If x=-1 and x=2 are extreme points of f(x)=α log|x|+β x^2+x, then

$$f(x)=\alpha log|x|+\beta {x}^{2}+x$$f(x)=αlog∣x∣+βx2+x

If $x<0$x<0

$$f⁡(x)=\alpha log(-x)+\beta {x}^{2}+x$$f⁡(x)=αlog(−x)+βx2+x

$$\therefore f{⁡}^{'}(x)=\frac{-\alpha }{(-x)}+2\beta x+1$$∴f⁡′(x)=(−x)−α​+2βx+1

If $x>0$x>0

$$f⁡(x)=\alpha \mathrm{log}x+\beta {x}^{2}+x$$f⁡(x)=αlogx+βx2+x

$$\therefore f{⁡}^{'}(x)=\frac{\alpha }{x}+2\beta x+1$$∴f⁡′(x)=xα​+2βx+1

$${ f⁡ }^{'} ( - 1 ) = - \alpha - 2 \beta + 1 = 0$$f⁡′(−1)=−α−2β+1=0

$${ f⁡ }^{'} ( 2 ) = \frac{ \alpha }{ 2 } + 4 \beta + 1 = 0$$f⁡′(2)=2α​+4β+1=0

$$2 { f⁡ }^{'} ( - 1 ) = - 2 \alpha - 4 \beta + 2 = 0$$2f⁡′(−1)=−2α−4β+2=0

and ${f}^{'}(2)=\frac{\alpha }{2}+4\beta +1=0$f′(2)=2α​+4β+1=0

Adding,

$$- \frac{ 3 \alpha }{ 2 } + 3 = 0$$−23α​+3=0

$$\therefore \alpha = 2$$∴α=2

$$\therefore \beta = \frac{ - 1 }{ 2 }$$∴β=2−1​