Natural Cubic Spline S on [0,2]

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A Natural Cubic Spline S on $[0,2]$ is defined by:

S(x)= $$S_0(x)=1+2x-x^3 \to 0 \leq x < 1 $$ $$S_1(x)=2+b(x-1)+c(x-1)^2+d(x-1)^3 \to 1 \leq x \leq 2$$

Find b,c and d

This question seems simple I am just not sure how to go about it. Can someone please guide me as to how to find the solution

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Spline of order $k$ should have continuous derivatives up to order $k-1$. In this case,

$$S_0(1) = S_1(1)$$ $$S_0'(1) = S_1'(1)$$ $$S_0''(1) = S_1''(1)$$

Since this is a natural cubic spline, the second derivative at the first node $(x=0)$ and at the last node $(x=2)$ should be $0$.

and then you can solve for $b, c, d$.