# Derivatives & the Slope of a graph

Given

## The Attempt at a Solution

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rock.freak667
Homework Helper
If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.

If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.
I know I found that it is increasing on the interval (-1,1), but that's wrong

rock.freak667
Homework Helper
How did you get that?

How did you get that?
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.

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Char. Limit
Gold Member
Now how about d and e?

Here's a clue...

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

Now how about d and e?

Here's a clue...

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

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Char. Limit
Gold Member
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?

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?
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....

Char. Limit
Gold Member
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.

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

Char. Limit
Gold Member
Well, that's strange...

That is the point where it's increasing.

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

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

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

Char. Limit
Gold Member
Oh...

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

Oh...

So it would be -1<x<1, x=/=(1/3) then...
huh?
oh are you saying because it's undefined there its (-1, 1/3) U (1/3, 1)???

vela
Staff Emeritus
Homework Helper
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.

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.
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

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Char. Limit
Gold Member
I do believe so.

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

vela
Staff Emeritus
Homework Helper