Chapter 4.4, Problem 46AYU

To determine

To calculate:

The following inequality equation:

  $\frac{x\left(x^2+1\right)\left(x-2\right)}{\left(x-15\right)\left(x+1\right)}\ge 0$

Answer to Problem 46AYU

The solution of the inequality equation $\frac{x\left(x^2+1\right)\left(x-2\right)}{\left(x-15\right)\left(x+1\right)}\ge 0$ is $x\ge 0,2,15$ .

Explanation of Solution

Given information:

  $\frac{x\left(x^2+1\right)\left(x-2\right)}{\left(x-15\right)\left(x+1\right)}\ge 0$

Calculation:

The given equation is:

  $\begin{array}{l}\frac{x\left(x^2+1\right)\left(x-2\right)}{\left(x-15\right)\left(x+1\right)}\ge 0\\ \frac{x\left(x^2+1\right)\left(x-2\right)}{\left(x-15\right)\left(x+1\right)}=0\\ x\left(x^2+1\right)\left(x-2\right)=0\\ x=0ORx-2=0OR\text{}x-15=0OR\text{}x+1=0\\ x=0x=2x=15x=-1\end{array}$

If

  $\begin{array}{l}\left(x^2+1\right)=0\\ x=\sqrt{-1}\\ x=i\end{array}$

So, this is not the solution of the given equation. The inequality value should be greater than or equal to zero, from the above solution $x<-1,0\le x\le 2,x>15$ .

Hence, the inequality is solved for $x\ge 0,2,15$ .