Another introductory calc problem

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Homework Help Overview

The discussion revolves around the relationship between the slope of the secant line between two points on a function f(x) and the average of the derivatives at those points. The original poster is curious about whether the equation involving the secant slope and the average of the derivatives holds true for general functions, beyond polynomial functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the validity of the relationship by suggesting the use of Taylor series expansions and considering counter-examples. Questions about the meaning of "average" and the applicability of polynomial proofs to arbitrary functions are raised.

Discussion Status

The discussion is ongoing, with participants providing insights and suggesting directions for further exploration. Some participants have noted that the relationship may not be exact and that the difference between the two sides of the equation increases with the distance between points a and b.

Contextual Notes

One participant mentions that they are new to calculus, having only been introduced to the subject four weeks prior, which may influence their understanding of the concepts discussed.

Fifty
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Given a function f(x), I'm wondering if the slope of the secant between points a, f(a) and b, f(b) is equal to the average of the derivative at a and b.

Mathematically, I want to know whether:

\frac{f(a) - f(b)}{a - b} = \frac{f'(a) + f'(b)}{2}

is true. I can see that it is for polynomial functions, but I can't seem to prove this relationship for a general function, f(x).

This isn't homework, I'm just curious. Thanks for the help!
 
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Have you learned about Taylor series expansions yet?

Chet
 
You could try to come up with a counter-example.
Also think about what "average" means.

If you have a proof for all polynomials, then can you apply the proof to any arbitrary sum of polynomials?
Can you write an arbitrary function as a sum of polynomials?
[Chester beat me to it]

Notes:
http://homepage.math.uiowa.edu/~idarcy/COURSES/25/2_7.pdf
 
Chestermiller said:
Have you learned about Taylor series expansions yet?

Chet
Nope. I was literally just introduced to calculus four weeks ago.

EDIT: It's late here, I'm going to have a look at this thread tomorrow (and the comments already posted).
 
In a nutshell ... any function f(x) can be written as a sum like this:
$$f(x)=a_0+a_1x+a_2x^2+\cdots $$

If you already have a proof that applies to polynomials - you can apply it to that.
 
Fifty said:
Nope. I was literally just introduced to calculus four weeks ago.

Your relation is not exact. It is a good approximation, but the difference between the two sides of the relationship increases with increasing distance between the two points a and b.

Chet
 
f(a)=f(b)-(b-a)f'(b)+\frac{(b-a)2}{2}f''(b) + ...
f(b)=f(a)+(b-a)f'(a)+\frac{(b-a)2}{2}f''(a) + ...
Subtracting:
2(f(b)-f(a))=(b-a)(f'(a)+f'(b))-\frac{(b-a)2}{2}(f''(b)-f''(a))+...
Or,
\frac{f(b)-f(a)}{b-a}=\frac{f'(a)+f'(b)}{2}-\frac{(b-a)}{4}(f''(b)-f''(a))+...
A better approximation is:
\frac{f(b)-f(a)}{b-a}≈f'\left(\frac{a+b}{2}\right)
This relation is exact if f is quadratic.

Chet
 

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