If the absolute maximum and absolute minimum | vedclass

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(A) Given function: $f(x) = x^3 - 2x^2 + x - 3$ on the interval $[0, 2]$.First, find the derivative: $f'(x) = 3x^2 - 4x + 1$.Set $f'(x) = 0$ to find c

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-$4$

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(A) Given function: $f(x) = x^3 - 2x^2 + x - 3$ on the interval $[0, 2]$.First, find the derivative: $f'(x) = 3x^2 - 4x + 1$.Set $f'(x) = 0$ to find critical points:$3x^2 - 3x - x + 1 = 0$$3x(x - 1) - 1(x - 1) = 0$$(3x - 1)(x - 1) = 0$So, $x = \frac{1}{3}$ and $x = 1$.Both critical points lie within the interval $[0, 2]$.Now, evaluate the function at the critical points and the endpoints:$f(0) = 0^3 - 2(0)^2 + 0 - 3 = -3$$f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 + \frac{1}{3} - 3 = \frac{1}{27} - \frac{2}{9} + \frac{1}{3} - 3 = \frac{1 - 6 + 9 - 81}{27} = \frac{-77}{27}$$f(1) = 1^3 - 2(1)^2 + 1 - 3 = 1 - 2 + 1 - 3 = -3$$f(2) = 2^3 - 2(2)^2 + 2 - 3 = 8 - 8 + 2 - 3 = -1$Comparing these values: $\{-3, \frac{-77}{27}, -3, -1\}$.The absolute maximum value $M = -1$.The absolute minimum value $m = -3$.Therefore, $M + m = -1 + (-3) = -4$.

If the absolute maximum and absolute minimum values of the function $f(x) = x^3 - 2x^2 + x - 3$ defined on $[0, 2]$ are $M$ and $m$ respectively, then $M + m =$

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