Need to find where the following function increase/decrease

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In summary, the student is trying to find the derivative of a function, but is having trouble understanding how to do it. They made a mistake in the numerator and are now trying to fix it.
  • #1
619snake
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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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  • #2
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:
 
  • #3
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](-6x3+ 12x2 + 18x - 30)/ (x2- 4x + 3)2[/tex]
 
  • #4
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.
 
  • #5
(-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
 
  • #6
ok... I checked the process... I found the mistake... now the equation stands like this:

(-18x^2 + 54x - 42)/(x^2 - 4x + 3)^2
 
  • #7
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?
 
  • #8
umm...
6x^4 - 36x^3 + 24x^2 + 18x - 18
 
  • #9
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?
 

1. What does it mean for a function to increase/decrease?

When we say a function is increasing, it means that as the input values increase, the output values also increase. In other words, the function is getting larger as the input values get larger. On the other hand, a decreasing function means that as the input values increase, the output values decrease. So, the function is getting smaller as the input values get larger.

2. How can I determine where a function is increasing/decreasing?

To determine where a function is increasing or decreasing, we can look at the slope of the function. If the slope is positive, the function is increasing, and if the slope is negative, the function is decreasing. We can also graph the function and see if it is sloping upwards or downwards.

3. What is the difference between a local maximum/minimum and an absolute maximum/minimum?

A local maximum or minimum is a point on a graph where the function has the highest or lowest value within a specific interval. An absolute maximum or minimum, on the other hand, is the highest or lowest value of the entire function, without any restrictions on the interval.

4. Can a function be both increasing and decreasing?

No, a function cannot be both increasing and decreasing at the same time. It can only be either increasing or decreasing. However, a function can have both increasing and decreasing intervals, meaning that it may increase in one interval and decrease in another.

5. How do I find the intervals where a function is increasing/decreasing?

To find the intervals where a function is increasing or decreasing, we can set the derivative of the function equal to zero and solve for the input values. These input values will give us the critical points of the function, and by plugging them into the original function, we can determine if the function is increasing or decreasing in that interval. We can also use a graph to identify the increasing and decreasing intervals.

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