To find whether the function f(x) is derivable at a point x = c we have to check that f ‘(c ^- ) = f ‘(c ⁺ ) = finite quantity, this condition must be fulfilled in order the function to be derivable.

As discussed above the top of this document where the description of the topic is given.

_{To prove the first two functions of f(x) to be derivable we must take the point x = - 2, around this point we will prove the derivability of this sub - functions because this point is common in them both.}

So by using the formula, f ‘(c) = , we get,

f ‘( - 2 ^- ) =

f ‘( - 2 ^- ) =

value of f(c) = - 1 because when we put exact value of x = c then we have to take the second sub function because it is only defined when x = - 2.

f ‘( - 2 ^- ) = = 2

if the limit would not have been simplified to a number before putting the limit then we would put x = - 2 - h,

f ‘( - 2 ⁺ ) =

f ‘( - 2 ⁺ ) = = 1

Therefore the function is not differentiable at x = - 2.

if the limit would not have been simplified to a number before putting the limit, then we would put x = - 2 + h.

Now we have to find the derivability at point x = 0, because at this point or near it, the next two sub-functions are defined and hence, we get,

f ‘(0 ^- ) =

f ‘(0 ^- ) =

putting x = 0 - h,

= = ∞

f ‘(0 ⁺ ) =

f ‘(0 ⁺ ) = = 1

as we can see that the value around a point is not the same, so we will say that the function is not derivable at x = 0.

At the end we conclude that at x = 0 and x = - 2 function is non differentiable.