If x = -1 and x = -2 are the extreme points of f(x) = αlog|x| + β x^2 + x then

At extreme points, the derivative equals zero.

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

The derivative of $\log|x|$log∣x∣ is $\frac{1}{x}$x1​

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

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Since $x = -1$x=−1 is an extreme point, $f'(-1) = 0$f′(−1)=0:

$$\frac{\alpha}{-1} + 2\beta(-1) + 1 = 0$$−1α​+2β(−1)+1=0

$$-\alpha - 2\beta + 1 = 0$$−α−2β+1=0

$\alpha + 2\beta = 1$α+2β=1 ...(Equation 1)

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Since $x = -2$x=−2 is an extreme point, $f'(-2) = 0$f′(−2)=0:

$$\frac{\alpha}{-2} + 2\beta(-2) + 1 = 0$$−2α​+2β(−2)+1=0

$$-\frac{\alpha}{2} - 4\beta + 1 = 0$$−2α​−4β+1=0

Multiplying by 2:

$$-\alpha - 8\beta + 2 = 0$$−α−8β+2=0

$\alpha + 8\beta = 2$α+8β=2 ...(Equation 2)

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From Equation 1: $\alpha + 2\beta = 1$α+2β=1

From Equation 2: $\alpha + 8\beta = 2$α+8β=2

Subtracting Equation 1 from Equation 2:

$$6\beta = 1$$6β=1

$$\beta = \frac{1}{6}$$β=61​

Substituting into Equation 1:

$$\alpha + 2\left(\frac{1}{6}\right) = 1$$α+2(61​)=1

$$\alpha + \frac{1}{3} = 1$$α+31​=1

$$\alpha = \frac{2}{3}$$α=32​

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Therefore, $\alpha = \frac{2}{3}$α=32​ and $\beta = \frac{1}{6}$β=61​