Application of Derivatives
85664
The minimum value of
\(f(x)=|x-1|+|x-2|+|x-3|\) is equal to
Explanation:
(B) : \(f(x)=|x-1|+|x-2|+|x-3|\)
\(f(x)=\left\{\begin{array}{ccc}6-3 x & , & x\lt 1 \\ 4-x & , & 1 \leq x\lt 2 \\ x & , & 2 \leq x\lt 3 \\ 3 x-6 & , & x \geq 3\end{array}\right.\)
\(f^{\prime}(x)=\left\{\begin{array}{cc}-3 & x\lt 1 \\ -1 & 1\lt x\lt 2 \\ 1 & 2\lt x\lt 3 \\ 3 & x>3\end{array}\right.\)
\(\because \mathrm{f}^{\prime}(\mathrm{x})\) is negative for \(\mathrm{x}\lt 2\) and positive for \(\mathrm{x}>2\).
\(\therefore \mathrm{f}(\mathrm{x})\) is minimum for \(\mathrm{x}=2\) \(\mathrm{f}(2)=2\)