$3 Per Year Web Hosting

Sunday, 28 September 2014

How to Find Inflection Points

In differential calculus, an inflection point is a point on a curve where the curvature changes sign (from plus to minus or from minus to plus). It is used in various disciplines, including engineering, economics, and statistics, to determine fundamental shifts in data. If you need to find the inflection points of a curve, scroll to Step 1.


Steps


Part 1: Understanding Inflection Points



  1. Understand concave functions. To understand inflection points, you need to distinguish between concave and convex functions. A concave function is a function in which no line that joins two points on its graph ever goes above the graph.





  2. Understand convex functions. A convex function is essentially the opposite of a concave function: it is a function in which no line that joins two points on its graph ever goes under the graph.





  3. Understand roots of a function. A root of a function is the point where the function equals zero.





    • If you were to graph a function, the roots would be the points at which the function hits the x-axis.




Part 2: Finding the Derivatives of a Function



  1. Find the first derivative of your function. Before you can find an inflection point, you’ll need to find derivatives of your function. The derivatives of the basic functions can be found in any calculus text; you’ll need to learn them before you can move on to more complex tasks. First derivatives are denoted as f ′(x). For polynomial expressions in the form axp + bx(p−1) + cx + d, the first derivative is apx(p−1) + b(p − 1)x(p−2) + c.





    • To illustrate, say you need to find the inflection point of the function f(x) = x3 +2x−1. Calculate the first derivative of that function like this:



      f ′(x) = (x3 + 2x − 1)′ = (x3)′ + (2x)′ − (1)′ = 3x2 + 2 + 0 = 3x2 + 2



  2. Find the second derivative of your function. The second derivative is the first derivative of the first derivative of the function, denoted f ′′(x).





    • In the example above, calculating the second derivative of the function would look like this:



      f ′′(x) = (3x2 + 2)′ = 2 × 3 × x + 0 = 6x



  3. Equalize the second derivative with zero. Set your second derivative equal to zero and solve the resulting equation. Your answer will be a possible inflection point.





    • In the example above, your calculation would look like this:



      f ′′(x) = 0

      6x = 0

      x=0



  4. Find the third derivative of your function. To see if your answer is actually an inflection point, find the third derivative, which is the first derivative of the second derivative of the function, denoted f ′′′(x).





    • In the example above, your calculation would look like this:



      f ′′′(x) = (6x)′ = 6




Part 3: Finding an Inflection Point



  1. Evaluate your third derivative. The standard rule for evaluating a possible inflection point is as follows: “If the third derivative doesn’t equal to zero, f ′′′(x) =/ 0, the possible inflection point is actually an inflection point.” Check your third derivative. If it does not equal zero, it is a real inflection point.





    • In the example above, your third derivative calculated to 6, not 0. It is, therefore, a real inflection point.



  2. Find the inflection point. The coordinate of the inflection point is denoted as (x,f(x)), where x is the point variable’s value at the inflection point and f(x) is the value of the function at the inflection point.





    • In the example above, recall that when you calculated the second derivative, you found that x = 0. Therefore, you need to find f(0) to determine your coordinates. Your calculation would look like this:



      f(0) = 03 +2×0−1 = −1.



  3. Note your coordinates. The coordinates of your inflection point are your x value and the value as calculated above.





    • In the example above, your inflection point coordinates are (0, -1).






Tips



  • The first derivative of a constant is always zero.


Sources and Citations







from How to of the Day http://ift.tt/1r4fwth

via Peter

No comments:

Post a Comment

$3 Per Year Web Hosting