Let p(x) be a real polynomial of least degree | vedclass

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(D) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$,$p^{\prime}(x)$ must have roots at $x=1$ and $x=3$. Thus,$p^{\prime}(x) = \

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$9$

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(D) Since $p(x)$ has a local maximum at $x=1$ and a local minimum at $x=3$,$p^{\prime}(x)$ must have roots at $x=1$ and $x=3$. Thus,$p^{\prime}(x) = \lambda(x-1)(x-3) = \lambda(x^2-4x+3)$. Integrating $p^{\prime}(x)$,we get $p(x) = \lambda(\frac{x^3}{3} - 2x^2 + 3x) + \mu$. Given $p(1) = 6$,we have $6 = \lambda(\frac{1}{3} - 2 + 3) + \mu = \frac{4}{3}\lambda + \mu$,which implies $18 = 4\lambda + 3\mu \quad \dots (i)$. Given $p(3) = 2$,we have $2 = \lambda(\frac{27}{3} - 2(9) + 3(3)) + \mu = \lambda(9 - 18 + 9) + \mu = \mu$. So,$\mu = 2$. Substituting $\mu = 2$ into equation $(i)$,$18 = 4\lambda + 3(2) \implies 18 = 4\lambda + 6 \implies 4\lambda = 12 \implies \lambda = 3$. Thus,$p^{\prime}(x) = 3(x-1)(x-3)$. Evaluating at $x=0$,$p^{\prime}(0) = 3(0-1)(0-3) = 3(-1)(-3) = 9$.

Let $p(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and a local minimum at $x=3$. If $p(1)=6$ and $p(3)=2$,then $p^{\prime}(0)$ is

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