Question: What is the slope of the curve y= (1/x-1) at x=2?

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Homework Help Overview

The discussion revolves around finding the slope of the curve defined by the function y = (1/x - 1) at the point x = 2. Participants are exploring the correct interpretation of the function and the method to compute the slope using calculus concepts.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the definition of the derivative to find the slope but seems to misinterpret the function. Some participants suggest rewriting the function for clarity and correctness. Others discuss the process of taking the derivative and evaluating it at a specific point.

Discussion Status

Participants are actively engaged in clarifying the function's definition and the correct approach to finding the slope. There is a mix of interpretations regarding the function, and some guidance has been offered on how to compute the derivative. However, there is no explicit consensus on the correct interpretation yet.

Contextual Notes

There is a noted confusion regarding the function's notation, with some participants questioning the original poster's representation of the equation. Additionally, there is a mention of the textbook's answer, which adds to the discussion about the correctness of the calculations presented.

rafay233
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Homework Statement


Question: Find the slope of the curve at the indicated point.

y= (1/x-1) at x=2


Homework Equations


Don't know what relevant questions mean.


The Attempt at a Solution


(y(x+h) - y(x))/(x+h)-h
( (1/x-1 + h) - (1/x-1) )/h
( (1/2-1 + h) - (1/2-1) )/h
( (1+h) - (1) )/h
h/h=1

the problem is the book says it should be -1.
Help anyone?
 
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You need to be careful when writing math online. What you wrote as y= (1/x-1) implies [itex]y = \frac{1}{x} - 1[/itex]. I suspect you meant to write [itex]y = \frac{1}{x-1}[/itex] as that function gives the correct answer.

You plugged in values wrong right from the start which is why your answer is wrong. Let's rewrite your function as [itex]f(x) = \frac{1}{x-1}[/itex]. Now you need to compute [itex]\displaystyle\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}[/itex]. Now, what does [itex]f(x+h)[/itex] equal? Remember, you need to plug in the value (x+h) where ever you see an x.
 
I made a mistake on the tex look at the next reply.
 
Well this question is fairly simple, all you have to do is take the derivative of the equation:

[tex]f(x)=\frac{1}{x-1}[/tex] and then calculate the limit:

[tex]\lim_{x\rightarrow2}f'(x)[/tex] and you should get the same answer that your textbook has.

Ps: the derivative is the slope of the tangent line at a certain point.
 
could you guys tell me what you are using to make the fractions look like that? So no one gets confused?
 
rafay233 said:
could you guys tell me what you are using to make the fractions look like that? So no one gets confused?

We are using a code called LaTeX, it makes all the math look pretty and readable.

I can give a quick little tutorial for fractions but to learn everything I'll redirect you to a website.

So, first of all, you must use TAGS to show that you want to type in code. The tags are
Code:
[itex][/itex]
. Everything between the tags will be put into code. What's nice about this forum is that if you see a pretty looking formula that was coded in LaTeX, you can right click on it and select Show Math As -> TeX commands to see what the code is. Try it out for ours above.

For basic fractions the code is
Code:
[itex]\frac{a}{b}[/itex]
which will look like: [itex]\frac{a}{b}[/itex]. Everything you put in the first set of brackets will go in the numerator, everything in the second will go in the denominator. You can even do fractions in fractions:
Code:
[itex]\frac{\frac{c}{d}}{b}[/itex]
= [itex]\frac{\frac{c}{d}}{b}[/itex].

To learn more if you want, try these two sites out:

http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Symbols
http://detexify.kirelabs.org/classify.html
 
Last edited by a moderator:
k thnks and btw i got the answere :D
 

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