# Homework Help: Derivatives & the Slope of a graph

1. Mar 28, 2010

### MitsuShai

1. The problem statement, all variables and given/known data
Given

2. Relevant equations

3. The attempt at a solution

Last edited: Mar 28, 2010
2. Mar 28, 2010

### rock.freak667

If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.

3. Mar 28, 2010

### MitsuShai

I know I found that it is increasing on the interval (-1,1), but that's wrong

4. Mar 28, 2010

### rock.freak667

How did you get that?

5. Mar 28, 2010

### MitsuShai

ok I re-did it again because I found an error
now I got (-1,3) for increasing
and (-infinity, -1) U (3, infinity) for decreasing
and I got it by doing the number line test (i think that's what it's called)
where I take all the criticals numbers and line them up and put test numbers in between them.

Last edited: Mar 28, 2010
6. Mar 28, 2010

### Char. Limit

Now how about d and e?

Here's a clue...

What is the shape of the graph when the derivative is zero?

7. Mar 28, 2010

### MitsuShai

it's a straight light, and on the original function that means that there's a point there where it is neither increasing or decreasing or a max or min.
I found an answer for them, but I'm not really confident
f(3)= error.....
f(-1)= 9/16
I'm suppose to get two numbers but I think I did something wrong again....I hate these problems....

Last edited: Mar 28, 2010
8. Mar 28, 2010

### Char. Limit

Hmm...

I'm sorry, but I don't get (-1,3) for increasing...

The derivative is this, as you have written, correct?

$$\frac{5(1-x)(x-1)}{(3x^2-10x+3)^2}$$

Ignore the denominator... where is the numerator zero? Where are each of the three components zero?

9. Mar 28, 2010

### MitsuShai

in the numerator: -1,1
denominator: 1/3, 3
dang it I redid it again and this time I got (-1,1) as increasing and I know that's wrong....

10. Mar 28, 2010

### Char. Limit

I said ignore the denominator... if the denominator is zero, your graph is really screwing up.

OK, so you know the graph of the derivative is zero at -1 and 1.

Thus, the minimum and the maximum must be at those two points.

11. Mar 28, 2010

### MitsuShai

That's what I did when I first started it and I got (-1,1) as increasing but that's wrong.....

12. Mar 28, 2010

### Char. Limit

Well, that's strange...

That is the point where it's increasing.

So I don't know why you got it wrong.

13. Mar 28, 2010

### vela

Staff Emeritus
There's a singularity at x=1/3, which is in (-1,1). That's probably why.

14. Mar 28, 2010

### MitsuShai

there's an error at 3 too, so does that mean I do count the denominator?
Even if I do count the denominator I get an error as the maximum

15. Mar 28, 2010

### Char. Limit

Oh...

So it would be -1<x<1, x=/=(1/3) then...

16. Mar 28, 2010

### MitsuShai

huh?
oh are you saying because it's undefined there its (-1, 1/3) U (1/3, 1)???

17. Mar 28, 2010

### vela

Staff Emeritus
Yes, exactly. The function isn't increasing at x=1/3 because it isn't defined there, so you have to split the interval as you have done.

18. Mar 28, 2010

### MitsuShai

ok, so I redid it over again and do you think these are correct:

b) (-1, 1/3) U (1/3, 1) --increasin interval
c) (-infinity, -1) U (1, 3) U (3, infinity) --decreasng interval
d) f(1)= 1/4 --max
e) f(-1)= 9/16 ---min

Last edited: Mar 28, 2010
19. Mar 28, 2010

### Char. Limit

I do believe so.

And that would solve your problem as well as allow you to solve d.

20. Mar 28, 2010

### vela

Staff Emeritus
What happens between x=-2 and x=-1?

21. Mar 28, 2010