Let f(x) be a polynomial of degree 3 such tha | vedclass

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(B) Since $f(x)$ is a polynomial of degree $3$,we can express it using Taylor expansion about $x=3$ as:$f(x) = f(3) + f'(3)(x-3) + \frac{f''(3)}{2!}(x

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$23$

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(B) Since $f(x)$ is a polynomial of degree $3$,we can express it using Taylor expansion about $x=3$ as:$f(x) = f(3) + f'(3)(x-3) + \frac{f''(3)}{2!}(x-3)^2 + \frac{f'''(3)}{3!}(x-3)^3$Given $f(3)=1$,$f'(3)=-1$,$f''(3)=0$,and $f'''(3)=12$,we substitute these values:$f(x) = 1 + (-1)(x-3) + \frac{0}{2}(x-3)^2 + \frac{12}{6}(x-3)^3$$f(x) = 1 - (x-3) + 2(x-3)^3$Now,differentiate $f(x)$ with respect to $x$:$f'(x) = -1 + 2 	imes 3(x-3)^2 = -1 + 6(x-3)^2$To find $f'(1)$,substitute $x=1$:$f'(1) = -1 + 6(1-3)^2 = -1 + 6(-2)^2 = -1 + 6(4) = -1 + 24 = 23$

Let $f(x)$ be a polynomial of degree $3$ such that $f(3)=1$,$f'(3) = -1$,$f''(3) = 0$ and $f'''(3)=12$. Then the value of $f'(1)$ is

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