If x = -1 and x = 2 are the extreme points of | vedclass

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(B) Given the function $y = a \log|x| + bx^2 + x$. The derivative is $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since $x = -1$ and $x = 2$ are extreme p

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$a = 2, b = -1/2$

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(B) Given the function $y = a \log|x| + bx^2 + x$. The derivative is $\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$. Since $x = -1$ and $x = 2$ are extreme points,the derivative must be zero at these points. For $x = -1$: $\frac{a}{-1} + 2b(-1) + 1 = 0 \implies -a - 2b + 1 = 0 \implies a + 2b = 1$. For $x = 2$: $\frac{a}{2} + 2b(2) + 1 = 0 \implies \frac{a}{2} + 4b + 1 = 0 \implies a + 8b = -2$. Subtracting the first equation from the second: $(a + 8b) - (a + 2b) = -2 - 1 \implies 6b = -3 \implies b = -1/2$. Substituting $b = -1/2$ into $a + 2b = 1$: $a + 2(-1/2) = 1 \implies a - 1 = 1 \implies a = 2$. Thus,$a = 2$ and $b = -1/2$.

If $x = -1$ and $x = 2$ are the extreme points of the function $y = a \log|x| + bx^2 + x$,then find the values of $a$ and $b$.

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