Thread: Determining derivative from first principles (using left and right hand derivatives)

I have a function ( which I don't know what you call it when it has 2 parts like this):


f(x) = { -1/2 (x^2 - 3) if x < -3
{ 3/2 ( x + 1 ) if x >= -3

I have to determine from first principles using the definition of the left-derivative and right derivative of f, whether the derivative of f is defined at x=-3

So for first equation
lim x-> -3 f(x) for both left and right limits using
f(x)-f(c)/(x-c) = -1/2 (x-3) = 3

and for second equation
lim x-> -3 f(x) for both left and right limits gives 3/2

Since the derivatives exist but are 2 different values, does this mean the function is not continuous at -3?

Thanks

First, you do NOT take left and right limits for both formulas. The first formula is only valid for x< -3 so you want to take the limit as x goes to -3 from below. the second formula is only valid for so you take the limit as x goes to -3 from above.

The function itself is clearly continuous: the limit from below is (-1/2)(9- 3)= -3 and the limit from above is (3/2)(-3+ 1)= (3/2)(-2)= -3 also. Since that joint value is the value of the function at x= -3, the function is continuous at -3.

You are also incorrect to say that "the derivatives exist but are two different values". Because the two one-sided limits are different, the limit itself does not exist and the function is not differentiable.

Thanks very much, I understand now that it is continuous because the left and right limits are equal, but when i use the first principles equation f(x)-f(c)/(x-c) on the 2nd equation it gives me

3/2(x+1) - (-3)
---------------
x - (-3)

= 3/2x+9/2 / (x+3)


= 3/2 (x+3) / ( x+3)


= 3/2

whereas doing the same for the first equation gives +3.. would this be correct way to justify that the function is not differentiable?

Yes, since the two one-sided limits are different, the limit itself, defining the derivative, does not exist and so the function is not differentiable at that point.