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Homework Help: Calculus I: Monotinic Functions and First Derivative Test

  1. Mar 13, 2010 #1

    Cod

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    1. The problem statement, all variables and given/known data
    Using the derivative f'(x) = (x-1)[tex]^{2}[/tex](x+2)[tex]^{2}[/tex] answer the following:

    a) What are the critical points of f?
    b) On what intervals is f increasing or decreasing?
    c) At what points, if any, does f assume local maximum and minimum values?


    2. Relevant equations
    n/a


    3. The attempt at a solution
    From strictly observing the derivative and using the quadratic equation to check, the critical points are x=1 and x=(-2). Once I pick a number within each of the intervals (-inf, -2), (-2, 1), and (1, inf) and implement it into the equation, I consistently get positives. By definition of Corollary 3: First Derivative Test for Monotonic Functions, the equation would be increasing on every interval; therefore, making a line. The problem is, when I graph the answer to check my solution, I notice a critical point at x=(-.05). Where did this CP come from and how do I figure it out algebraically?


    Any guidance would be greatly appreciated.
     
  2. jcsd
  3. Mar 13, 2010 #2

    Mark44

    Staff: Mentor

    You should be able to tell by nothing more than inspection that the critical numbers are 1 and -2.
    You're just substituting values into the derivative function, not implementing anything.
    True, the derivative is positive everywhere except at x = 1 and x = -2, which means that the slope of the function f is >= 0, but that doesn't make the graph of f a line. For example, if f(x) = ln(x), f'(x) = 1/x > 0 for all x > 0, but the graph of the natural log function is not a line.
    The only critical numbers are 1 and -2. Your value of -.05 might have come from your mistaken assumption that the graph of the underlying function was a straight line. It isn't.
     
    Last edited: Mar 13, 2010
  4. Mar 13, 2010 #3
    What function did you plot?

    BTW, a very quick way to see that the first derivative is everywhere nonnegative is that it's a product of squares.
     
  5. Mar 13, 2010 #4

    Cod

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    Tinyboss, I plotted the derivative that was provided: [(x-1)^2][(x+2)^2].

    Mark44, thanks for the explainations; however, I'm still failing to understand how the derivative is decreasing over the interval (-.5, 1). I realize that if I found the original function it may be easier, but according to the textbook and my instructor, we should be able to get all the answers strictly from the first derivative. Sorry if I'm frustrating you, I just can't get my nugget wrapped around this question.

    Thanks again for the help y'all.
     
  6. Mar 13, 2010 #5
    Why shouldn't the derivative decrease on that interval? As long as the derivative is positive, the original function will be increasing.
     
  7. Mar 13, 2010 #6

    Cod

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    I thought for the function to be decreasing along an interval, the derivative had to be less than 0 when you substitute any value from within that interval. Isn't that correct? For example, if I evaluate at f(0), I still get a positive even though the function is decreasing on the interval (-.5, 1). Shouldn't I get a negative since the function is decreasing along that interval?
     
  8. Mar 13, 2010 #7

    Mark44

    Staff: Mentor

    With this problem, you really have keep your terminology clear, or you will confuse yourself and anyone trying to understand you. You are given a function f' that is already the derivative of some unknown function f. If you start talking about the derivative, are you talking about f' or are you talking about its derivative, f''?

    You said in another post that you graphed y = f'(x) = (x + 2)^2(x - 1)^2. The zeroes of this graph are at -2 and 1. The y-values on this graph are zero at x = - 2 and x = 1, and are positive everywhere else. This graph (y = f'(x)) is decreasing on (-inf, -2), increasing on (-2, ~-.5), decreasing on (~-.5, 1), and increasing on (1, inf). The increasing and decreasing here tell you about the sign of the derivative of the derivative, i.e., f''.

    The sign of the derivative tells you where the underlying (and unknown) function f is increasing and decreasing. Since the graph of y = f'(x) is greater than or equal to zero for all x, the graph of y = f(x) is increasing on (-inf, -2), (-2, 1), and (1, inf).
     
  9. Mar 13, 2010 #8

    Cod

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    That makes sense Mar44. Sorry I was using confusing verbage, I didn't realize I was. Thanks for the explaination and "putting up with me".
     
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