## Precalculus: Mathematics for Calculus, 7th Edition

The solution is $(-\infty,-1)\cup[3,\infty-)$ The graph is:
$\dfrac{x-3}{x+1}\ge0$ Find the intervals. The factor of the numerator is $x-3$ and the factor of the denominator is $x+1$. Set them equal to $0$ and solve for $x$: $x-3=0$ $x=3$ $x+1=0$ $x=-1$ The factors are zero when $x=-1,3$. These two numbers divide the real line into the following intervals: $(-\infty,-1)$ $,$ $(-1,3)$ $,$ $(3,\infty)$ Elaborate a diagram, using test points to determine the sign of each factor in each interval: (refer to the attached image below) It can be seen from the diagram that the intervals $(-\infty,-1)$ and $(3,\infty)$ satisfy the inequality. The endpoint $-1$ does not satisfy the inequality because it makes the denominator equal to $0$ but the endpoint $3$ does satisfy the inequality, because of the symbol $\ge$ The solution is $(-\infty,-1)\cup[3,\infty)$