If y=a log x+b x^2+x has its extreme value at | vedclass

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(A) Given the function $y=a \log x+b x^2+x$. Find the derivative with respect to $x$: $\frac{d y}{d x}=\frac{a}{x}+2 b x+1$. Since the function has ex

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$\frac{3}{2}$

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(A) Given the function $y=a \log x+b x^2+x$. Find the derivative with respect to $x$: $\frac{d y}{d x}=\frac{a}{x}+2 b x+1$. Since the function has extreme values at $x=-1$ and $x=2$,the derivative must be zero at these points. At $x=-1$: $\frac{a}{-1}+2 b(-1)+1=0 \Rightarrow -a-2 b+1=0 \Rightarrow a+2 b=1$ (Equation $i$). At $x=2$: $\frac{a}{2}+2 b(2)+1=0 \Rightarrow \frac{a}{2}+4 b+1=0 \Rightarrow a+8 b=-2$ (Equation $ii$). Subtracting Equation $i$ from Equation $ii$: $(a+8 b)-(a+2 b)=-2-1 \Rightarrow 6 b=-3 \Rightarrow b=-\frac{1}{2}$. Substitute $b=-\frac{1}{2}$ into Equation $i$: $a+2(-\frac{1}{2})=1 \Rightarrow a-1=1 \Rightarrow a=2$. Therefore,$a+b=2+(-\frac{1}{2})=\frac{3}{2}$.

If $y=a \log x+b x^2+x$ has its extreme value at $x=-1$ and $x=2$,then the value of $a+b$ is

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