Application of Yager's Theorem

In order to graph a function, it is very useful to find any sources of symmetry, as this simplifies the process of graphing, by allowing us to focus on one section of the graph, knowing that other parts will replicate the findings of that section.

Yager's Theorem, as mentioned in the previous post, extends the usual tools of finding odd and even functions, by allowing for instances where a source of symmetry is not the usual y-axis or origin.

To determine whether a function has an axes of symmetry, substitute x + a into the function and equate this to the function where -x + a is substituted.

Example 1

          For (x - 1)²

          If f(x + a) = f(-x + a)

Substituting f(x) = (x - 1)² gives:

         (x - 1 + a)² = (-x - 1 + a)²

x² -x + ax - x + 1 - a + ax - a + a² = x² + x - ax + x + 1 - a - ax - a + a²

                     -2x + 2ax -2x + 2ax = 0

/2x):                          -1 + a -1 + a = 0

                                                    a =1

Therefore, (x - 1)² has an axes of symmetry at x = 1

Example 2

          For x³

          If g(x + a) = g(-x + a)

Substituting g(x) = x³ gives:

               (x + a)³ = (-x + a)³

x³ + 3x².a + 3xa² + a³ = -x³ + 3x².a - 3xa² + a³       (Binomial Theorem)

                  2x³ + 6xa² = 0

/2x):               x² + 3a² = 0

                                  a = √(-x²/3)

Therefore, to satisfy the formula, the distance: a, must vary and hence, x³ does not have an axes of symmetry.