Question

Show that `|c|<1` ?

Given `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`

Answer 1

Solved.

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

• `f(1)f(-1)<0`

According to Bolzano Theorem (generalization)

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

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

• If `c>=1` then `f(x)!=0` if `x` `in` `(-oo,c)uu(c,+oo)`

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

CONTRADICTION!

• If `c<=-1` then `f(x)!=0` if `x` `in` `(-oo,c)uu(c,+oo)`

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`