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If the absolute maximum and absolute minimum values of the function. $f(x)=x^3-2 x^2+x-3$ defined on $[0,2]$ are $\mathrm{M}$ and $\mathrm{m}$ respectively, then $\mathrm{M}+\mathrm{m}=$
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The correct answer is:
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$f(x)=x^3-2 x^2+x-3$
$$
f^{\prime}(x)=3 x^2-4 x+1
$$
$$
\begin{aligned}
& f^{\prime}(x)=0 \Rightarrow 3 x^2-4 x+1=0 \\
& \Rightarrow 3 x^2-3 x-x+1=0 \Rightarrow(3 x-1)(x-1)=0 \\
& \therefore \quad x=\frac{1}{3}, 1
\end{aligned}
$$
$\begin{aligned} & f^{\prime \prime}(x)=6 x-4 \\ & f^{\prime \prime}\left(\frac{1}{3}\right) < 0 \Rightarrow \text { Maxima } \\ & f^{\prime \prime}(1)>0 \Rightarrow \text { Minima } \\ & f(0)=-3 \text { and } f(2)=8-8+2-3=-1 \\ & f\left(\frac{1}{3}\right)=\frac{1}{27}-\frac{2}{9}+\frac{1}{3}-3=\frac{-77}{27} \\ & f(1)=1-2+1-3=-3 \\ & \therefore \quad \text { Absolute } \max =-1=M \\ & \quad \text { Absolute } \min =-3=m \\ & \therefore \quad M+m=-3-1=-4 .\end{aligned}$
$$
f^{\prime}(x)=3 x^2-4 x+1
$$
$$
\begin{aligned}
& f^{\prime}(x)=0 \Rightarrow 3 x^2-4 x+1=0 \\
& \Rightarrow 3 x^2-3 x-x+1=0 \Rightarrow(3 x-1)(x-1)=0 \\
& \therefore \quad x=\frac{1}{3}, 1
\end{aligned}
$$
$\begin{aligned} & f^{\prime \prime}(x)=6 x-4 \\ & f^{\prime \prime}\left(\frac{1}{3}\right) < 0 \Rightarrow \text { Maxima } \\ & f^{\prime \prime}(1)>0 \Rightarrow \text { Minima } \\ & f(0)=-3 \text { and } f(2)=8-8+2-3=-1 \\ & f\left(\frac{1}{3}\right)=\frac{1}{27}-\frac{2}{9}+\frac{1}{3}-3=\frac{-77}{27} \\ & f(1)=1-2+1-3=-3 \\ & \therefore \quad \text { Absolute } \max =-1=M \\ & \quad \text { Absolute } \min =-3=m \\ & \therefore \quad M+m=-3-1=-4 .\end{aligned}$
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