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Need help, URGENT maths question: (1 Viewer)

InteGrand

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Thanks! Why didn't I think of that?!
If we use L'Hospital's rule, we'll end up having to assume that f' is continuous at a after the differentiation step in order to take the limit as h -> 0 of the resulting expression. But we aren't given this fact, so we can't assume it, so can't really use L'Hospital like this.
 

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If we use L'Hospital's rule, we'll end up having to assume that f' is continuous at a after the differentiation step in order to take the limit as h -> 0 of the resulting expression. But we aren't given this fact, so we can't assume it, so can't really use L'Hospital like this.
 

InteGrand

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We used that assumption in the last equality. I.e. assuming that f'(a + ph) will tend to f'(a) as h -> 0. This essentially is using the assumption that f' is continuous at a (by definition of continuity).

(In general it could happen that f'(a + t) has no limit even as t -> 0.)
 

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We used that assumption in the last equality. I.e. assuming that f'(a + ph) will tend to f'(a) as h -> 0. This essentially is using the assumption that f' is continuous at a (by definition of continuity).

(In general it could happen that f'(a + t) has no limit even as t -> 0.)
But if the question states that f is differentiable at x=a, can't we assume that f' is continuous at a?
 

InteGrand

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leehuan

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I love how I'm seeing questions that I already asked InteGrand.
 

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InteGrand

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That last Q. is essentially just based on using the higher-order derivative test ( https://en.wikipedia.org/wiki/First_derivative_test#Higher-order_derivative_test ). So you look for places where the first and second derivatives are simultaneously 0 for a horizontal point of inflexion, and out of these points, the points where the first non-zero derivative at the point is an odd derivative (e.g. third or fifth derivative) will be a horizontal point of inflexion.
 

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That last Q. is essentially just based on using the higher-order derivative test ( https://en.wikipedia.org/wiki/First_derivative_test#Higher-order_derivative_test ). So you look for places where the first and second derivatives are simultaneously 0 for a horizontal point of inflexion, and out of these points, the points where the first non-zero derivative at the point is an odd derivative (e.g. third or fifth derivative) will be a horizontal point of inflexion.
Sorry, but I still don't get the question. Because when I differentiate it, I get something very complex, and it is very hard to solve it. Is there a simpler way to do this?

In order for this to be true, we need g to be of fixed sign (and integrable of course), rather than simply non-zero. (If g is given to be continuous though, then g(x) =/= 0 suffices.)

A proof of that result is given here: https://en.wikipedia.org/wiki/Mean_...rst_Mean_Value_Theorem_for_Definite_Integrals .
Is there any way this proof connects to this question I answered before it, because it says if I generalise the proof below, I will be able to answer the question.

Just another question:



Is this right, or am I missing something
 

InteGrand

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Sorry, but I still don't get the question. Because when I differentiate it, I get something very complex, and it is very hard to solve it. Is there a simpler way to do this?



Is there any way this proof connects to this question I answered before it, because it says if I generalise the proof below, I will be able to answer the question.
You could also have answered the first Q. in a similar way to the Wikipedia proof linked. Then the Wikipedia proof becomes a generalisation of the first proof.
 

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You could also have answered the first Q. in a similar way to the Wikipedia proof linked. Then the Wikipedia proof becomes a generalisation of the first proof.
Right, thanks. Do you mind (I know this sounds stupid, but) also giving a proof of the question below, because I get the derivative test, but when I differentiate it I get something very complex, so its hard to determine the points of inflexion.

Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?

I also have another question:
 

leehuan

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I asked my calculus tutor that one











As does the other case



And then sub them back into the above equations.

Note: One of them does yield the answers at the back of the textbook

And he was way too lazy to test for P.O.I.
 

1008

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I asked my calculus tutor that one











As does the other case



And then sub them back into the above equations.

Note: One of them does yield the answers at the back of the textbook

And he was way too lazy to test for P.O.I.
Thanks :) I have another question, though:


 
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