Need to find where the following function increase/decrease

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Homework Help Overview

The discussion revolves around the function f(x)=(3x^2-6x)/(x^2-4x+3) and the goal of determining where this function is increasing or decreasing. Participants are exploring the use of derivatives, specifically applying the Quotient Rule, to analyze the behavior of the function.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss finding intercepts and asymptotes, and they attempt to compute the derivative using the Quotient Rule. There are questions about the correctness of the derivative and the simplification process, with some participants expressing uncertainty about how to proceed with the derivative test.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's calculations and attempting to clarify the correct form of the derivative. There is a focus on ensuring the accuracy of the numerator in the derivative expression, and some participants are revisiting their calculations to identify mistakes.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is an emphasis on understanding the derivative's implications for the function's behavior without reaching a final conclusion.

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Homework Statement



f(x)=(3x^2-6x)/(x^2-4x+3)

The Attempt at a Solution


I already have the intercepts: x= 0, x= 2, y= 0
Horizontal Asymptote: y= 3
[STRIKE]Vertical Asymptote: x=[/STRIKE]

Derivative using Quotient Rule:
f'(x)=(6x-6)/(x^(2)-4x+3)-((3x^(2)-6x)(2x-4))/((x^(2)-4x+3)^(2))

I'm stuck in the process of the derivative test... got no idea how to perform it :confused:
 
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619snake said:

Homework Statement



f(x)=(3x^2-6x)/(x^2-4x+3)

The Attempt at a Solution


I already have the intercepts: x= 0, x= 2, y= 0
Horizontal Asymptote: y= 3
[STRIKE]Vertical Asymptote: x=[/STRIKE]

Derivative using Quotient Rule:
f'(x)=(6x-6)/(x^(2)-4x+3)-((3x^(2)-6x)(2x-4))/((x^(2)-4x+3)^(2))
That should be
f'(x)=[(6x-6)(x^2-4x+3) - (3x^2-6x)(2x-4)]/((x^2-4x+3)^2)

The denominator is never negative, so you should be able to tell where this function is increasing (f'(x) > 0) and decreasing (f'(x) < 0) by determining the intervals where the numerator is positive or negative.

Multiply the pairs of factors in the numerator and combine like terms. You should end up with a quadratic expression in the numerator that, with any luck, you can factor.



619snake said:
I'm stuck in the process of the derivative test... got no idea how to perform it :confused:
 
Ok, if I multiply the pairs of factors and combine like terms I get something like this... correct me if I'm mistaken:

f'(x)=[tex](-6x<sup>3</sup>+ 12x<sup>2</sup> + 18x - 30)/ (x<sup>2</sup>- 4x + 3)<sup>2</sup>[/tex]
 
Can you rewrite that so that it's easier to read? Don't mix [ sup] tags in with LaTeX.

There shouldn't be an x3 term, so I think you made a mistake.
 
(-6x^3 + 12x^2 + 18x + 3) / (x^2 - 4x + 3)^2

that's what I got.. but I'm not sure
I'll do the process again to see if I made a mistake
 
ok... I checked the process... I found the mistake... now the equation stands like this:

(-18x^2 + 54x - 42)/(x^2 - 4x + 3)^2
 
That's not what I'm getting. The denominator is fine, but you have mistakes in the numerator.

The numerator is
6(x - 1)(x2 - 4x + 3) - 6x(x - 2)(x - 2)
= 6(x3 - 5x2 + 7x - 3) - 6x(x2 - 4x + 4)

If you expand those pairs of factors and collect like terms, what do you get?
 
umm...
6x^4 - 36x^3 + 24x^2 + 18x - 18
 
No, that's not even close.
Starting with 6(x - 1)(x2 - 4x + 3) - 6x(x - 2)(x - 2)
= 6(x3 - 5x2 + 7x - 3) - 6x(x2 - 4x + 4)

show me step by step what you're doing.
 
  • #10
Mark44 said:
Starting with 6(x - 1)(x2 - 4x + 3) - 6x(x - 2)(x - 2)
= 6(x3 - 5x2 + 7x - 3) - 6x(x2 - 4x + 4)

Ok, after this, I did the following (hope I get it right now)

= 6(x^3 - 5x^2 + 7x - 3) - 6x^3 + 24x^2 - 24x
= 6x^3 - 30x^2 + 42x - 18 - 6x^3 + 24x^2 - 24x
= -6x^2 -18x - 18
 
  • #11
That's better, but not perfect. It should be -6x^2 + 18x - 18, which can be factored to -6(x^2 - 3x + 3).

So f'(x) = [-6(x^2 - 3x + 3)]/[(x^2 - 4x + 3)^2]

The denominator is always >= 0, so the sign of f'(x) is controlled by -6(x^2 - 3x + 3). When is f'(x) > 0 and when is f'(x) < 0?
 

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