Question

 2. Prove that g(x) = ( 3x + 1 ) / x has a root on [ - 1 , 1 ]

Answer 1

ANSWER

First note, that the function $g(x)=\frac{3x+1}{x}$ is defined on $[-1,0) \bigcup (0,1]$ . On each of this intervals, the function is continuous .If $x>0$ , then $g(x)>0.$ On the set $[-1,0)$ , the values of the function are both positive and negative. For example, $g(-1)=\frac{3\cdot (-1)+1}{-1}=2>0$ , $g(-0.1)=\frac{3\cdot (-0.1)+1}{-0.1}=-7<0$ . On the segment $[-1,-0.1]$ the function is continuous and $g(-1)<0<g(-0.1)$ . By the Intermediate Value Theorem , there exists a number $c\in(-1,-0.1)$ such that $g(c)=0.$

Note: the root of the equation $\frac{3x+1}{x}=0$ is easy of calculate by solving the equation $3x+1=0$ .$c=-\frac{1}{3}$ .