Sketching Inverse Functions
$begingroup$
I have to find the inverse of the function $f(x) = frac{1}{x}$, where $x geq 3$. However in the answer, the coordinates seem to be given as $(3,1)$. Shouldn't it be $(3,frac{1}{3})$? Or is it that since it's a sketch, it's just an estimation of how the functions would look like and so it doesn't have to be 'accurate'?
I've just started doing these and I'm a bit confused on how I'm supposed to sketch these reflections.
functions
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
$endgroup$
add a comment |
$begingroup$
I have to find the inverse of the function $f(x) = frac{1}{x}$, where $x geq 3$. However in the answer, the coordinates seem to be given as $(3,1)$. Shouldn't it be $(3,frac{1}{3})$? Or is it that since it's a sketch, it's just an estimation of how the functions would look like and so it doesn't have to be 'accurate'?
I've just started doing these and I'm a bit confused on how I'm supposed to sketch these reflections.
functions
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
$endgroup$
add a comment |
$begingroup$
I have to find the inverse of the function $f(x) = frac{1}{x}$, where $x geq 3$. However in the answer, the coordinates seem to be given as $(3,1)$. Shouldn't it be $(3,frac{1}{3})$? Or is it that since it's a sketch, it's just an estimation of how the functions would look like and so it doesn't have to be 'accurate'?
I've just started doing these and I'm a bit confused on how I'm supposed to sketch these reflections.
functions
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
$endgroup$
I have to find the inverse of the function $f(x) = frac{1}{x}$, where $x geq 3$. However in the answer, the coordinates seem to be given as $(3,1)$. Shouldn't it be $(3,frac{1}{3})$? Or is it that since it's a sketch, it's just an estimation of how the functions would look like and so it doesn't have to be 'accurate'?
I've just started doing these and I'm a bit confused on how I'm supposed to sketch these reflections.
functions
functions
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
edited Dec 20 '18 at 4:24
mathpadawan
1,888422
1,888422
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
asked Dec 20 '18 at 4:03
lubaba ahmedlubaba ahmed
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).
Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
$endgroup$
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).
Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
$endgroup$
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
add a comment |
$begingroup$
When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).
Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
$endgroup$
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
add a comment |
$begingroup$
When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).
Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
$endgroup$
When doing the inverse of a function the x-values and y-values are swapped. This would make sense why your answer would be (3,1) rather than (3,1/3).
Another thing that is important to remember is that the inverse is relfected about the line y=x. Just something to keep in mind!
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
answered Dec 20 '18 at 4:17
JoshPuhachJoshPuhach
11
11
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
add a comment |
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
(3,1) is given as the coordinate of the original function, i.e. f(x) = 1/x, how could it's coordinates be such?
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:22
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
What is the question exactly asking? Are you supposed to sketch the graph or just determine whether it is an inverse? The inverse point of (3,1) is (1,3) because the inverse is merely swapping the x's and y's of every point in the function.
$endgroup$
– JoshPuhach
Dec 20 '18 at 4:27
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
$begingroup$
It's asking me to sketch the inverses keeping in mind the domain. I'm confused about the coordinates as f(3) = 1/3 not 1. I understand that you'd have to reflect the graph about the line y=x, but the coordinates don't seem to match.
$endgroup$
– lubaba ahmed
Dec 20 '18 at 4:29
add a comment |
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