The range of the real valued function f(x) = | vedclass

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(B) Given the function $f(x) = |x-2| + |x-3|$.We analyze the function in three intervals: $x \leq 2$,$2 < x < 3$,and $x \geq 3$.For $x \leq 2$,$f(x) =

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$[1, \infty)$

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(B) Given the function $f(x) = |x-2| + |x-3|$.We analyze the function in three intervals: $x \leq 2$,$2 For $x \leq 2$,$f(x) = -(x-2) - (x-3) = -x + 2 - x + 3 = -2x + 5$. Since $x \leq 2$,$-2x \geq -4$,so $f(x) \geq -4 + 5 = 1$.For $2 For $x \geq 3$,$f(x) = (x-2) + (x-3) = 2x - 5$. Since $x \geq 3$,$2x \geq 6$,so $f(x) \geq 6 - 5 = 1$.Combining these,the minimum value of the function is $1$ and it takes all values greater than or equal to $1$.Thus,the range is $[1, \infty)$.Therefore,option $(B)$ is correct.

The range of the real valued function $f(x) = |x-2| + |x-3|$ is

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