Let P(x) be a polynomial of degree 2,with P(2 | vedclass

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(B) Using the Taylor series expansion of $P(x)$ about $x=2$:$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$Given $P(2)=-1,

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$-0.002$

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(B) Using the Taylor series expansion of $P(x)$ about $x=2$:$P(x) = P(2) + P^{\prime}(2)(x-2) + \frac{P^{\prime \prime}(2)}{2!}(x-2)^2$Given $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$:$P(x) = -1 + 0(x-2) + \frac{2}{2}(x-2)^2$$P(x) = -1 + (x-2)^2$Now,substitute $x=1.001$:$P(1.001) = -1 + (1.001-2)^2$$P(1.001) = -1 + (-0.999)^2$$P(1.001) = -1 + 0.998001$$P(1.001) = -0.001999$Rounding to the nearest provided option,we get $-0.002$.

Let $P(x)$ be a polynomial of degree $2$,with $P(2)=-1, P^{\prime}(2)=0, P^{\prime \prime}(2)=2$. Then $P(1.001)$ is

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