Let f :R → R be defined by

Then f (– 1) + f (2) + f (4) is

Given that,

Now,

f(-1) = 3(-1) = -3 [since -1<1 and f(x) = 3x for x≤ 1]

f(2) = 2² = 4 [since 2<3 and f(x) = x² for 1<x≤ 3]

f(4) = 2(4) = 8 [since 4>3 and f(x) = 2x for x>3]

∴ f (– 1) + f (2) + f (4) = -3+4+8 = 9