Test the following equation for symmetry: `y=5/(x^2+0.3)` Please show all your work.

Expert Answers
txmedteach eNotes educator| Certified Educator

There are a few main types of symmetry we look for. Three primary forms of symmetry are those that display the following relationships (see links below):

1) `f(-x) = -f(x)` This is "odd" symmetry

2) `f(-x) = f(x)` This is "even" symmetry

3) `f(x) = f^-1(x)` This symmetry is found when a function is its own inverse.

We're going to test the function `y(x)` for each of these symmetries:

Test for "odd" and "even" symmetry:

To test for odd symmetry, we check to see whether `y(-x) = -y(x)`.

Let's check:

`y(-x) = 5/((-x)^2+0.3)=5/(x^2+0.3)`

Clearly, `y(-x) != -y(x)`.  In fact, we just demonstrated that `y(-x) = y(x)`

This function is not odd. However, we just proved that it has even symmetry.

Test for inverse symmetry:

To test whether the function has this sort of symmetry, we check whether reversing the variables changes the function.

Let's switch them:

`x = 5/(y^2+0.3)`

Let's now solve for  by multiplying through by `y^2+0.3`

`x(y^2+0.3) = 5`

`xy^2 + 0.3x = 5`

`xy^2 = 5 - 0.3x`

`y^2 = 5/x - 0.3`

`y = sqrt(5/x - 0.3)`

Clearly, the inverse of the function is not equivalent to the function, itself.

Therefore, we have only one type of symmetry in this case: even symmetry.

Hope that helps!