[an error occurred while processing this directive]
Massey logo
Home > College of Sciences > Institute of Fundamental Sciences >
Maths First > Online Maths Help > Calculus > Differentiation > The Sign of the Derivative > Points of Inflection
SEARCH
MASSEY
MathsFirst logo College of Science Brandstrip
[an error occurred while processing this directive]

 

The Sign of the Derivative

Recall the following information: Let f(x) be a function and assume that for each value of x, we can calculate the slope of the tangent to the graph y = f(x)  at x. This slope depends on the value of x that we choose, and so is itself a function. We call this function the derivative of f(x) and denote it by f ´ (x).

 

   

The derivative of f(x) at the point x is equal to the slope of the tangent
to y = f(x) at x.

   


Points of Inflection

As we saw on the previous page, if a local maximum or minimum occurs at a point then the derivative is zero (the slope of the function is zero or horizontal). It is not, however, true that when the derivative is zero we necessarily have a local maximum or minimum. There is a third possibility. With a maximum we saw that the function changed from increasing to decreasing at that point. With a minimum it changed from decreasing to increasing. If the function has zero slope at a point, but is either increasing on either side of the point or decreasing on either side of the point we call that a point of inflection.

                              


These basic properties of the point of inflection are summarized in the following table.

Behaviour Graphs

Derivative
(slope of tangent)

at point slightly to the
left of the POI x0

Derivative
(slope of tangent)

at POI x0

Derivative
(slope of tangent)

at point slightly to the
right of the POI x0

Point of Inflection

Type I

f ´(x0) > 0


(positive, increasing)

f ´(x) = 0


(zero)

f ´(x0+) > 0


(positive, increasing)

Point of Inflection

Type II

f ´(x0) < 0


(negative, decreasing)

f ´(x) = 0
(zero)

f ´(x0+) < 0


(negative, decreasing)



Exercise

Look carefully at the points on the graph of y = f(x) where the tangent becomes horizontal. f'(x) = 0 at these points and the points are called stationary points.
Some of these stationary points are local maximums or minimums. Some are neither. These are the points of inflection.


Exercise

Look closely at each of the four examples provided in the applet below. Where does the curve y = f (x) have points of inflection? What are the values of f '(x) at these points?
Enter in some of your own curves and see if they have points of inflection and if so where. When you enter your own function, you must use the symbols + for add, for subtract, * for multiply,  / for divide, and ^ to raise to a power. You can also use various mathematical functions: sin, cos, tan, sec, cot, csc, arcsin, arccos, arctan, exp, ln, log2, log10, abs, sqrt and cubert. (Here, "abs" is the absolute value function, "sqrt" is the square root function and "cubert" is the cube root function.)


Make sure you understand the following connections between the two graphs. For a turning point (x0,y0) to be a point of inflection:

  •      f '(x0) = 0
  •      and f '(x0) must have the same sign for points close to but either side of x = x0.

Can't see the above java applets? Click here to see how to enable Java on your web browser. (These applets are based on free Java applets from JavaMath )

This gives a method for finding the points of inflection for a function. See later for the preferred method.

  1. Differentiate the function, f (x), to obtain f '(x).
  2. Solve the equation f '(x) = 0 for x to get the values of x at minima or maxima or points of inflection.
  3. For each x value:
    • Determine the value of f '(x) for values a little smaller and a little larger than the x value.
    • Decide whether you have a minimum/maximum or a point of inflection.
    • Calculate the value of the function at the x value for the point of inflection.

Example

Find the point of inflection on the curve of y = f(x) = 2x3 − 6x2 + 6x − 5

First, the derivative f '(x) = 6x2 − 12x + 6

Solve f '(x) = 0 = 6x2 − 12x + 6 = 6(x2 − 2x + 1) = 6(x − 1)2

There is just one solution, x = 1.

At a value of x just below 1, the value of f '(x) is positive e.g. f '(½) = 6(1 - ½)2 = 1½
At a value of x just above 1, the value of f '(x) is positive. e.g. f '(1½) = 6(1 -1 ½)2= 1½

At both sides of x = 1, the derivative is positive. There is a point of inflection when x = 1.

f(1) = -3, so (1,-3) is a point of inflection on the curve.

Exercise

To see some worked examples, get a new exercise and immediately click show answer until you are confident.

You may wish to use your computer's calculator for some of these.

Find the point of inflection on the graph of

f(x) = x4 + x3 + x2 + x +

Working:

Inflection point

In all of the examples seen so far, the first derivative is zero at a point of inflection but this is not always the case. To see points of inflection treated more generally, look forward into the material on the second derivative, concavity and points of inflection.

<< Maximum and Minimum |  Differentiation Index  |  The Second Derivative >>

 

   Contact Us | About Massey University | Sitemap | Disclaimer | Last updated: November 21, 2012     © Massey University 2003