The minimum value of the function f(x)=2|x-1| | vedclass

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(B) Given, $f(x) = 2|x-1| + |x-2|$.We analyze the function in different intervals:Case $1$: If $x < 1$, then $f(x) = -2(x-1) - (x-2) = -2x + 2 - x + 2

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$1$

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(B) Given, $f(x) = 2|x-1| + |x-2|$.We analyze the function in different intervals:Case $1$: If $x Case $2$: If $1 \leq x Case $3$: If $x \geq 2$, then $f(x) = 2(x-1) + (x-2) = 2x - 2 + x - 2 = 3x - 4$. Since the slope is $3$, the function is increasing in this interval.Comparing these, the function decreases until $x=1$ and then starts increasing.Therefore, the minimum value occurs at $x=1$.$f(1) = 2|1-1| + |1-2| = 2(0) + |-1| = 0 + 1 = 1$.

The minimum value of the function $f(x)=2|x-1|+|x-2|$ is

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