Solve the given nonlinear inequality. Write the solution set using interval notation. Graph the solution set.$\frac{x(x-1)}{x+5}\ge 0$

Key Concepts: Inequalities, Interval Notation, Graphing Explanation: To solve this inequality, we need to determine the sign of the expression $\frac{x(x-1)}{x+5}$ for different intervals of $x$. We can do this by plotting the critical points of the expression, which are $x=0$, $x=1$, and $x=-5$. These points divide the real line into four intervals: $(-\infty ,-5)$, $(-5,0)$, $(0,1)$, and $(1,\infty )$. We can pick a test value in each interval and evaluate the sign of the expression for that value. Alternatively, we can observe that the numerator of the expression changes sign at $x=0$ and $x=1$, while the denominator changes sign at $x=-5$. The full solution set is obtained by considering all possible combinations of signs for the numerator and denominator. Step by Step Solution: Step 1. Plot the critical points $x=-5$, $x=0$, and $x=1$ on the number line. Step 2. Pick a value in the interval $(-\infty ,-5)$, e.g. $x=-6$, and evaluate the expression $\frac{x(x-1)}{x+5}$. We have $\frac{(-6)(-6-1)}{(-6)+5}=-42<0$, so this interval is part of the solution set. Step 3. Pick a value in the interval $(-5,0)$, e.g. $x=-1$, and evaluate the expression. We have $\frac{(-1)(-1-1)}{(-1)+5}=\frac{2}{3}>0$, so this interval is not part of the solution set. Step 4. Pick a value in the interval $(0,1)$, e.g. $x=\frac{1}{2}$, and evaluate the expression. We have $\frac{\frac{1}{2}(\frac{1}{2}-1)}{\frac{1}{2}+5}=-\frac{3}{22}<0$, so this interval is part of the solution set. Step 5. Pick a value in the interval $(1,\infty )$, e.g. $x=2$, and evaluate the expression. We have $\frac{2(2-1)}{2+5}=\frac{2}{7}>0$, so this interval is not part of the solution set. Step 6. Thus the solution set is $(-\infty ,-5)\cup (0,1)$. Step 7. The interval notation for the solution set is $\boxed{(-\infty ,-5)\cup (0,1)}$. Step 8. The graph of the solution set is shown below:\n [graph will be added as an image url] Final Answer: (-\infty, -5)\cup(0,1)