Reversing Derivatives: Finding the Original Function and Point of Derivation

Click For Summary

Homework Help Overview

The discussion revolves around finding the original function and the point of derivation from given limits that represent derivatives. The subject area includes calculus, specifically the concept of derivatives and their definitions.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of a derivative and its application to the given limits. There is uncertainty about how to reverse-engineer the original function from the derivative limits without integrating. Some participants suggest using the limit definition of a derivative, while others question the assumptions made about the values of a and the function itself.

Discussion Status

The discussion is active, with participants sharing their thoughts on the relationship between the limits and the original function. Some guidance is offered regarding the interpretation of the limits, but there is no explicit consensus on the values of a or the exact form of f(x). Multiple interpretations are being explored, particularly regarding the first limit.

Contextual Notes

Participants are grappling with the challenge of deriving the original function from the limits provided, and there is mention of potential confusion regarding the values of a and the nature of the functions involved. The discussion reflects a learning environment where assumptions and definitions are being questioned.

skateza
Messages
45
Reaction score
0

Homework Statement


Each of the limits represent a derivative f'(a). Find f(x) and a.
1) lim x-> 0 [(6^x) - 1]/x
2) lim x-> 0 [((4+h)^(-2)) - 0.0625]/h

2. The attempt at a solution

i'm not sure what I'm suppose to be solving for, they give us a derivative, or a partial derivative that's not completely solved and they want us to go backward to find the original function and the number that gave us the derivative they have. How do we do this without integrating.
 
Physics news on Phys.org
skateza said:

Homework Statement


Each of the limits represent a derivative f'(a). Find f(x) and a.
1) lim x-> 0 [(6^x) - 1]/x
2) lim x-> 0 [((4+h)^(-2)) - 0.0625]/h

2. The attempt at a solution

i'm not sure what I'm suppose to be solving for, they give us a derivative, or a partial derivative that's not completely solved and they want us to go backward to find the original function and the number that gave us the derivative they have. How do we do this without integrating.

How about using the definition of a derivative...
 
Okay but i never even thought of that...
lim x -> 0 [f(a+x)-f(a)]/x = lim x -> 0 [(6^x) - 1]/x

so does that mean f(a+x) = 6^x, and f(a) = 1

than to find f(x) you do f(a+x)-f(a) = 6^x - 1

than a = 6^a = 2 than you take the log of that to find a?
 
skateza said:
Okay but i never even thought of that...
lim x -> 0 [f(a+x)-f(a)]/x = lim x -> 0 [(6^x) - 1]/x

so does that mean f(a+x) = 6^x, and f(a) = 1

than to find f(x) you do f(a+x)-f(a) = 6^x - 1

than a = 6^a = 2 than you take the log of that to find a?
You never even thought of that? The problem said each was a derivative? Don't you associate derivatives with limits?

No, a is not 6^a and certainly not 2 (where did you get the 2 from?). What is 6^0?
Now what do you think f(x) and a are?

For the second problem you might want to calculate (1/4)2 as a decimal number.
 
i was being sarcastic, of course that's the first thing i did...

i think on the bottom of my last post i meant to put

f(a+x) = 6^x, and f(a) = 1 Therefore, f(x) = 6^x - 1
if you stuff a in for x, to get f(a), you get it f(a) = 6^x - 1
Since f(a) = 1
1 = 6^x -1
6^x = 2
do a log to find a

OR... do i take the limit somewhere in there to find 6^0 = 1 and it'll all make sense...
 
What about f(0 + x) = 6^x (suggestive notation... hope that rings a bell)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
26
Views
4K
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
16
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
Replies
3
Views
2K