Find the number of points of extremum of f(x) = |x2 – 1| + |x3 – 1| + |x5 – 1|
We have f(x) = |x2 – 1| + |x3 – 1| + |x5 – 1|
Clearly, the number of points of extremum is 3 at x = 1/2, 0, 1
