+0  
 
0
35
2
avatar

Solve the inequality \(\frac{1}{x - 1} - \frac{4}{x - 2} + \frac{4}{x - 3} - \frac{1}{x - 4} < \frac{1}{30}.\)

 Feb 27, 2019
 #1
avatar+203 
-1

x=0.375485 or x=−8.913063 or x=15.537578(Makes both sides equal)

 

x=1 or x=2 or x=4 or x=4(Makes left denominator equal to 0)

 

Check intervals in between critical points. (Test values in the intervals to see if they work.)

 

x<−8.913063(Works in original inequality)

 

−8.913063

 

0.375485

 

1

 

2

 

4

 

x>15.537578(Works in original inequality)

 

Answer:

x<−8.913063 or 0.375485 15.537578

 Feb 27, 2019
 #2
avatar+21842 
+1

Solve the inequality

 

\(\large{\dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} < \dfrac{1}{30}} \)

 

 

\(\small{ \begin{array}{|rcll|} \hline \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} &<& \dfrac{1}{30} \quad | \quad x\ne 1,\ x\ne 2,\ x\ne 3,\ x\ne 4 \\\\ \dfrac{1}{x - 1} - \dfrac{4}{x - 2} + \dfrac{4}{x - 3} - \dfrac{1}{x - 4} -\dfrac{1}{30} &<& 0 \\\\ -\dfrac{x^4-10x^3+5x^2+100x+84}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \\\\ \boxed{\small{x^4-10x^3+5x^2+100x+84 = (x - 7)(x - 6)(x + 1)(x + 2)}} \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{30(x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot 30 \\\\ -\dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &<& 0 \quad | \quad \cdot(-1)! \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)}{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \quad | \quad \cdot \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2) \Big((x - 1)(x - 2)(x - 3)(x - 4)\Big)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \dfrac{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)^2(x - 2)^2(x - 3)^2(x - 4)^2 }{ (x - 1)(x - 2)(x - 3)(x - 4)} &>& 0 \\\\ \mathbf{(x - 7)(x - 6)(x + 1)(x + 2)(x - 1)(x - 2)(x - 3)(x - 4)} & \mathbf{>}& \mathbf{0} \\ \hline \end{array} } \)

 

\(\begin{array}{|r|c|c|c|c|c|c|c|c|} \hline &(-\infty,-2) & (-2,-1) & (-1,1) & (1,2) & (2,3) & (3,4) & (4,6) & (6,7) & (7,\infty) \\ \hline x+2 & -& +& +& +& +& +& +& +& +\\ \hline x+1 & -& -& +& +& +& +& +& +& +\\ \hline x-1 & -& -& -& +& +& +& +& +& +\\ \hline x-2 & -& -& -& -& +& +& +& +& +\\ \hline x-3 & -& -& -& -& -& +& +& +& +\\ \hline x-4 & -& -& -& -& -& -& +& +& +\\ \hline x-6 & -& -& -& -& -& -& -& +& +\\ \hline x-7 & -& -& -& -& -& -& -& -& +\\ \hline \text{sign} & \\ \text{of} & \\ \text{all} & \mathbf{\large{+}} & -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}}& -& \mathbf{\large{+}} \\ & \mathbf{>0} & & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0}& & \mathbf{>0} \\ \hline \end{array}\)

 

\(\mathbf{(-\infty,-2) \cup (-1,1) \cup (2,3) \cup (4,6) \cup (7,\infty)}\)

 

laugh

 Feb 27, 2019

11 Online Users