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limits of different functions


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#1 bahozkaleez

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Posted 1 January 2017 - 06:29 PM

I am currently reading a book on calculus and I have come across a problem which I can't solve. I do feel like the answer is something simple. Please note that I am fairly new to calculus.

 

mathz.PNG

 

Thank you.


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#2 studiot

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Posted 1 January 2017 - 08:07 PM

What is your definition of f'(x) ?

 

Are you sure the question does not say

 

If f(x) is differentiable at xo then prove that.....etc ?

 

Consider the differentiability of f(x) = |x| and of f(x) =x2 at xo = 0

 

 

 


Edited by studiot, 1 January 2017 - 09:13 PM.

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#3 Function

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Posted 1 January 2017 - 08:57 PM

What is your definition of f'(x) ?

 

Are you sure the question does not say

 

If f(x) is differentiable at xo then prove that.....etc ?

 

Consider f(x) = |x| at xo = 0

 

I've lost all my maths skills (except for statistics), but we always denoted the derivative function of f(x) as f'(x).


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So I guess we are who we are for a lot of reasons. And maybe we'll never know most of them.

But even if we don't have the power to choose where we come from, we can still choose where we go from there.

We can still do things. And we can try to feel okay about them.

 

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#4 studiot

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Posted 1 January 2017 - 09:14 PM

 

I've lost all my maths skills (except for statistics), but we always denoted the derivative function of f(x) as f'(x).

 

Yes, that's right the derived function is f'(x).


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#5 mathematic

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Posted 2 January 2017 - 01:22 AM

Place  -f(x_0)+f(x_0) in the middle of the numerator.  It is obvious what follows.


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#6 Sriman Dutta

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Posted 2 January 2017 - 06:07 AM

By defining f(x)=x^2+3x or anything else, you can put its value in the equation and get the desired results.


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#7 zztop

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Posted 2 January 2017 - 06:50 AM

I am currently reading a book on calculus and I have come across a problem which I can't solve. I do feel like the answer is something simple. Please note that I am fairly new to calculus.

 

attachicon.gifmathz.PNG

 

Thank you.

\frac{f(x_0+\Delta x)-f(x_0-\Delta x)}{2 \Delta x}=\frac{f(x_0+\Delta x)-f(x_0)+f(x_0)-f(x_0-\Delta x)}{2 \Delta x}

 

"mathematic" beat me to it


Edited by zztop, 2 January 2017 - 06:53 AM.

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#8 Sriman Dutta

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Posted 2 January 2017 - 06:59 AM

Hey zztop, you forget the limit.

;-)


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We are what we repeatedly do.
Excellence is then not an act but a habit.
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#9 bahozkaleez

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Posted 3 January 2017 - 09:10 AM

\frac{f(x_0+\Delta x)-f(x_0-\Delta x)}{2 \Delta x}=\frac{f(x_0+\Delta x)-f(x_0)+f(x_0)-f(x_0-\Delta x)}{2 \Delta x}

 

"mathematic" beat me to it

Thank you, this really helped me.


Edited by bahozkaleez, 3 January 2017 - 09:10 AM.

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