# Show that |c|<1 ?

## Given $f$ continuous , $f : \mathbb{R} \to \mathbb{R}$ If $f \left(x\right) \ne 0$ for $x$$\in$$\left(- \infty , c\right) \cup \left(c , + \infty\right)$ and $f \left(1\right) f \left(- 1\right) < 0$ then show that $| c | < 1$

Jan 18, 2018

Solved.

$f$ is continuous in $\mathbb{R}$ and so $\left[- 1 , 1\right] \subseteq \mathbb{R}$ .

• $f \left(1\right) f \left(- 1\right) < 0$

According to Bolzano Theorem (generalization)

$\exists {x}_{0}$$\in$$\left(- 1 , 1\right) : f \left({x}_{0}\right) = 0$

Supposed $| c | \ge 1$ $\iff$ $c \ge 1$ or $c \le - 1$

• If $c \ge 1$ then $f \left(x\right) \ne 0$ if $x$$\in$$\left(- \infty , c\right) \cup \left(c , + \infty\right)$

However, $f \left({x}_{0}\right) = 0$ with ${x}_{0}$$\in$$\left(- 1 , 1\right)$ $\implies$ -1 < ${x}_{0}$ $< 1 \le c$ $\implies$ ${x}_{0}$$\in$$\left(- \infty , c\right)$

• If $c \le - 1$ then $f \left(x\right) \ne 0$ if $x$$\in$$\left(- \infty , c\right) \cup \left(c , + \infty\right)$

However, $f \left({x}_{0}\right) = 0$ with ${x}_{0}$$\in$$\left(- 1 , 1\right)$ $\implies$

$c \le - 1$ $<$ ${x}_{0} < 1$ $\implies$ ${x}_{0}$$\in$$\left(c , + \infty\right)$

$| c | < 1$