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Xdmakdmadsak
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I have a differentiated a term and ended up with

x^2+1 all over x^2. I need to work out the turning points or stationary points of this equation. I have equated this to 0 but end up with x^2=-1 which gives me no values what are the turning points?

RDKGames
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RDKGames
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(Original post by Xdmakdmadsak)
I have a differentiated a term and ended up with

x^2+1 all over x^2. I need to work out the turning points or stationary points of this equation. I have equated this to 0 but end up with x^2=-1 which gives me no values what are the turning points?

RDKGames
Either you differentiated it wrong, or there aren’t any.
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nabsers
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Hey!

Did you differentiate the function correctly?
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Xdmakdmadsak
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original function was x^2-1 all over x. RDKGames
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RDKGames
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(Original post by Xdmakdmadsak)
original function was x^2-1 all over x. RDKGames
That has no stationary points.
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Xdmakdmadsak
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is this function an injection or surjection? RDKGames
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RDKGames
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(Original post by Xdmakdmadsak)
is this function an injection or surjection? RDKGames
What do you think? Refer to definitions
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Xdmakdmadsak
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(Original post by RDKGames)
What do you think? Refer to definitions
Its not an injection if two elements map onto the same element?
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RDKGames
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(Original post by Xdmakdmadsak)
Its not an injection if two elements map onto the same element?
If two DIFFERENT elements map onto the same 'element' then its not injective.
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RDKGames
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(Original post by Xdmakdmadsak)
so this function will be surjective. So what would the inverse function be and how would you calculate that?
Inverse functions exist only if the function itself is injective. (or more strictly, it must be injective and surjective)
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Xdmakdmadsak
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How would you workout the general expression using undetermined coefficients of this f(m)
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Xdmakdmadsak
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(Original post by Xdmakdmadsak)
RDKGames
Have you learnt about it?? Just try to employ it here in the exact same fashion.
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Xdmakdmadsak
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(Original post by RDKGames)
Have you learnt about it?? Just try to employ it here in the exact same fashion.
I have but i cant remember how to start it
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Xdmakdmadsak
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(Original post by RDKGames)
Have you learnt about it?? Just try to employ it here in the exact same fashion.
help needed
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RDKGames
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(Original post by Xdmakdmadsak)
I have but i cant remember how to start it
Look at your notes!

Denote f(x) = x^2 - 6x +3

Define

f(0) + f(1) + f(2) + f(3) + \ldots + f(n) = A_0 + A_1 n + A_2 n^2 + \ldots (*)

\implies f(0) + f(1) + f(2) + f(3) + \ldots + f(n+1) = A_0 + A_1 (n+1) + A_2 (n+1)^2 + \ldots  (**)

Then (**) - (*) means

f(n+1) = A_1 + A_2[(n+1)^2 - n^2] + A_3 [(n+1)^3 - n^3] +  \ldots

Now compare coefficients.
Last edited by RDKGames; 1 month ago
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Xdmakdmadsak
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`

(Original post by RDKGames)
Look at your notes!

Denote f(x) = x^2 - 6x +3

Define

f(0) + f(1) + f(2) + f(3) + \ldots + f(n) = A_0 + A_1 n + A_2 n^2 + \ldots (*)

\implies f(0) + f(1) + f(2) + f(3) + \ldots + f(n+1) = A_0 + A_1 (n+1) + A_2 (n+1)^2 + \ldots  (**)

Then (**) - (*) means

f(n+1) = A_1 + A_2[(n+1)^2 - n^2] + A_3 [(n+1)^3 - n^3] +  \ldots

Now compare coefficients.
How can you compare coefficeints to that?
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RDKGames
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(Original post by Xdmakdmadsak)
`


How can you compare coefficeints to that?
Whats f(n+1) as a polynomial?

Then simplify the RHS to a similar form (ignoring the terms past the 3 dots) and compare the coefficients of n^2, n, and const term.
Also reason why the rest of the A_i terms must be zero.
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Xdmakdmadsak
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is the quadratic also know as homogeneous differential equation or is that something else?
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RDKGames
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(Original post by Xdmakdmadsak)
is the quadratic also know as homogeneous differential equation or is that something else?
Something else. A differential equation involves derivatives first of all... we don’t have that here.
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