I have a homework question to prove that if $f(x)$ is continuous in $\mathbb R$ then exists $x$ such that $f(x)f(x+1) \ge 0$.
I am failing to see why this is true and how I can prove this.
Can someone please help me out? Thanks :)
I have a homework question to prove that if $f(x)$ is continuous in $\mathbb R$ then exists $x$ such that $f(x)f(x+1) \ge 0$.
I am failing to see why this is true and how I can prove this.
Can someone please help me out? Thanks :)
If $f(x)=0$ for some $x$, then it is trivial. Otherwise, $f>0$ or $f<0$ always, by continuity. In this case take any $x$.
Addendum: the continuity assumption cannot be lifted; for example, take $f(x)=(-1)^{\lfloor x \rfloor}$.
Addendum #2: the example of $f(x)=\sin(\pi x)$ shows that it is possible for the set of $x$'s satisfying the requirement to have zero measure (although it is always at least countably infinite).
Suppose we have that /The product is negative /For some $x\in \mathbb{R}$.
Then the signs of both /Pieces of product above /Can not be the same.
Then by IVT/For this $x$ and $x+1$/Root lies in-between.
Let that root be $r$/Product at $r$,$r+1$/Zero, QED.