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Thread: Help with first derivative test

  1. #1
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    Help with first derivative test

    Alright, I've been thinking of going back to school and have been reviewing some calculus I took years ago. I'm having a bit of a problem with a First Derivative Test question and hoping someone can help me out. As I understand the first derivative test I am to take the derivative of a given function f(x), set the result to zero, and find the value(s) of x that makes f'(x) = 0. these are the critical points where the graph of f(x) *might* change direction. so I have this seemingly harmless function to work on:

    f(x) = x3 - (3/2)x2 + 2x - 3

    I can derive that pretty easily: f'(x) = 3x2 - 3x + 2

    when I set this to zero I get 3x2 - 3x + 2 = 0 -> x2 - x + 2/3 = 0

    Now my troubles begin, I can't factor that directly, if I use the quadratic formula I get a negative in the square root, that leaves me with completing the square. I can do that but am not sure what to do with that in order to find the critical points. When I go back to the intent of the problem I am really to find where f(x) increases and where it decreases. The critical points are supposed to tell me where directional changes occur. But I recognize that since this f(x) is a cubic it increases everywhere. If I did not know that I'd be lost. Can someone help me find my way?
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  2. #2
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    Re: Help with first derivative test

    The first derivative test examines a function's monotonic properties -- in other words, where the function is increasing or decreasing. If you graph your function, you will observe that the function is always increasing. Since the function never changes from increasing to decreasing (and vice versa), no maximum or minimum is achieved.
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  3. #3
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    Re: Help with first derivative test

    Completing the square gives you x^2- x+ 1/4- 1/4+2/3= (x- 1/2)^2+ 5/12. Just as you saw when you got a negative number under the square root using the quadratic formula, that is never 0. There are no critical points, the derivative is always positive, and the graph is always increasing just as you say.
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  4. #4
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    Re: Help with first derivative test

    Thanks HallsofIvy. I had completed the square and had gotten what you show. so, to be clear, there are no critical points because the derivative cannot be solved for zero? or because the derivative is positive. (i suppose had it been negative that would have implied an always decreasing graph?)

    thanks,
    scott
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  5. #5
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    Re: Help with first derivative test

    Yes.
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