leehuan
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Because L'Hopitals comes after that stuff in the MATH1151 bookThanks! Why didn't I think of that?!
Because L'Hopitals comes after that stuff in the MATH1151 bookThanks! 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.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.
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).
But if the question states that f is differentiable at x=a, can't we assume that f' is continuous at a?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.)
No. There exist functions that are differentiable somewhere but not continuously differentiable there. An example of such a function is found here: https://en.wikipedia.org/wiki/Differentiable_function#Differentiability_classes .But if the question states that f is differentiable at x=a, can't we assume that f' is continuous at a?
In continuation to this question, I have another:Just another question:
Is this right, or am I missing something
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.)In continuation to this question, I have another:
Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?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 .
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?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.
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.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 .
Just another question:
Is this right, or am I missing something
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.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.
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.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.
Thanks InteGrand, I get the proof on Wikipedia. But does that connect to my previous proof at all?
I also have another question:
Thanks I have another question, though: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:
I think the same Q got asked and answered here: http://community.boredofstudies.org...271/laters-maths-help-thread.html#post7133125 .Thanks I have another question, though:
Thanks InteGrand! I had this another question:I think the same Q got asked and answered here: http://community.boredofstudies.org...271/laters-maths-help-thread.html#post7133125 .
I believe this Q. was asked here before, and answered by Sy123: http://community.boredofstudies.org...1437/intermediate-value-theorem-question.html .Thanks InteGrand! I had this another question:
I get the first part, but can't do the second one...