If you flip it in the line y=x then it is inverted. So it has an inverse. But that's not right. Why?

will'o'wisp2
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 05102017 20:14

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AHappyStudent
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 05102017 20:24
(Original post by will'o'wisp2)
If you flip it in the line y=x then it is inverted. So it has an inverse. But that's not right. Why? 
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 05102017 20:41
(Original post by AHappyStudent)
You haven't specified what f(x)17 is, but I assume that it's going to be a manytoone function. The inverse would be onetomany, which functions cannot be 
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 05102017 20:45
(Original post by will'o'wisp2)
no, it's f(x)=17 my mistake
The inverse of this function would therefore be one to many, meaning it is not a function. Therefore f(x) = 17 has no inverse 
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 05102017 20:49
(Original post by Protostar)
Then yes  this function is many to one, as every value of x will give 17 for f(x).
The inverse of this function would therefore be one to many, meaning it is not a function. Therefore f(x) = 17 has no inverse
so how do i know if something rank like this here is a one to one or not? 
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 05102017 20:57
(Original post by will'o'wisp2)
fantastic, so it has to be injective/ one to one to have in inverse
so how do i know if something rank like this here is a one to one or not?
I hardly expect they’d give something as disgusting as that but perhaps you could tell from the equation.
Usually they’ll give you an equation that can be graphed easily, or you can tell the shape of the graph from the equation. For example, you know that an equation is going to be manytoone if it’s highest power is x^{2}, or if they’re a trigonometric function graph or similar  but the easiest way of telling I think is to just do a quick sketch if you can. Like I said, I highly doubt they’d give you something like what you’ve written there  it’d be something you’d have an idea of how it looks. 
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 05102017 20:59
(Original post by will'o'wisp2)
fantastic, so it has to be injective/ one to one to have in inverse
so how do i know if something rank like this here is a one to one or not? 
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 05102017 21:03
(Original post by Protostar)
Yes  this is criteria for an inverse function to exist
I hardly expect they’d give something as disgusting as that but perhaps you could tell from the equation.
Usually they’ll give you an equation that can be graphed easily, or you can tell the shape of the graph from the equation. For example, you know that an equation is going to be manytoone if it’s highest power is x^{2}, or if they’re a trigonometric function graph or similar  but the easiest way of telling I think is to just do a quick sketch if you can. Like I said, I highly doubt they’d give you something like what you’ve written there  it’d be something you’d have an idea of how it looks.
i don't know how to draw that
(Original post by AHappyStudent)
Just prove that for for two values a and b, f(a) = f(b). This could be done with a simple example or possibly a sketch
but then do i just guess the values by trial and error? 
etothepiiplusone
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 05102017 21:08
Perhaps also by showing there is a noninflexion critical value of the function? i.e a local minimum or local maximum?

AHappyStudent
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 05102017 21:10
(Original post by will'o'wisp2)
so there's the horizontal line test which tells if it's injective but what about something fairly normal like this?
i don't know how to draw that
yo lol sketch it for me xD
but then do i just guess the values by trial and error?
Could do but for a function like that it's going to be near impossible.
Maybe you could say k = f(x), then find solutions for x. This wouldn't be easy at all for a function like this either.
Maybe you could get away with some clever argument that means that there must be a case where f(a) = f(b)
Finding a point where dx/dx = 0 and d^{2}y / dx^{2} != 0 (is not equal to zero) would justify a solution as a minima or maxima must overlap on itself. 
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 05102017 21:12
(Original post by etothepiiplusone)
Perhaps also by showing there is a noninflexion critical value of the function? i.e a local minimum or local maximum?
also
isn't the 2nd diff showing a max or a min? so what does that tell me?
(Original post by AHappyStudent)
That looks like a nasty one to sketch
Could do but for a function like that it's going to be near impossible.
Maybe you could say k = f(x), then find solutions for x. This wouldn't be easy at all for a function like this either.
Maybe you could get away with some clever argument that means that there must be a case where f(a) = f(b)
Finding a point where dx/dx = 0 and d^{2}y / dx^{2} != 0 (is not equal to zero) would justify a solution as a minima or maxima must overlap on itself. 
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 05102017 21:15
(Original post by will'o'wisp2)
so there's the horizontal line test which tells if it's injective but what about something fairly normal like this?
i don't know how to draw that
yo lol sketch it for me xD
but then do i just guess the values by trial and error?
which is equal to (2)^{2} + x^{4}
If you’re including all values of x (not just > 0), to my knowledge at least, you’d know that this is many to one, as the highest power of x is a multiple of 2 
I hate maths
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 05102017 21:16
(Original post by will'o'wisp2)
so there's the horizontal line test which tells if it's injective but what about something fairly normal like this?
i don't know how to draw that
yo lol sketch it for me xD
but then do i just guess the values by trial and error? 
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 05102017 21:17
(Original post by Protostar)
This can be rewritten as (2  x^{2})^{2}
which is equal to (2)^{2} + x^{4}
If you’re including all values of x (not just > 0), to my knowledge at least, you’d know that this is many to one, as the highest power of x is a multiple of 2
(Original post by I hate maths)
I assume you mean but it doesn't matter much.You can easily see that the function is even (symmetrical about the yaxis) because , so it's many to one. 
AHappyStudent
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 05102017 21:19
(Original post by will'o'wisp2)
what is this?
also
isn't the 2nd diff showing a max or a min? so what does that tell me?
so like it has to curve? but what about 1/x graphs?
For it to be many to one, there has to be two distinct points along the curve at the same yvalue. 1/x isn't an example of this as there is no point where 1/a = 1/b for a != b. 
etothepiiplusone
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 05102017 21:31
A similar argument follows for local minima.
Local maximum and local minima occur when f'(x)=0, but f'(x)=0 can imply an inflexion point, which doesn't guarantee onetooneness.
, and you are right in the use of the second derivative to confirm this.
I don't know much about maths of this level and was mostly posting to see if I was correct. Someone who does know may wish to verify or clear my point up. 
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 05102017 21:37
(Original post by AHappyStudent)
If the 2nd diff isn't equal to zero then the point isn't a point of inflection, which is a case where the graph wouldn't have to be manytoone. Take y = x^3 as an example.
For it to be many to one, there has to be two distinct points along the curve at the same yvalue. 1/x isn't an example of this as there is no point where 1/a = 1/b for a != b.
ah so 1/x is a one to one then
(Original post by etothepiiplusone)
Yes, so where the function goes up and comes back down again, anywhere (local maximum), it will have to bump into some points twice so is not onetoone.
A similar argument follows for local minima.
Local maximum and local minima occur when f'(x)=0, but f'(x)=0 can imply an inflexion point, which doesn't guarantee onetooneness.
, and you are right in the use of the second derivative to confirm this.
I don't know much about maths of this level and was mostly posting to see if I was correct. Someone who does know may wish to verify or clear my point up. 
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 06102017 09:27
(Original post by will'o'wisp2)
so if i do the second diff and it ain't 0 then it's a one to one?
ah so 1/x is a one to one then
i see
1st diff = 0 and 2nd diff = zero : Doesn't provide a suitable example to say that the graph is many to one
1st diff = 0 and 2nd diff != zero : Provides an example that shows that the function must be many to one 
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 07102017 01:17
(Original post by AHappyStudent)
No, the other way around. If the 2nd diff is zero then the point you have tested is a point of inflection, so does not provide evidence for it being manytoone
1st diff = 0 and 2nd diff = zero : Doesn't provide a suitable example to say that the graph is many to one
1st diff = 0 and 2nd diff != zero : Provides an example that shows that the function must be many to one
guess i'll leave this diff stuff out if i doesn't provide solid proof 
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 07102017 13:57
(Original post by will'o'wisp2)
fantastic, so it has to be injective/ one to one to have in inverse
so how do i know if something rank like this here is a one to one or not?
Edit: I have no idea where I was going with this. I should've really explained that is not onetoone as its derivative is neither strictly positive nor strictly negative, then perhaps using the product rule to conclude that the function is not onetoone.
Basically, a continuous function is onetoone if its derivative is strictly positive or strictly negative.Last edited by _gcx; 07102017 at 18:30.
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