Show that $| c | < 1$ $|c|<1$ ?

Question

Given $f$ $f$ continuous , $f:RR->RR$
If $f(x)!=0$ for $x$ $in$ $(-oo,c)uu(c,+oo)$ and

$f(1)f(-1)<0$

then show that $|c|<1$

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Answer 1 · 2018-01-18T15:56:34

Solved.

$f$ is continuous in $RR$ and so $[-1,1]subeRR$ .

According to Bolzano Theorem (generalization)

$EE x_0$ $in$ $(-1,1): f(x_0)=0$

Supposed $|c|>=1$ $<=>$ $c>=1$ or $c<=-1$

However, $f(x_0)=0$ with $x_0$ $in$ $(-1,1)$ $=>$ $-1 <$ $x_0$ $<1<=c$ $=>$ $x_0$ $in$ $(-oo,c)$

CONTRADICTION!

However, $f(x_0)=0$ with $x_0$ $in$ $(-1,1)$ $=>$

$c<=-1$ $<$ $x_0<1$ $=>$ $x_0$ $in$ $(c,+oo)$

CONTRADICTION!

Therefore,

$|c|<1$