What is the Instantaneous Rate of Change at X=2 for the Function F(x) = 4/(x-1)?

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Homework Help Overview

The discussion revolves around finding the instantaneous rate of change (IRC) of the function F(x) = 4/(x-1) at the point x=2. Participants are exploring the application of the limit definition of the derivative.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss setting up the limit definition of the derivative and express confusion regarding the simplification of their expressions. There are attempts to manipulate the algebraic forms to isolate terms and clarify the setup.

Discussion Status

The discussion includes various attempts to simplify the expression for the limit, with some participants providing corrections to earlier posts. There is an indication of productive engagement, though no consensus on the final result is reached within the discussion.

Contextual Notes

Participants are working under the constraints of homework rules, focusing on the limit definition without providing a complete solution. There are indications of potential misunderstandings in the algebraic manipulation of terms.

wearingthin
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Homework Statement



Find the irc at X=2

F(x) = [tex]\frac{4}{x-1}[/tex]



Homework Equations



[tex]\stackrel{lim}{h\rightarrow0}[/tex][tex]\frac{F(a+h)-F(a)}{h}[/tex]



The Attempt at a Solution



I end up with [tex]\frac{4}{h+h^{2}}[/tex] or some other form of an h on the bottom. Is there something else i can do to it??
 
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[tex]\frac{4}{(2+h)-1}[/tex] - [tex]\frac{4}{2-1}[/tex]

all over h

is how i set it up originally
 
and then

[tex]\frac{4(1)}{(1+h)1}[/tex] - [tex]\frac{4(1+4)}{1(1+h)}[/tex]

still all over h
 
which becomes


[tex]\frac{h}{1+h}[/tex]

all over h

right??
 
You made the mistake in your third post; that 4 in the very top right should be an h

wearingthin said:
and then

[tex]\frac{4(1)}{(1+h)1}[/tex] - [tex]\frac{4(1+4)}{1(1+h)}[/tex]

still all over h

First factor out the four so it's easier to work with[tex]\frac{\frac{4}{h+1} - \frac{4}{1}*\frac{h + 1}{h + 1}}{h} = \frac{4\left(\frac{1}{h + 1} - \frac{h + 1}{h + 1}\right)}{h}[/tex]
then see if you can cancel that h in the denominator.
 
IRC = -4

You're a genius! Thank you


-Matt
 

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