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Derivative sketching

  1. May 19, 2007 #1
    hey, i was wondering what were the general rules for derivative sketching such as T.P's going to X-ints etc. Thnx
  2. jcsd
  3. May 19, 2007 #2

    Gib Z

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    I know that X-ints means the x intercepts, but what does T.P stand for? Sorry, mental black.

    Its just a sketch, so it doesn't have to be perfect, just the critical points correct and some what correctly shaped.

    Say we have some function f. At all the values where the graph has a turning point, or changes from negative to positive/positive to negative gradient, that means its derivative at that point is equal to zero. Any points where the tangent is vertical, the derivative is approaching infinity.

    Any intervals where the slope is positive, the derivative is positive. Where the slope is negative, derivative is negative.

    EDIT: Ahh i see now, Turning Points :) Well yea there you go
  4. May 19, 2007 #3
    ty, i also found this if anyone was having a similiar problem

    y = f(x) - y = f ’(x)
    Max Point - x-intercept (above [x-axis]-below[x-axis])
    Min Point - x-intercept (below [x-axis]-above[x-axis])
    Inflection Point - Turning Point
  5. May 19, 2007 #4

    Gib Z

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    I have no idea at all what that post means.
  6. May 19, 2007 #5
    Figure out where the critical points are, which are the points where the first derivative is zero. Of course when the first derivative is zero you have a maximum or a minimum. You can determine if the crit point is a max or a min based on how the first derivative is CHANGING, which is the second derivative. Find the second derivative and evaluate it at the critical point and if it is negative (decreasing first derivative) then it is obviously a maximum and if it is positive then it is obviously a minimum. If the second derivative is ZERO when evaluated at a critical point then you have an inflection point that happens to have a first derivative of zero!

    Now no you know where the peaks and valleys are. Now figure out where the inflection points are by setting the second derivative to zero and solving. Remember the inflection point is the point at which the function switches from concave up to concave down.

    Also, you can find the x and y intercepts of the function.

    All of these clues help you construct the graph.
  7. May 19, 2007 #6
    thankyou .

    i cant set dy/dx to 0 and solve etc as I'm not given a function, only a graph of the function, which then needs to be translated into, f'(x), f''(X) f'''(x) if possible.
    Last edited: May 19, 2007
  8. May 19, 2007 #7
    Gib Z if u were wondering what it means, the max point x-coord on a graph of f(x) will be the x-int of f'(x) where the graph goes from above the x-axis (where the f(x) is +ve) to below the x-axis (where f(x) is -ve).
  9. May 19, 2007 #8
    well then you work backwards. find the zeros of the function, the maxima and minima, and the inflection points and you can determine the polynomial (if it's a polynomail) that fits the data.
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