The Obvious solution is f(x) = -g(x).

But consider,

let g be any real valued,continuous  function that satisfies f + g =0 for at least one value of x (1.4)

Let H(x) =  ( f + g)dx

By LMVT for H over (1,4),

H'(c) = [H(4) - H(1)]/3

But H'(c) = f + g

if f + g =0 for at least one c  (1.4),

Then H'(c) = 0 for at least one c  (1.4),

implies , a solution for g can be got if ,<Though this is not the only solution>

H(4) - H(1) =0

implies _[1]^[4]( f + g)dx = 0

implies

 

_[1]^[4](g)dx = -3

Hence any function that satisfies _[1]^[4](g)dx = -3

is one class of  solution.

example g(x) = (-2x/5)

That solves the problem!