[BlindMath] Fwd: Post Scriptum:Re: Average Rate of Change problem

[Sabra Ewing]

None of you are reading the function it correctly. The variables are X1 and X2 I believe. We will have to clarify what the function is before we do this. We also need to know if the problem is supposed to be linear or not. We can do this by making points, but I believe a problem like this is supposed to be one year. If we are doing X squared though, it is going to be a parabola. I think. This is pre-calculus, not pre-algebra, so I doubt it is as easy as plugging in the numbers for X. Because you do not do in precalculus, I doubt they want you to get the derivative of this. Also, a lot of you are using these strange Curly bracket things in your work, and it makes it very hard to read. You are not riding in a markup language. Some of you should remember that. I'm sorry to be blunt, but just saying. We normally use parentheses for the types of things you are using the bracket for. A bracket is supposed to indicate that you do something first before the parentheses.

Sabra Ewing

> On Aug 19, 2018, at 10:25 AM, Doug and Molly Miron via BlindMath <blindmath at nfbnet.org> wrote:
> 
> P. S. Elise
> 
> Before I got out of bed this morning, it occurred to me that I should have said, as some others have since, that (rate of change)=(slope of graph)=derivative=df/dx=f'(x).  This is for a function, like yours, which is continuous over an interval of a continuous variable, no breaks or jumps.  Averaging over a data set of discrete or sampled values is not quite the same thing, especially if the underlying process is not uniform, or evenly spaced on the independent variable.  I see that in several replies they have derived the same general result I did.  If one substitutes your expression into this result and do the algevbraic reduction, the result is avg. slope=x_2+x_1-1.  if x_2=x_1=x the result is 2x-1, the derivative of f(x) at x.----Doug
> 
> -------- Forwarded Message --------
> Subject:    Re: [BlindMath] Average Rate of Change problem
> Date:    Sat, 18 Aug 2018 09:49:24 -0500
> From:    Doug and Molly Miron <mndmrn at hbci.com>
> To:    FElise Berkley via BlindMath <blindmath at nfbnet.org>
> 
> 
> 
> Good day Elise,
> 
> 
> The average of a function over an interval is the integral of the
> function divided by the length of the interval.  In this case, the
> function being averaged is the derivative of f(x), so the integral of
> the derivative is the original vunction.  Therefore, the answer is
> (f(x_2)-f(x_1))/(x_2-x_1).  I don't know the rules for your pre-calculus
> course so I don't know if you're allowed to use this reasoning and have
> to take some other, more complicated, approach such as a
> finite-difference approximation.
> 
> 
> Regards,
> 
> Doug Miron
> 
> 
>> On 8/18/2018 6:54 AM, Elise Berkley via BlindMath wrote:
>> Hello, mathematicians.
>> In my precalc class, we are studying "average rate of change." I am so
>> stuck and I am asking for help with this problem.
>>    Find the average rate of change of f(x) = x2 – x + 4 from x_1  = 2 to x_2=6 .
>> If anyone can help me with this, I would greatly appreciate it. Thanks!
>> Elise Berkley
>> 
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