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F(2+t) = | t² + 4t + 4 - 4 |
        = |  t (t+4) |
f(2)  = 0
f(2+t) - f(2)           = |t² - 4t|
--------------              ----------
   t                            t

limit as t -> 0+  and 0-   have different values
So at t = 0 and x=2, derivative does not exist

1 5 1
Thank you but Why did we take (t+2=x)?
to be able to prove by way of : definition of derivative in terms of limits
Thank you
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