Expert: Socrates - 4/21/2009

Question
I have a piecewise function:

f(x) (-6x^2+4x   for x<0
    (3x^2-2     for x > or equal to 0


and according to the definition of derivative, to compute f'(0), we need to compute the left-hand limit
lim x-> 0-
and the right-hand limit
lim x-> 0+

but when I plug in (-1) for the LHL of -6x^2+4x and (0 or 1) for 3x^2-2 I don't get the right answer and from my understanding is the numbers closest to 0, from the left and right hand side.. what am i doing wrong?


Answer
The function you have defined is neither continuous nor differentiable at 0.

The limit for f(x) as x approaches 0 from the left will be -6x^2+4x at x=0 , so this left hand limit for f(x) is 0

The limit for f(x) as x approaches 0 from the right will be 3x^2-2 at x=0 , so this right hand limit for f(x) is -2


The left and right limits are not the same , so the function f is not continuous at 0. Since it is not continuous , it has no derivative at x=0 either.

The function will , however ,  have left and right hand derivatives at 0.

For the left hand derivative at x = 0 , put 0 in for x in
(-6x^2+4x)' = -12x+4 and get 4

For the right hand derivative at x = 0 , put 0 in for x in
(3x^2-2)' = 6x and get 0

The different values for the left and right derivative at 0 also shows that f(x) has no two sided or regular derivative at 0