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

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(A) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.First,find the derivative $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.Since $x = -1$ and $x = 2$ are

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

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(A) Given $f(x) = \alpha \log |x| + \beta x^2 + x$.First,find the derivative $f'(x) = \frac{\alpha}{x} + 2\beta x + 1$.Since $x = -1$ and $x = 2$ are extreme points,$f'(x) = 0$ at these points.For $x = -1$: $\frac{\alpha}{-1} + 2\beta(-1) + 1 = 0 \Rightarrow -\alpha - 2\beta + 1 = 0 \Rightarrow \alpha + 2\beta = 1$ (Equation $1$).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 $2$).Subtracting Equation $1$ from Equation $2$: $(\alpha + 8\beta) - (\alpha + 2\beta) = -2 - 1 \Rightarrow 6\beta = -3 \Rightarrow \beta = -\frac{1}{2}$.Substitute $\beta = -\frac{1}{2}$ into Equation $1$: $\alpha + 2(-\frac{1}{2}) = 1 \Rightarrow \alpha - 1 = 1 \Rightarrow \alpha = 2$.Thus,$(\alpha, \beta) = (2, -\frac{1}{2})$.

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

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