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

## Given $f : \mathbb{R} \to \mathbb{R}$, continuous with $f \left(a\right) + f \left(b\right) + f \left(c\right) = 0$ , $a , b , c$ $\in$$\mathbb{R}$

Dec 12, 2017

Check below.

#### Explanation:

Got it now.

For $f \left(a\right) + f \left(b\right) + f \left(c\right) = 0$

We can either have

• $f \left(a\right) = 0$ and $f \left(b\right) = 0$ and $f \left(c\right) = 0$ 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 \left(a\right) =$$- f \left(b\right)$
That means $f \left(a\right) f \left(b\right) < 0$

$f$ continuous in $\mathbb{R}$ and so $\left[a , b\right] \subseteq \mathbb{R}$

According to Bolzano's theorem there is at least one ${x}_{0}$$\in$$\mathbb{R}$ so $f \left({x}_{0}\right) = 0$

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

Eventually $f$ has at least one root in $\mathbb{R}$

Dec 12, 2017

See below.

#### Explanation:

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

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

$f \left(a\right) f \left(b\right) < 0$
$f \left(a\right) f \left(c\right) < 0$
$f \left(b\right) f \left(c\right) < 0$

will be true, otherwise

$f \left(a\right) f \left(b\right) > 0 , f \left(a\right) f \left(c\right) > 0 , f \left(b\right) f \left(c\right) > 0$

will imply that

$f \left(a\right) > 0 , f \left(b\right) > 0 , f \left(c\right) > 0$ or $f \left(a\right) < 0 , f \left(b\right) < 0 , f \left(c\right) < 0$.

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

Now if one of $f \left({x}_{i}\right) f \left({x}_{j}\right) > 0$ by continuity, exists a $\zeta \in \left({x}_{i} , {x}_{j}\right)$ such that $f \left(\zeta\right) = 0$