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## Homework Statement

Given

## Homework Equations

## The Attempt at a Solution

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- Thread starter MitsuShai
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- #1

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Given

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- #2

rock.freak667

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- #3

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

- #4

rock.freak667

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How did you get that?

- #5

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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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- #6

Char. Limit

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Now how about d and e?

Here's a clue...

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

Here's a clue...

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

- #7

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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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- #8

Char. Limit

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

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

[tex]\frac{5(1-x)(x-1)}{(3x^2-10x+3)^2}[/tex]

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

- #9

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

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

[tex]\frac{5(1-x)(x-1)}{(3x^2-10x+3)^2}[/tex]

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....

- #10

Char. Limit

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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

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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.....

- #12

Char. Limit

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That is the point where it's increasing.

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

- #13

vela

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

- #14

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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

- #15

Char. Limit

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Oh...

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

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

- #16

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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)???

- #17

vela

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- #18

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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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- #19

Char. Limit

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I do believe so.

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

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

- #20

vela

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

- #21

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

nothing important, that was a typo, how about this for an answer:

(-infinity, -1) U (1, 3) U (3, infinity) --decreasng interval

- #22

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Thanks guys the answers were right :)

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