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Derivatives & the Slope of a graph

  1. Mar 28, 2010 #1
    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. jcsd
  3. Mar 28, 2010 #2

    rock.freak667

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    If I remember correctly, f(x) is an increasing function for f'(x) > 0 and a decreasing function for f'(x)< 0.
     
  4. Mar 28, 2010 #3
    I know I found that it is increasing on the interval (-1,1), but that's wrong
     
  5. Mar 28, 2010 #4

    rock.freak667

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    How did you get that?
     
  6. Mar 28, 2010 #5
    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
  7. Mar 28, 2010 #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?
     
  8. Mar 28, 2010 #7
    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
  9. Mar 28, 2010 #8

    Char. Limit

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    Hmm...

    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?
     
  10. Mar 28, 2010 #9
    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....
     
  11. Mar 28, 2010 #10

    Char. Limit

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    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.
     
  12. Mar 28, 2010 #11
    That's what I did when I first started it and I got (-1,1) as increasing but that's wrong.....
     
  13. Mar 28, 2010 #12

    Char. Limit

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    Well, that's strange...

    That is the point where it's increasing.

    So I don't know why you got it wrong.
     
  14. Mar 28, 2010 #13

    vela

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    There's a singularity at x=1/3, which is in (-1,1). That's probably why.
     
  15. Mar 28, 2010 #14
    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
     
  16. Mar 28, 2010 #15

    Char. Limit

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    Oh...

    So it would be -1<x<1, x=/=(1/3) then...
     
  17. Mar 28, 2010 #16
    huh?
    oh are you saying because it's undefined there its (-1, 1/3) U (1/3, 1)???
     
  18. Mar 28, 2010 #17

    vela

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    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.
     
  19. Mar 28, 2010 #18
    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
  20. Mar 28, 2010 #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.
     
  21. Mar 28, 2010 #20

    vela

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