(x)$ are identical functions
C) $f(x)cdot g(x)=1 forall x>0$
D) $frac1f(x)cdot g(x)=1 forall x>0$
I don't have the solution but as per the answer key Only A is the correct statement , B,C,D are incorrect statement .
My Approach for B let $t(x)=frac1f(x)$ , now the question is whether $t(x)$ & $g(x)$ are identical function, my thought would be that they are identical function because for identical function we need to check domain and range on $t(x)$ and not on its reciprocal.But on contrary in the ANSWER Key this is mentioned as INCORRECT.
Regarding C and D I don't know why it is incorrect.
functions
edited 1 hour ago
zipirovich
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
If $f(x)=fracxln x$ & $g(x)=fracln xx$. Then identify the correct statement.
A) $frac1g(x)$ and $f(x)$ are identical functions
B) $frac1f(x)$ and $g(x)$ are identical functions
C) $f(x)cdot g(x)=1 forall x>0$
D) $frac1f(x)cdot g(x)=1 forall x>0$
I don't have the solution but as per the answer key Only A is the correct statement , B,C,D are incorrect statement .
My Approach for B let $t(x)=frac1f(x)$ , now the question is whether $t(x)$ & $g(x)$ are identical function, my thought would be that they are identical function because for identical function we need to check domain and range on $t(x)$ and not on its reciprocal.But on contrary in the ANSWER Key this is mentioned as INCORRECT.
Regarding C and D I don't know why it is incorrect.
functions
functions
edited 1 hour ago
zipirovich
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
edited 1 hour ago
zipirovich
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
edited 1 hour ago
zipirovich
10.2k11630
edited 1 hour ago
zipirovich
10.2k11630
edited 1 hour ago
zipirovich
10.2k11630
10.2k11630
asked 2 hours ago
Samar Imam Zaidi
1,303316
asked 2 hours ago
Samar Imam Zaidi
1,303316
asked 2 hours ago
Samar Imam Zaidi
1,303316
1,303316
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.
Consider the function $g(x)=dfracln xx$. The domain of this function is $x>0$, i.e. $xin(0,+infty)$. (I hope you understand why.)
Consider the function $G(x)=dfrac1g(x)$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $xin(0,+infty)$, and exclude all the points where $g(x)=dfracln xx=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=dfrac1g(x)$ is $xin(0,1)cup(1,+infty)$.
Similarly, you can find the domains of $f(x)$, $dfrac1f(x)$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.
As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.
answered 1 hour ago
zipirovich
10.2k11630
add a comment |Â
up vote
3
down vote
The important point is that $ln 1=0$, so you can't divide by it. In A you get a $ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x gt 0$, but $f(1)$ is not so $frac 1f(1)$ is not either. C and D both fail for $x=1$ for the same reason.
answered 1 hour ago
Ross Millikan
281k23191358
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.
Consider the function $g(x)=dfracln xx$. The domain of this function is $x>0$, i.e. $xin(0,+infty)$. (I hope you understand why.)
Consider the function $G(x)=dfrac1g(x)$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $xin(0,+infty)$, and exclude all the points where $g(x)=dfracln xx=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=dfrac1g(x)$ is $xin(0,1)cup(1,+infty)$.
Similarly, you can find the domains of $f(x)$, $dfrac1f(x)$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.
As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.
answered 1 hour ago
zipirovich
10.2k11630
add a comment |Â
up vote
2
down vote
accepted
You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.
Consider the function $g(x)=dfracln xx$. The domain of this function is $x>0$, i.e. $xin(0,+infty)$. (I hope you understand why.)
Consider the function $G(x)=dfrac1g(x)$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $xin(0,+infty)$, and exclude all the points where $g(x)=dfracln xx=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=dfrac1g(x)$ is $xin(0,1)cup(1,+infty)$.
Similarly, you can find the domains of $f(x)$, $dfrac1f(x)$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.
As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.
answered 1 hour ago
zipirovich
10.2k11630
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.
Consider the function $g(x)=dfracln xx$. The domain of this function is $x>0$, i.e. $xin(0,+infty)$. (I hope you understand why.)
Consider the function $G(x)=dfrac1g(x)$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $xin(0,+infty)$, and exclude all the points where $g(x)=dfracln xx=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=dfrac1g(x)$ is $xin(0,1)cup(1,+infty)$.
Similarly, you can find the domains of $f(x)$, $dfrac1f(x)$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.
As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.
answered 1 hour ago
zipirovich
10.2k11630
You're on the right track: the key to this question is to identify the domains of all the functions involved. Here are some of them.
Consider the function $g(x)=dfracln xx$. The domain of this function is $x>0$, i.e. $xin(0,+infty)$. (I hope you understand why.)
Consider the function $G(x)=dfrac1g(x)$. This function, since it involves $g(x)$, is only defined when $g(x)$ is defined; and moreover, we can't have $g(x)=0$. So we take the domain of $g(x)$, i.e. $xin(0,+infty)$, and exclude all the points where $g(x)=dfracln xx=0$. There's only one such point, $x=1$. Therefore, the domain of $G(x)=dfrac1g(x)$ is $xin(0,1)cup(1,+infty)$.
Similarly, you can find the domains of $f(x)$, $dfrac1f(x)$, etc. And you should see that the domains of the two functions in (A) are the same, but in (B) they are not.
As for (C) and (D), again it's about the domains. For example, in (C), the expression $f(x)cdot g(x)$ is undefined at $x=1$ because $f(x)$ is undefined at $x=1$. So the statement is not true at $x=1$ (the left-hand side is undefined, and as such it's not equal to $1$), and thus we can't say that it holds for all $x>0$.
answered 1 hour ago
zipirovich
10.2k11630
answered 1 hour ago
zipirovich
10.2k11630
answered 1 hour ago
zipirovich
10.2k11630
answered 1 hour ago
zipirovich
10.2k11630
10.2k11630
add a comment |Â
add a comment |Â
up vote
3
down vote
The important point is that $ln 1=0$, so you can't divide by it. In A you get a $ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x gt 0$, but $f(1)$ is not so $frac 1f(1)$ is not either. C and D both fail for $x=1$ for the same reason.
answered 1 hour ago
Ross Millikan
281k23191358
add a comment |Â
up vote
3
down vote
The important point is that $ln 1=0$, so you can't divide by it. In A you get a $ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x gt 0$, but $f(1)$ is not so $frac 1f(1)$ is not either. C and D both fail for $x=1$ for the same reason.
answered 1 hour ago
Ross Millikan
281k23191358
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The important point is that $ln 1=0$, so you can't divide by it. In A you get a $ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x gt 0$, but $f(1)$ is not so $frac 1f(1)$ is not either. C and D both fail for $x=1$ for the same reason.
answered 1 hour ago
Ross Millikan
281k23191358
The important point is that $ln 1=0$, so you can't divide by it. In A you get a $ln x$ in both denominators, so both sides of the equation are undefined at $x=1$. In B, $g(x)$ is nicely defined for all $x gt 0$, but $f(1)$ is not so $frac 1f(1)$ is not either. C and D both fail for $x=1$ for the same reason.
answered 1 hour ago
Ross Millikan
281k23191358
answered 1 hour ago
Ross Millikan
281k23191358
answered 1 hour ago
Ross Millikan
281k23191358
answered 1 hour ago
Ross Millikan
281k23191358
281k23191358
add a comment |Â
add a comment |Â
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