Question

Show that `f` has at least one root in `RR` ?

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

Answer 1

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_0` `in` `RR` 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`

Answer 2

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`