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Many thanks.

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- Thread starter Wicketer
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- #1

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Many thanks.

- #2

maajdl

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Try to see if you can built a primitive F(c) such that F(c²+1)=F(c) .

If you can find one, then f is its derivative.

I guess it might be useful to observe that c²+1 > c .

Defining F in the interval [0,1] might be all that is necessary.

Will F be continuous, will its derivative be continuous, ???

Did you give the full statement of the question?

Try to graph such a function.

Give it a try.

If you can find one, then f is its derivative.

I guess it might be useful to observe that c²+1 > c .

Defining F in the interval [0,1] might be all that is necessary.

Will F be continuous, will its derivative be continuous, ???

Did you give the full statement of the question?

Try to graph such a function.

Give it a try.

Last edited:

- #3

HallsofIvy

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- #4

disregardthat

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But, it is easily seen that no such function can exist, because the derivative of F(x^2+1) is equal to the derivative of F(x), which means that 2xf(x^2+1) = f(x). Plugging in x = 0 shows that f(0) = 0. Assuming you mean non-zero as in f can not take the value 0.

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