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(B) Since $f(x)$ is a polynomial of degree $3$,$f^{\prime \prime}(x)$ is a linear function. Given $f^{\prime}(x)$ has a critical point at $x=1$,$f^{\p

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$3$

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(B) Since $f(x)$ is a polynomial of degree $3$,$f^{\prime \prime}(x)$ is a linear function. Given $f^{\prime}(x)$ has a critical point at $x=1$,$f^{\prime \prime}(1)=0$. Thus,$f^{\prime \prime}(x) = \lambda(x-1)$.Integrating $f^{\prime \prime}(x)$,we get $f^{\prime}(x) = \frac{\lambda x^2}{2} - \lambda x + C$.Since $f(x)$ has a critical point at $x=-1$,$f^{\prime}(-1) = 0$. Substituting $x=-1$ gives $\frac{\lambda}{2} + \lambda + C = 0$,so $C = -\frac{3\lambda}{2}$.Thus,$f^{\prime}(x) = \frac{\lambda x^2}{2} - \lambda x - \frac{3\lambda}{2}$.Integrating $f^{\prime}(x)$,we get $f(x) = \frac{\lambda x^3}{6} - \frac{\lambda x^2}{2} - \frac{3\lambda x}{2} + d$.Using $f(1) = -6$: $\frac{\lambda}{6} - \frac{\lambda}{2} - \frac{3\lambda}{2} + d = -6 \Rightarrow -\frac{11\lambda}{6} + d = -6 \Rightarrow -11\lambda + 6d = -36 \dots (i)$.Using $f(-1) = 10$: $-\frac{\lambda}{6} - \frac{\lambda}{2} + \frac{3\lambda}{2} + d = 10 \Rightarrow \frac{5\lambda}{6} + d = 10 \Rightarrow 5\lambda + 6d = 60 \dots (ii)$.Subtracting $(i)$ from $(ii)$: $16\lambda = 96 \Rightarrow \lambda = 6$.Substituting $\lambda = 6$ into $(ii)$: $5(6) + 6d = 60 \Rightarrow 30 + 6d = 60 \Rightarrow d = 5$.So,$f(x) = x^3 - 3x^2 - 9x + 5$.$f^{\prime}(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$.Setting $f^{\prime}(x) = 0$ gives critical points $x=3$ and $x=-1$.$f^{\prime \prime}(x) = 6x - 6$. At $x=3$,$f^{\prime \prime}(3) = 18 - 6 = 12 > 0$,so $f(x)$ has a local minima at $x=3$.

Let $f(x)$ be a polynomial of degree $3$ such that $f(-1)=10$,$f(1)=-6$,$f(x)$ has a critical point at $x=-1$,and $f^{\prime}(x)$ has a critical point at $x=1$. Then $f(x)$ has a local minima at $x=$

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