Question

Solve graphically.

Express answers in both inequality and interval notation.

$\frac{x+3}{x-3}\ge 0$

Answer 1

$\frac{x+3}{x-3}\ge o$

$\frac{x-(-3)}{x-3}\ge o$

critical point are

$x-(-3)=0$ $x-3=0$

$x=-3$ $x=3$

$\therefore$ For $x\ge -3$

$xϵ[-3,\alpha )$