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Help with interpreting a derivative of a given function geometrically.

  1. Jun 27, 2014 #1
    This is one of the the things I did not quite master in my calculus 1 course last semester. I understand for a function to be different on a point a. It must be defined at point a n not have any cusp or appear vertically tangent.

    My question is for a general function. How to I sketch it's derivative?

    What I do know: horizontal tangents are very important because these points will lie on the x-axis. Pay attention to see where the line is decreasing and increasing. Do not understand what to do with the increasing and decreasing portion. My book Stewart does not really explain this portion of derivatives in terms I can understand. Sorry for this elementary question.
  2. jcsd
  3. Jun 27, 2014 #2
    Also it must be continuous at point a
  4. Jun 27, 2014 #3


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    If, at a given point, f is increasing, then f' is positive there. If f is decreasing, f' is negative. In order to transition from "increasing" to "decreasing" the function must "level off" so the derivative there is 0 as you say- to go from positive or negative the derivative must pass through 0. If a function is "convex upward" then the second derivative is positive which means that the first derivative is increasing. Conversely, if a function is "concave downward" the derivative is decreasing.

    (You added "Also it must be continuous at point a". It is not clear to me whether you intended "it" to mean the function or the derivative. If a function is not continuous at a, it cannot be differentiable so the question is moot. While the derivative of a differentiable function is not necessarily continuous, it can be shown that it must satisfy the "intermediate value property": If f'(a)= X and f'(b)= Y then f' must take on every value between X and Y at some point between a and b. That why, even if f' is not continuous, if f changes from increasing to decreasing, so that f' changes from positive to negative, f' must be 0 there.)
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