Mixing
Calculating inverse - what about assumptions?
$begingroup$
Can someone verify this and answer my questions? I've chosen simple function on purpose. I've also added my paper from an exam at the very end.
Find inverse function $f^{-1}$ to function $f(x)= frac{x+1}{x-3}$. Show the domain of both $f$ and $f^{-1}$.
Domain of $f$:
$x-3 neq 0$
$x neq 3$
$D_{f} = mathbb{R}backslash{3}$
Finding inverse:
$y = frac{x+1}{x-3}$
$x longleftrightarrow y$
$x = frac{y+1}{y-3} quad quad /*(y-3)quad quad$ QUESTION 1
$xy-3x=y+1$
$xy-y=3x+1$
$y(x-1)=3x+1 quad quad /:(x-1) quad quad$ QUESTION 2
$y=frac{3x+1}{x-1}$
$f^{-1}(x) = frac{3x+1}{x-1}$
Domain of inverse:
$x = mathbb{R}backslash{1}$
I tagged "question 1" and "question 2" above.
Question 1: when multiplying both sides by $(y-3)$, should I make an assumption that $y-3 > 0 Rightarrow y > 3$? Because what if y was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 2: when dividing both sides by $(x-1)$, should I make an assumption that $x - 1 > 0 Rightarrow x > 1$? Because what if I'm dividing both sides by $(x-1)$ and $(x-1)$ was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 3: how is it possible that the assumptions from Question 1 and/or Question 2 don't affect the final domain of $f^{-1}(x)$?
Side note: For example here (link to my paper: https://i.imgur.com/zkJxAOD.jpg ) I was told to make an assumptions. I am really confused now when should I make an assumptions and when not to...
Thanks for any help.
real-analysis algebra-precalculus inverse inverse-function
edited Jan 27 at 22:16
user376343
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
$endgroup$
add a comment |
$begingroup$
Can someone verify this and answer my questions? I've chosen simple function on purpose. I've also added my paper from an exam at the very end.
Find inverse function $f^{-1}$ to function $f(x)= frac{x+1}{x-3}$. Show the domain of both $f$ and $f^{-1}$.
Domain of $f$:
$x-3 neq 0$
$x neq 3$
$D_{f} = mathbb{R}backslash{3}$
Finding inverse:
$y = frac{x+1}{x-3}$
$x longleftrightarrow y$
$x = frac{y+1}{y-3} quad quad /*(y-3)quad quad$ QUESTION 1
$xy-3x=y+1$
$xy-y=3x+1$
$y(x-1)=3x+1 quad quad /:(x-1) quad quad$ QUESTION 2
$y=frac{3x+1}{x-1}$
$f^{-1}(x) = frac{3x+1}{x-1}$
Domain of inverse:
$x = mathbb{R}backslash{1}$
I tagged "question 1" and "question 2" above.
Question 1: when multiplying both sides by $(y-3)$, should I make an assumption that $y-3 > 0 Rightarrow y > 3$? Because what if y was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 2: when dividing both sides by $(x-1)$, should I make an assumption that $x - 1 > 0 Rightarrow x > 1$? Because what if I'm dividing both sides by $(x-1)$ and $(x-1)$ was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 3: how is it possible that the assumptions from Question 1 and/or Question 2 don't affect the final domain of $f^{-1}(x)$?
Side note: For example here (link to my paper: https://i.imgur.com/zkJxAOD.jpg ) I was told to make an assumptions. I am really confused now when should I make an assumptions and when not to...
Thanks for any help.
real-analysis algebra-precalculus inverse inverse-function
edited Jan 27 at 22:16
user376343
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
$endgroup$
add a comment |
$begingroup$
Can someone verify this and answer my questions? I've chosen simple function on purpose. I've also added my paper from an exam at the very end.
Find inverse function $f^{-1}$ to function $f(x)= frac{x+1}{x-3}$. Show the domain of both $f$ and $f^{-1}$.
Domain of $f$:
$x-3 neq 0$
$x neq 3$
$D_{f} = mathbb{R}backslash{3}$
Finding inverse:
$y = frac{x+1}{x-3}$
$x longleftrightarrow y$
$x = frac{y+1}{y-3} quad quad /*(y-3)quad quad$ QUESTION 1
$xy-3x=y+1$
$xy-y=3x+1$
$y(x-1)=3x+1 quad quad /:(x-1) quad quad$ QUESTION 2
$y=frac{3x+1}{x-1}$
$f^{-1}(x) = frac{3x+1}{x-1}$
Domain of inverse:
$x = mathbb{R}backslash{1}$
I tagged "question 1" and "question 2" above.
Question 1: when multiplying both sides by $(y-3)$, should I make an assumption that $y-3 > 0 Rightarrow y > 3$? Because what if y was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 2: when dividing both sides by $(x-1)$, should I make an assumption that $x - 1 > 0 Rightarrow x > 1$? Because what if I'm dividing both sides by $(x-1)$ and $(x-1)$ was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 3: how is it possible that the assumptions from Question 1 and/or Question 2 don't affect the final domain of $f^{-1}(x)$?
Side note: For example here (link to my paper: https://i.imgur.com/zkJxAOD.jpg ) I was told to make an assumptions. I am really confused now when should I make an assumptions and when not to...
Thanks for any help.
real-analysis algebra-precalculus inverse inverse-function
edited Jan 27 at 22:16
user376343
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
$endgroup$
Can someone verify this and answer my questions? I've chosen simple function on purpose. I've also added my paper from an exam at the very end.
Find inverse function $f^{-1}$ to function $f(x)= frac{x+1}{x-3}$. Show the domain of both $f$ and $f^{-1}$.
Domain of $f$:
$x-3 neq 0$
$x neq 3$
$D_{f} = mathbb{R}backslash{3}$
Finding inverse:
$y = frac{x+1}{x-3}$
$x longleftrightarrow y$
$x = frac{y+1}{y-3} quad quad /*(y-3)quad quad$ QUESTION 1
$xy-3x=y+1$
$xy-y=3x+1$
$y(x-1)=3x+1 quad quad /:(x-1) quad quad$ QUESTION 2
$y=frac{3x+1}{x-1}$
$f^{-1}(x) = frac{3x+1}{x-1}$
Domain of inverse:
$x = mathbb{R}backslash{1}$
I tagged "question 1" and "question 2" above.
Question 1: when multiplying both sides by $(y-3)$, should I make an assumption that $y-3 > 0 Rightarrow y > 3$? Because what if y was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 2: when dividing both sides by $(x-1)$, should I make an assumption that $x - 1 > 0 Rightarrow x > 1$? Because what if I'm dividing both sides by $(x-1)$ and $(x-1)$ was negative or $0?$ Then the sign would change... Why should I make or why should I not make such assumption?
Question 3: how is it possible that the assumptions from Question 1 and/or Question 2 don't affect the final domain of $f^{-1}(x)$?
Side note: For example here (link to my paper: https://i.imgur.com/zkJxAOD.jpg ) I was told to make an assumptions. I am really confused now when should I make an assumptions and when not to...
Thanks for any help.
real-analysis algebra-precalculus inverse inverse-function
real-analysis algebra-precalculus inverse inverse-function
edited Jan 27 at 22:16
user376343
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
edited Jan 27 at 22:16
user376343
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
edited Jan 27 at 22:16
user376343
3,9584829
edited Jan 27 at 22:16
user376343
3,9584829
edited Jan 27 at 22:16
user376343
3,9584829
3,9584829
asked Jan 27 at 17:29
wenoweno
32511
asked Jan 27 at 17:29
wenoweno
32511
asked Jan 27 at 17:29
wenoweno
32511
32511
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Q1: No, because this is an equation, not an inequality. More details: $yneq 3$ is assumed since $(y-3)$ is in the denominator. So we can freely multiply by $(y-3).$
Q2: Similarly, we are no signs to care about. Though when dividing by $(x-1)$ we should make assumption $xneq 1.$
Q3: The final domain is affected, because $1$ is excluded.
answered Jan 27 at 22:20
user376343user376343
3,9584829
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
votes
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active
oldest
votes
$begingroup$
Q1: No, because this is an equation, not an inequality. More details: $yneq 3$ is assumed since $(y-3)$ is in the denominator. So we can freely multiply by $(y-3).$
Q2: Similarly, we are no signs to care about. Though when dividing by $(x-1)$ we should make assumption $xneq 1.$
Q3: The final domain is affected, because $1$ is excluded.
answered Jan 27 at 22:20
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
Q1: No, because this is an equation, not an inequality. More details: $yneq 3$ is assumed since $(y-3)$ is in the denominator. So we can freely multiply by $(y-3).$
Q2: Similarly, we are no signs to care about. Though when dividing by $(x-1)$ we should make assumption $xneq 1.$
Q3: The final domain is affected, because $1$ is excluded.
answered Jan 27 at 22:20
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
Q1: No, because this is an equation, not an inequality. More details: $yneq 3$ is assumed since $(y-3)$ is in the denominator. So we can freely multiply by $(y-3).$
Q2: Similarly, we are no signs to care about. Though when dividing by $(x-1)$ we should make assumption $xneq 1.$
Q3: The final domain is affected, because $1$ is excluded.
answered Jan 27 at 22:20
user376343user376343
3,9584829
$endgroup$
Q1: No, because this is an equation, not an inequality. More details: $yneq 3$ is assumed since $(y-3)$ is in the denominator. So we can freely multiply by $(y-3).$
Q2: Similarly, we are no signs to care about. Though when dividing by $(x-1)$ we should make assumption $xneq 1.$
Q3: The final domain is affected, because $1$ is excluded.
answered Jan 27 at 22:20
user376343user376343
3,9584829
answered Jan 27 at 22:20
user376343user376343
3,9584829
answered Jan 27 at 22:20
user376343user376343
3,9584829
answered Jan 27 at 22:20
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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