The function is defined as `h(x)=x^3+1` on `(-oo,1]` and `h(x)=3x` on `(1,oo)` . We are asked to find `h'(1)` .

First note that the function is defined at 1, h(1)=2. However the function has a jump discontinuity at x=1. The limit from the left is 2, while the limit from the right is 3.

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**Since the function is not continuous at 1, there is no derivative at 1.**

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** Note that if you were to algebraically differentiate, you would get `h'(x)=3x` from either side, resulting in `h'(1)=3` , but this is incorrect. A derivative can only exist at a point if the function is continuous at that point. This is a necessary, but not sufficient, condition.

The function is defined as h(x)= x^3+1 if x <= 1 and 3x if x>1. To determine h'(1), it can be seen that h(x) = 3x.

h'(1) = (3x)' = 3

**The value of h'(1) = 3.**