If y=a log x+b x^2+x has its extremum values | vedclass

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(B) Given the function $y=a \log x+b x^2+x$. First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the functio

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$a=2, b=-\frac{1}{2}$

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(B) Given the function $y=a \log x+b x^2+x$. First,find the derivative with respect to $x$: $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since the function has extremum values at $x=-1$ and $x=2$,the derivative must be zero at these points. For $x=-1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ (Equation $i$). For $x=2$: $\frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (Equation $ii$). Subtracting Equation $i$ from Equation $ii$: $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$. Substituting $b = -\frac{1}{2}$ into Equation $i$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$. Thus,the values are $a=2$ and $b=-\frac{1}{2}$.

If $y=a \log x+b x^2+x$ has its extremum values at $x=-1$ and $x=2$,then

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