Question

solve with the test - point method.

$w>\frac{w - 2}{w - 1}$

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

Explanation:

Step1: Rearrange the inequality

Move all terms to one - side: $w-\frac{w - 2}{w - 1}>0$. Get a common denominator: $\frac{w(w - 1)-(w - 2)}{w - 1}>0$. Expand the numerator: $\frac{w^{2}-w - w + 2}{w - 1}>0$, which simplifies to $\frac{w^{2}-2w + 2}{w - 1}>0$. The discriminant of the quadratic function $y = w^{2}-2w + 2$ is $\Delta=(-2)^{2}-4\times2=-4<0$, and the coefficient of $w^{2}$ is positive, so $w^{2}-2w + 2>0$ for all real $w$. So the inequality is equivalent to $\frac{1}{w - 1}>0$.

Step2: Find the critical point

Set the denominator equal to zero. For $y=\frac{1}{w - 1}$, the critical point is $w = 1$.

Step3: Test the intervals

We have two intervals to test: $(-\infty,1)$ and $(1,\infty)$. For the interval $(-\infty,1)$, let's choose $w = 0$. Then $\frac{1}{0 - 1}=-1<0$. For the interval $(1,\infty)$, let's choose $w = 2$. Then $\frac{1}{2 - 1}=1>0$.

Answer:

$(1,\infty)$