The sum of all the local minimum values of th | vedclass

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(C) Given $f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1) \quad \dots(i)$Differentiating with respect to $x$:$f^{\prime}(x

Vedclass · Accepted Answer

$-27$

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(C) Given $f(x)=x^{3}-3 x^{2}-\frac{3}{2} f^{\prime \prime}(2) x+f^{\prime \prime}(1) \quad \dots(i)$Differentiating with respect to $x$:$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2} f^{\prime \prime}(2) \quad \dots(ii)$$f^{\prime \prime}(x)=6 x-6 \quad \dots(iii)$From $(iii)$,$f^{\prime \prime}(2)=6(2)-6=6$ and $f^{\prime \prime}(1)=6(1)-6=0$.Substituting $f^{\prime \prime}(2)=6$ into $(ii)$:$f^{\prime}(x)=3 x^{2}-6 x-\frac{3}{2}(6) = 3 x^{2}-6 x-9$.Setting $f^{\prime}(x)=0$ for critical points:$3(x^{2}-2 x-3)=0 \Rightarrow 3(x-3)(x+1)=0 \Rightarrow x=3, -1$.Using the second derivative test:$f^{\prime \prime}(-1)=6(-1)-6=-12 $f^{\prime \prime}(3)=6(3)-6=12 > 0$ (Local minima at $x=3$).Substituting $f^{\prime \prime}(2)=6$ and $f^{\prime \prime}(1)=0$ into $(i)$:$f(x)=x^{3}-3 x^{2}-9 x$.The local minimum value is $f(3)=3^{3}-3(3^{2})-9(3) = 27-27-27 = -27$.

The sum of all the local minimum values of the twice differentiable function $f: R \rightarrow R$ defined by $f(x)=x^{3}-3 x^{2}-\frac{3 f^{\prime \prime}(2)}{2} x+f^{\prime \prime}(1)$ is:

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