Need to find where the following function increase/decrease

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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)=(-6x<sup>3</sup>+ 12x<sup>2</sup> + 18x - 30)/ (x<sup>2</sup>- 4x + 3)<sup>2</sup>
 
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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