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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.

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