Ok one of my feeble attempts at making up a problem (Atleast I have not encountered one like this). Lets see how quickly this gets destroyed.

 

A function f(x) is such that ₁⁴ f(x) dx = 3. Find a function g(x) such that f(x)+g(x) has a zero in (1,4).

Don't assume f(x).

 

 I am looking for a one-size-fits-all g(x) for all f(x) which satisfy the given condition 

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!

good. that was what i was looking for.

 

If _a^b f(x) = 0, then by mean value theorem, there exists c(a,b) so that f'(c) = 0

 

Another way to understand is that if _a^b f(x) = 0, then f(x) neither be +ve for all x in the interval nor -ve for all x. So there's got to be x in the interval for which it is zero.