The minimum value of f(x)=|x−6|+|x+3|+|x−8|+|x+4|+|x−3|,x∈R is
Open in App
Solution
f(x)=|x−6|+|x+3|+|x−8|+|x+4|+|x−3|,x∈R Critical points are 6,−3,8,−4,3 Minimum value occurs at median of −4,−3,3,6,8 Median =(5+12)thterm=3 ∴ Minimum value =|3−6|+|3+3|+|3−8|+|3+4|+|3−3|=21