If x=-1 and x=2 are extreme points of f(x)=al | vedclass

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(C) Given $f(x)=\alpha \log |x|+\beta x^2+x$.The derivative is $f^{\prime}(x)=\frac{\alpha}{x}+2\beta x+1$.Since $x=-1$ and $x=2$ are extreme points,$

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$\alpha=2, \beta=-\frac{1}{2}$

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(C) Given $f(x)=\alpha \log |x|+\beta x^2+x$.The derivative is $f^{\prime}(x)=\frac{\alpha}{x}+2\beta x+1$.Since $x=-1$ and $x=2$ are extreme points,$f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.For $x=-1$: $\frac{\alpha}{-1}+2\beta(-1)+1=0 \Rightarrow -\alpha-2\beta+1=0 \Rightarrow \alpha+2\beta=1$ (Equation $i$).For $x=2$: $\frac{\alpha}{2}+2\beta(2)+1=0 \Rightarrow \frac{\alpha}{2}+4\beta+1=0 \Rightarrow \alpha+8\beta=-2$ (Equation $ii$).Subtracting $(i)$ from (ii): $(\alpha+8\beta)-(\alpha+2\beta)=-2-1 \Rightarrow 6\beta=-3 \Rightarrow \beta=-\frac{1}{2}$.Substituting $\beta=-\frac{1}{2}$ into $(i)$: $\alpha+2(-\frac{1}{2})=1 \Rightarrow \alpha-1=1 \Rightarrow \alpha=2$.Thus,$\alpha=2$ and $\beta=-\frac{1}{2}$.

If $x=-1$ and $x=2$ are extreme points of $f(x)=\alpha \log |x|+\beta x^2+x$,then

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