If f(x)=a log |x|+b x^2+x has extreme values | vedclass

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(B) Given,$f(x)=a \log |x|+b x^2+x$.The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.Since the function has extreme values at $x=-1$ and $x=2

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$\left(2, -\frac{1}{2}\right)$

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(B) Given,$f(x)=a \log |x|+b x^2+x$.The derivative is $f^{\prime}(x) = \frac{a}{x} + 2bx + 1$.Since the function has extreme values at $x=-1$ and $x=2$,we have $f^{\prime}(-1)=0$ and $f^{\prime}(2)=0$.For $x=-1$: $f^{\prime}(-1) = \frac{a}{-1} + 2b(-1) + 1 = 0 \Rightarrow -a - 2b + 1 = 0 \Rightarrow a + 2b = 1$ $(i)$.For $x=2$: $f^{\prime}(2) = \frac{a}{2} + 2b(2) + 1 = 0 \Rightarrow \frac{a}{2} + 4b + 1 = 0 \Rightarrow a + 8b = -2$ (ii).Subtracting $(i)$ from (ii): $(a + 8b) - (a + 2b) = -2 - 1 \Rightarrow 6b = -3 \Rightarrow b = -\frac{1}{2}$.Substituting $b = -\frac{1}{2}$ into $(i)$: $a + 2(-\frac{1}{2}) = 1 \Rightarrow a - 1 = 1 \Rightarrow a = 2$.Thus,the ordered pair $(a, b) = \left(2, -\frac{1}{2}\right)$.

If $f(x)=a \log |x|+b x^2+x$ has extreme values at $x=-1$ and $x=2$,then the ordered pair $(a, b)=$

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