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Why can f(x)-17 have no inverse?

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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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?


You haven't specified what f(x)-17 is, but I assume that it's going to be a many-to-one function. The inverse would be one-to-many, which functions cannot be
Original post by AHappyStudent
You haven't specified what f(x)-17 is, but I assume that it's going to be a many-to-one function. The inverse would be one-to-many, which functions cannot be


no, it's f(x)=17 my mistake
Original post by will'o'wisp2
no, it's f(x)=17 my mistake


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 :smile:
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 :smile:


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 f(x)=x5+8x4−9x2−2sin2xf(x)=\dfrac{x^5 + 8x^4 -9x^2 -2}{sin^2 x} is a one to one or not?
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 f(x)=x5+8x4−9x2−2sin2xf(x)=\dfrac{x^5 + 8x^4 -9x^2 -2}{sin^2 x} is a one to one or not?


Yes - this is criteria for an inverse function to exist :smile:

I hardly expect they’d give something as disgusting as that :lol: 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 many-to-one if it’s highest power is x2, 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.
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 f(x)=x5+8x4−9x2−2sin2xf(x)=\dfrac{x^5 + 8x^4 -9x^2 -2}{sin^2 x} is a one to one or not?


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
Original post by Protostar
Yes - this is criteria for an inverse function to exist :smile:

I hardly expect they’d give something as disgusting as that :lol: 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 many-to-one if it’s highest power is x2, 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.


so there's the horizontal line test which tells if it's injective but what about something fairly normal like this?

f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt 4-x^2 } 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

yo lol sketch it for me xD

but then do i just guess the values by trial and error?
is something rank like this here f(x)=x5+8x4−9x2−2sin2xf(x)=\dfrac{x^5 + 8x^4 -9x^2 -2}{sin^2 x} is a one to one or not?


Perhaps also by showing there is a non-inflexion critical value of the function? i.e a local minimum or local maximum?
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?

f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt 4-x^2 } 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?


That looks like a nasty one to sketch :tongue:

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 d2y / dx2 != 0 (is not equal to zero) would justify a solution as a minima or maxima must overlap on itself.
Original post by etothepiiplusone
Perhaps also by showing there is a non-inflexion critical value of the function? i.e a local minimum or local maximum?

what is this?
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 :tongue:

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 d2y / dx2 != 0 (is not equal to zero) would justify a solution as a minima or maxima must overlap on itself.

so like it has to curve? but what about 1/x graphs?
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?

f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt 4-x^2 } 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?


This can be rewritten as (-2 - x2)-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 will'o'wisp2
so there's the horizontal line test which tells if it's injective but what about something fairly normal like this?

f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt 4-x^2 } 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?


I assume you mean f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt{4-x^2}} but it doesn't matter much.You can easily see that the function is even (symmetrical about the y-axis) because f(x)=f(−x)f(x)=f(-x), so it's many to one.
Original post by Protostar
This can be rewritten as (-2 - x2)-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

i see
Original post by I hate maths
I assume you mean f(x)=1−4−x2f(x)=\dfrac{1}{-\sqrt{4-x^2}} but it doesn't matter much.You can easily see that the function is even (symmetrical about the y-axis) because f(x)=f(−x)f(x)=f(-x), so it's many to one.

ah ok thanks a bunch
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?


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 many-to-one. 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 y-value. 1/x isn't an example of this as there is no point where 1/a = 1/b for a != b.
Original post by will'o'wisp2
what is this?
also
isn't the 2nd diff showing a max or a min?


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 one-to-one.

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 one-to-one-ness.
, 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.
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 many-to-one. 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 y-value. 1/x isn't an example of this as there is no point where 1/a = 1/b for a != b.

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
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 one-to-one.

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 one-to-one-ness.
, 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.


i see
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


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 many-to-one

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
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 many-to-one

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


but if the first diff is 0 then is the second diff not automatically 0 anyway?

guess i'll leave this diff stuff out if i doesn't provide solid proof :colonhash:
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 f(x)=x5+8x4−9x2−2sin2xf(x)=\dfrac{x^5 + 8x^4 -9x^2 -2}{sin^2 x} is a one to one or not?


<snip>

Edit: I have no idea where I was going with this. I should've really explained that x5+8x4−9x2−2x^5 + 8x^4 - 9x^2 - 2 is not one-to-one as its derivative is neither strictly positive nor strictly negative, then perhaps using the product rule to conclude that the function csc⁡2x(x5+8x4−9x2−2)\csc^2 x\left(x^5+8x^4 - 9x^2 - 2\right) is not one-to-one.

Basically, a continuous function ff is one-to-one if its derivative is strictly positive or strictly negative.
(edited 6 years ago)

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