2. If $x$ is a real number, find the minimum value of $|x+1|+2|x-5|+|2 x-7|+\left|\frac{x-11}{2}\right|$.
(1 mark)若 $x$ 為實數, 求 $|x+1|+2|x-5|+|2 x-7|+\left|\frac{x-11}{2}\right|$ 的最小值 。
(1 分)
2. $\frac{45}{4}$
2. It suffices to minimise $2|x+1|+4|x-5|+4\left|x-\frac{7}{2}\right|+|x-11|$ (which is 4 times the expression in the question), i.e. to find the minimum total distance from $x$ to the 11 numbers: $-1,-1, \frac{7}{2}, \frac{7}{2}, \frac{7}{2}, \frac{7}{2}, 5,5,5,5,11$.
By the triangle inequality, we have $|x-a|+|x-b| \geq|(x-a)+(b-x)|=b-a$ for any $a \leq b$, and equality holds whenever $x$ lies between $a$ and $b$. From this, we try to pair up the numbers as follows: $(-1,11),(-1,5),\left(\frac{7}{2}, 5\right),\left(\frac{7}{2}, 5\right),\left(\frac{7}{2}, 5\right)$, leaving a single number $\frac{7}{2}$. From the above discussion, the total distance from $x$ to each pair is minimised when $x$ lies between them. Hence we can achieve the minimum by minimising the distance from $x$ to $\frac{7}{2}$ while making sure that $x$ lies between each pair of numbers. Clearly, this can be achieved by taking $x=\frac{7}{2}$. It follows that the answer is $\left|\frac{7}{2}+1\right|+2\left|\frac{7}{2}-5\right|+\left|2 \times \frac{7}{2}-7\right|+\left|\frac{\frac{7}{2}-11}{2}\right|=\frac{45}{4}$.

Remark. The total distance from $x$ to a given set of numbers is minimised when $x$ is the median of the set of numbers.