Poorly worded. It is equal to left and right IFF differentiable.
Y = absolute value of x
at x = 0
the symmetric derivative equals 0
The right hand derivative equals positive 1
The left hand derivative equals negative 1
The symmetric derivative is between the left and right derivative
Is there a case where the symmetric derivative is not between the left and right hand derivative?
To be more accurate I should have said it is equal to left and right IFF it is differentiable AT THAT POINT.
i.e. it must have equal left and right derivatives at that point, otherwise the symmetric derivative simply does not exist.
That function would lack a well-defined derivative (at that point).
But tbh, I concede I could be wrong. This is not my wheelhouse anymore.
Because 1 / 0 is undefined or does not exist
The right and left derivative of
y = 1 / ( x * x )
do not exist or are undefined when x = 0
but the symmetric derivative exists and is equal to 0
OK, sure you can deliberately choose to use a discontinuity as an example of an exception.
Your original question made it seem like you were asking about normal continuous functions, but you were setting up a gotcha the whole time. Boo!
I am just saying that the symmetric derivitive is not the same as the derivitive that you get if the left and right derivatives are equal
I want to know when the left and right derivatives are not equal but are both defined and real numbers
Is the symmetric derivative always between the right and left derivative
Now Wikipedia seems to say it is the average of the right and left derivative when both such derivatives exist
Which would mean it is between the two
But I am unsure
So what I want to know is
if the right derivative exists and the symmetric derivative exists and the left derivative exists
But the left derivative does not equal the right derivative
Will the symmetric derivative always be between the right and left derivatives in such a case
If the right and left derivatives are real numbers