Inverse functions help.
Hi
In the following question the second paragraph of the solution says, WLOG, a<b, then f(a)>f(b). Could anyone explain where f(a)> f(b) comes from? I understand rest of the solution, just not this part
Thank you!
In the following question the second paragraph of the solution says, WLOG, a<b, then f(a)>f(b). Could anyone explain where f(a)> f(b) comes from? I understand rest of the solution, just not this part
Thank you!
Original post by DFranklin
Because by the way you're setting things up, f(a) = b and f(b) = a.
Ah, I see. Thank you!
What they've done algebraically is more intuitive in a geometric sense. (at least to me!)
If a function meets it's inverse, the intersection MUST lie on the line reflection.
So we have 3 equations which we can solve 'simultaneously' :
1. y = f(x)
2. y = f^-1 (x)
3. y = x .... since this is the line of reflection.
so, when solving f(x) intersect with inverse, we can choose instead to solve
either
f(x) = x
or f^-1(x) = x
which will always give the same solutions.
hth
If a function meets it's inverse, the intersection MUST lie on the line reflection.
So we have 3 equations which we can solve 'simultaneously' :
1. y = f(x)
2. y = f^-1 (x)
3. y = x .... since this is the line of reflection.
so, when solving f(x) intersect with inverse, we can choose instead to solve
either
f(x) = x
or f^-1(x) = x
which will always give the same solutions.
hth
Original post by begbie68
What they've done algebraically is more intuitive in a geometric sense. (at least to me!)
If a function meets it's inverse, the intersection MUST lie on the line reflection.
So we have 3 equations which we can solve 'simultaneously' :
1. y = f(x)
2. y = f^-1 (x)
3. y = x .... since this is the line of reflection.
so, when solving f(x) intersect with inverse, we can choose instead to solve
either
f(x) = x
or f^-1(x) = x
which will always give the same solutions.
hth
If a function meets it's inverse, the intersection MUST lie on the line reflection.
So we have 3 equations which we can solve 'simultaneously' :
1. y = f(x)
2. y = f^-1 (x)
3. y = x .... since this is the line of reflection.
so, when solving f(x) intersect with inverse, we can choose instead to solve
either
f(x) = x
or f^-1(x) = x
which will always give the same solutions.
hth
Is this always true? It does not hold for f(x) = -1/x, f(x)= -x^3.
The first is self-inverse, with no points of intersection on the line y=x.
The second meets its inverse at (-1,1), (0,0), (1,-1), 2 of which do not lie on the line y=x
Original post by BobbJo
Is this always true? It does not hold for f(x) = -1/x, f(x)= -x^3.
The first is self-inverse, with no points of intersection on the line y=x.
The second meets its inverse at (-1,1), (0,0), (1,-1), 2 of which do not lie on the line y=x
The first is self-inverse, with no points of intersection on the line y=x.
The second meets its inverse at (-1,1), (0,0), (1,-1), 2 of which do not lie on the line y=x
Yes, of course.
I'd only tried it with situations that work!
The situations you describe have y = -x as a line of reflection & this will work as the '3rd simultaneous'
Obviously f(x) = a-x is a self- inverse, and will be coincident for all x, and all a.
I'll have to rethink ... for now, not always work, but often .... and certainly in the case in OP ...
(edited 5 years ago)
Original post by BobbJo
Is this always true? It does not hold for f(x) = -1/x, f(x)= -x^3.
The first is self-inverse, with no points of intersection on the line y=x.
The second meets its inverse at (-1,1), (0,0), (1,-1), 2 of which do not lie on the line y=x
The first is self-inverse, with no points of intersection on the line y=x.
The second meets its inverse at (-1,1), (0,0), (1,-1), 2 of which do not lie on the line y=x
Hi,
Thank you for those examples.
The second function you provided has inverse which can be obtained by reflecting the function in the line y=-x instead of usual y=x, couldn't figure out why. Could you enlighten? Thank you!
Original post by Quantum Horizon1
Hi,
Thank you for those examples.
The second function you provided has inverse which can be obtained by reflecting the function in the line y=-x instead of usual y=x, couldn't figure out why. Could you enlighten? Thank you!
Thank you for those examples.
The second function you provided has inverse which can be obtained by reflecting the function in the line y=-x instead of usual y=x, couldn't figure out why. Could you enlighten? Thank you!
y=-x^3
-y=x^3
x=(-y)^1/3
therefore f^-1(x)=-x^1/3. This is obtained by the usual method of reflecting into the line y=x.
-x^(1/3)=-x^3
x=0,1,-1
https://www.desmos.com/calculator/zqfmgrmnrj
Here is a graph
Original post by Quantum Horizon1
Hi,
Thank you for those examples.
The second function you provided has inverse which can be obtained by reflecting the function in the line y=-x instead of usual y=x, couldn't figure out why. Could you enlighten? Thank you!
Thank you for those examples.
The second function you provided has inverse which can be obtained by reflecting the function in the line y=-x instead of usual y=x, couldn't figure out why. Could you enlighten? Thank you!
it happens with both of BobbJo's examples.
I think it's because to arrive at y = -x^3, we've reflected the graph of y=x^3 in the line x=0 OR we've reflected in the line y=0, and so we can reflect the line y=x in the same lines.
In other words, each of those original graphs have rotational symmetry order 2 about origin.
nice question. Where did you get it from if you dont mind me asking?
Original post by Rohan77642
nice question. Where did you get it from if you dont mind me asking?
It was from Art of Problem Solving website.
Original post by begbie68
What they've done algebraically is more intuitive in a geometric sense. (at least to me!)
If a function meets it's inverse, the intersection MUST lie on the line reflection.
If a function meets it's inverse, the intersection MUST lie on the line reflection.
Even if correct (which it is for f strictly increasing), the question setter obviously felt it needed justification, so I don't think it's helpful to post something that just assumes it.
Note that if we define f(x) = 1-x (x rational), 2-x (x irrational), then f is self-inverse everywhere, but clearly the graph of f is two distinct "lines" (in quotes because each is missing an infinite number of points).
(edited 5 years ago)
Original post by DFranklin
Note that if we define f(x) = 1-x (x rational), 2-x (x irrational), then f is self-inverse everywhere, but clearly the graph of f is two distinct "lines" (in quotes because each is missing an infinite number of points).
Note that if we define f(x) = 1-x (x rational), 2-x (x irrational), then f is self-inverse everywhere, but clearly the graph of f is two distinct "lines" (in quotes because each is missing an infinite number of points).
... but we're unable to strictly graph that function (?), since it's infinitely non-continuous...
In the same way would shouldn't graph nth term = An + b, even though the points lie on a straight line.
Original post by begbie68
... but we're unable to strictly graph that function (?), since it's infinitely non-continuous...
In the same way would shouldn't graph nth term = An + b, even though the points lie on a straight line.
In the same way would shouldn't graph nth term = An + b, even though the points lie on a straight line.
It's a function. You can make a graph of a non-continuous function (a graph is defined as the set of points (x, f(x)), which doesn't require f to be cts).
It's not exactly a "nice" function, but it's a clear counter-example.
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