Solve Limits & Derivatives Homework: y=-x+5

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Homework Help Overview

The discussion revolves around finding the equation of the tangent line to the curve defined by the function y=(x-1)/(x-2) at the point (3,2). Participants are exploring the concept of derivatives and tangent lines in the context of limits and calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various attempts to find the derivative of the function to determine the slope of the tangent line. There are mentions of using limits to define the slope and the challenges faced due to not having learned the quotient rule yet. Some express confusion about the correct application of the definitions provided for derivatives.

Discussion Status

There are multiple approaches being explored, including the use of limit definitions for derivatives. Some participants have offered guidance on how to proceed with the calculations, while others are still grappling with the concepts and the necessary formulas. The discussion is ongoing, with no explicit consensus reached.

Contextual Notes

Participants note that their class has not yet covered the quotient rule, which is relevant to finding the derivative of the given function. The constraints of the homework assignment and the current curriculum are influencing the discussion.

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Homework Statement


Find an equation of the tangent line to the curve at the given point.

y=(x-1)/(x-2), (3,2)


Homework Equations


The Attempt at a Solution


I've had several attempts at this, all of them are wrong. My solution book says the answer should be y=-x+5

One attempt I made is:
lim as h->0 [(3+h-1)/(3+h-2) - (x-1)/(x-2)]/h

lim as h->0 [(h+2)/(h+1) - (x-1)/(x-2)]/h

and I lost track of where I wrote the next step to this attempt. I've already attempted this 4 times.
 
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you just need to use the general definition of a line:

(y-p)=m(x-r)

where (p,r) is the given point. and m is the slope.

To find the slope do you need to find the derivative and then evaluate for x=3.

But you don't need to use the definition of a derivative. just take the derivative of (x-1)/(x-2) using the quotient rule and then plug-in x=3 and you got your slope and plug in the point and slope into the formula and you'll get y=-x+5
 
bob1182006 said:
you just need to use the general definition of a line:

(y-p)=m(x-r)

where (p,r) is the given point. and m is the slope.

To find the slope do you need to find the derivative and then evaluate for x=3.

But you don't need to use the definition of a derivative. just take the derivative of (x-1)/(x-2) using the quotient rule and then plug-in x=3 and you got your slope and plug in the point and slope into the formula and you'll get y=-x+5

Thanks for the help.

The only problem I'm having currently is finding the derivative of (x-1)/(x-2). At this point in time my class hasn't learned the quotient rule yet, I checked the index and we don't learn it until the next chapter.

The only formula's that we were given in this section of this chapter is:

The tangent line to the curve y=f(x) at the point P(a,f(a)) is the line through P with slope:
m=lim as h->0 [f(a+h)-f(a)]/h

and

m=lim as x->a [f(x)-f(a)]/(x-a)

There's also a formula for instantaneous rate of change, but I don't think I need to use that for this problem.

I was able to do other problems like this, but they were stuff like y=x^2 .
 
Last edited:
o i see, well then you do have to use one of those definitions of the slope

[tex]lim_{h->0} \frac{\frac{h+2}{h+1}-\frac{x-1}{x-2}}{h}[/tex]

from here find a common denominator, to combine the fraction into 1 and you should get h to cancel.
 

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