Finding the Slope of a Function: A Scientific Approach

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Homework Help Overview

The discussion revolves around finding the slope of the function g(x) = x/(x-2) at the point (3,3). Participants are exploring the application of derivative concepts and methods.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the limit definition of the derivative and the quotient rule. There are attempts to simplify expressions and clarify the correct approach to finding the slope.

Discussion Status

Some participants have provided guidance on the methods available for finding the slope, including the definition of the derivative and the quotient rule. There is an acknowledgment of algebraic challenges in the attempts made. One participant indicates they have resolved their confusion and no longer need assistance.

Contextual Notes

There is mention of imposed homework rules that restrict the use of certain methods, such as the quotient rule, which may affect how participants approach the problem.

TommG
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I have to find the slope of the function

g(x) = x/(x-2), (3,3)

my attempt

[(3+h)/((3+h)-2)] - [(3)/(3-2)] [itex]\div[/itex] h

got rid of (3+h) and 3
[(1/-2) -(1/-2)] [itex]\div[/itex] h
0/h

answer in book is -2
 
Last edited:
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TommG said:
I have to find the slope of the function

g(x) = x/x-2, (3,3)

my attempt

[(3+h)/((3+h)-2)] - [(3)/(3-2)] [itex]\div[/itex] h

got rid of (3+h) and 3
[(1/-2) -(1/-2)] [itex]\div[/itex] h
0/h

answer in book is -2
##g(x)=\frac{x}{x-2}## right?
Can't you use the quotient rule?
 
adjacent said:
##g(x)=\frac{x}{x-2}## right?
Can't you use the quotient rule?

yes g(x)= x/(x-2)

I don't think I can use the quotient rule. Don't you need a limit? I wasn't given a limit only a function.
 
TommG said:
yes g(x)= x/(x-2)

I don't think I can use the quotient rule. Don't you need a limit? I wasn't given a limit only a function.
No.
You can solve it in two ways.
  1. The definition of the derivative of a function
  2. Quotient rule
The definition of the derivative of a function ##f## with respect to x: ##\frac{\text{d}f}{\text{d}x}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}##

The quotient rule states that ##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}## Where u is the numerator and v is the denominator.

The quotient rule is much easier.
 
TommG said:
I have to find the slope of the function

g(x) = x/(x-2), (3,3)

my attempt

[(3+h)/((3+h)-2)] - [(3)/(3-2)] [itex]\div[/itex] h

got rid of (3+h) and 3
[(1/-2) -(1/-2)] [itex]\div[/itex] h
0/h

answer in book is -2

##{x \over x-2} \not = {1 \over -2}##.

You may want to review fractions and what manipulations are allowed.
 
adjacent said:
No.
You can solve it in two ways.
  1. The definition of the derivative of a function
  2. Quotient rule
The definition of the derivative of a function ##f## with respect to x: ##\frac{\text{d}f}{\text{d}x}=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}##

The quotient rule states that ##(\frac{u}{v})'=\frac{u'v-uv'}{v^2}## Where u is the numerator and v is the denominator.

The quotient rule is much easier.

Ok then I have to use the first option. Not allowed to use the second option yet.

this is the definition I have to use

The definition of the derivative of a function ##f## with respect to x: ##\frac{\text{d}f}{\text{d}x}=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}##
 
TommG said:
I have to find the slope of the function

g(x) = x/(x-2), (3,3)

my attempt

[(3+h)/((3+h)-2)] - [(3)/(3-2)] [itex]\div[/itex] h

got rid of (3+h) and 3
[(1/-2) -(1/-2)] [itex]\div[/itex] h
0/h

answer in book is -2
Your attempt:$$\frac{\frac{3+h}{(3+h)-2}-\frac{3}{3-2}}{h}$$ is correct.
You only have some algebra problems in simplifying that. Try again.

If you use that definition,you will get the derivative with respect to x. If you use 3 instead of x, you will get derivative of the function at x=3. That's what you did there in your attempt
 
Last edited:
thank all of you who helped I have figured it out. I do not need help anymore.
 
TommG said:
thank all of you who helped I have figured it out. I do not need help anymore.
Happy to know that :smile:
 

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