Let x=-1 and x=2 be the critical points of th | vedclass

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(A) $f(x) = x^3 + ax^2 + b \ln|x| + 1$$f'(x) = 3x^2 + 2ax + \frac{b}{x}$Since $x=-1$ and $x=2$ are critical points,$f'(-1) = 0$ and $f'(2) = 0$.$f'(-1

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$21.1$

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(A) $f(x) = x^3 + ax^2 + b \ln|x| + 1$$f'(x) = 3x^2 + 2ax + \frac{b}{x}$Since $x=-1$ and $x=2$ are critical points,$f'(-1) = 0$ and $f'(2) = 0$.$f'(-1) = 3 - 2a - b = 0 \implies 2a + b = 3$$f'(2) = 12 + 4a + \frac{b}{2} = 0 \implies 8a + b = -24$Subtracting the equations: $6a = -27 \implies a = -4.5$Substituting $a$: $2(-4.5) + b = 3 \implies -9 + b = 3 \implies b = 12$So,$f(x) = x^3 - 4.5x^2 + 12 \ln|x| + 1$.In the interval $[-2, -0.5]$,we check critical points and endpoints.$f'(x) = 3x^2 - 9x + \frac{12}{x} = \frac{3(x^3 - 3x^2 + 4)}{x} = \frac{3(x+1)(x-2)^2}{x}$.In $[-2, -0.5]$,$f'(x) = 0$ at $x = -1$.$f(-1) = -1 - 4.5 + 12 \ln(1) + 1 = -4.5$.$f(-2) = -8 - 4.5(4) + 12 \ln(2) + 1 = -8 - 18 + 1 + 12(0.7) = -25 + 8.4 = -16.6$.$f(-0.5) = -0.125 - 4.5(0.25) + 12 \ln(0.5) + 1 = -0.125 - 1.125 + 1 - 12(0.7) = -0.25 - 8.4 = -8.65$.$M = -4.5$ and $m = -16.6$.$|M+m| = |-4.5 - 16.6| = |-21.1| = 21.1$.

Let $x=-1$ and $x=2$ be the critical points of the function $f(x)=x^3+ax^2+b \ln|x|+1, x \neq 0$. Let $m$ and $M$ respectively be the absolute minimum and the absolute maximum values of $f$ in the interval $\left[-2, -\frac{1}{2}\right]$. Then $|M+m|$ is equal to (Take $\ln 2 \approx 0.7$):

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