Question

solve using the test - point method.
\frac{w^{2}-w - 20}{w - 14}geq0

the solution set is
(type your answer in interval notation.)

Explanation:

Step1: Factor the numerator

Factor $w^{2}-w - 20=(w - 5)(w+4)$. So the inequality becomes $\frac{(w - 5)(w + 4)}{w-14}\geq0$.

Step2: Find the critical points

Set the numerator and denominator equal to zero. For the numerator $(w - 5)(w + 4)=0$, we get $w=5$ and $w=-4$. For the denominator $w - 14=0$, we get $w = 14$. The critical points are $w=-4,5,14$.

Step3: Test the intervals

The intervals are $(-\infty,-4)$, $(-4,5)$, $(5,14)$ and $(14,\infty)$.

• For $w<-4$, let $w=-5$. Then $\frac{(-5 - 5)(-5 + 4)}{-5-14}=\frac{(-10)(-1)}{-19}=-\frac{10}{19}<0$.
• For $-40$.
• For $5
• For $w>14$, let $w = 15$. Then $\frac{(15 - 5)(15 + 4)}{15-14}=\frac{(10)\times(19)}{1}=190>0$.

Answer:

$[-4,5]\cup(14,\infty)$