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The minimum value of the function $f(x)=2|x-1|+|x-2|$ is
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Verified Answer
The correct answer is:
1
Given,
$$
\begin{array}{l}
\quad f(x)=2|x-1|+|x-2| \\
\Rightarrow \quad f(x)=\left\{\begin{array}{c}
-2(x-1)-(x-2), x < 1 \\
2(x-1)-(x-2), 1 \leq x < 2 \\
2(x-1)+(x-2), x \geq 2 \\
-3 x+4, \quad x < 1 \\
x_{1} \\
3 x-4, \quad 1 \leq x < 2 \\
x \geq 2
\end{array}\right.
\end{array}
$$
$f^{\prime}(x)=\left\{\begin{array}{cc}-3 & x < 1 \\ 1 & 1 \leq x < 2 \\ 3 & x \geq 2\end{array}\right.$
So, $f(x)$ will be minimum at $x=1$ and the minimum value is 1 .
$$
\begin{array}{l}
\quad f(x)=2|x-1|+|x-2| \\
\Rightarrow \quad f(x)=\left\{\begin{array}{c}
-2(x-1)-(x-2), x < 1 \\
2(x-1)-(x-2), 1 \leq x < 2 \\
2(x-1)+(x-2), x \geq 2 \\
-3 x+4, \quad x < 1 \\
x_{1} \\
3 x-4, \quad 1 \leq x < 2 \\
x \geq 2
\end{array}\right.
\end{array}
$$
$f^{\prime}(x)=\left\{\begin{array}{cc}-3 & x < 1 \\ 1 & 1 \leq x < 2 \\ 3 & x \geq 2\end{array}\right.$
So, $f(x)$ will be minimum at $x=1$ and the minimum value is 1 .
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