Show that ff has at least one root in RR ?

Given f:RR->RR, continuous with f(a)+f(b)+f(c)=0 , a,b,c inRR

2 Answers
Dec 12, 2017

Check below.

Explanation:

Got it now.

For f(a)+f(b)+f(c)=0

We can either have

  • f(a)=0 and f(b)=0 and f(c)=0 which means that f has at least one root, a ,b ,c

  • One of the two numbers at least to be opposite between them

Let's suppose f(a)=-f(b)
That means f(a)f(b)<0

f continuous in RR and so [a,b]subeRR

According to Bolzano's theorem there is at least one x_0inRR so f(x_0)=0

Using Bolzano's theorem in other intervals [b,c] ,[a,c] will lead to the same conclusion.

Eventually f has at least one root in RR

Dec 12, 2017

See below.

Explanation:

If one of f(a), f(b),f(c) equals zero, there we have a root.

Now supposing f(a) ne 0, f(b) ne 0, f(c) ne 0 then at least one of

f(a)f(b) < 0
f(a)f(c) < 0
f(b)f(c) < 0

will be true, otherwise

f(a)f(b) > 0, f(a)f(c) > 0, f(b)f(c) > 0

will imply that

f(a) > 0, f(b) > 0, f(c) > 0 or f(a) < 0, f(b) < 0, f(c) < 0.

In each case the result for f(a)+f(b)+f(c) could not be null.

Now if one of f(x_i)f(x_j) > 0 by continuity, exists a zeta in (x_i,x_j) such that f(zeta) = 0