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We are to determine a unique polynomial $f(x)$ of degree ≤ 3 such that $$ f(x_0) = 1, \quad f'(x_0) = 2, \quad f(x_1) = 2, \quad f'(x_1) = 3, $$ where $x_1 - x_0 = h$. --- ### Step 1: Assume a cubic polynomial Let $$ f(x) = A + B(x-x_0) + C(x-x_0)^2 + D(x-x_0)^3 $$ Then its derivative is $$ f'(x) = B + 2C(x-x_0) + 3D(x-x_0)^2 $$ --- ### Step 2: Apply conditions at $x_0$ At $x = x_0$, $(x - x_0) = 0$: 1. $f(x_0) = A = 1$ 2. $f'(x_0) = B = 2$ So polynomial reduces to $$ f(x) = 1 + 2(x-x_0) + C(x-x_0)^2 + D(x-x_0)^3 $$ --- ### Step 3: Apply conditions at $x_1 = x_0 + h$ Let $u = x - ________ ______ ___ ________ ____ _________.
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