List the critical values of the related function. Then solve the inequality.
-$$\frac { x } { x ^ { 2 } + 3 x - 4 } + \frac { 2 } { x ^ { 2 } - 16 } \leq \frac { 2 x } { x ^ { 2 } - 5 x + 4 }$$
A) $- 5 - \sqrt { 23 } , - 4 , - 5 + \sqrt { 23 } , 1,4 ; [ - 5 - \sqrt { 23 } , - 4 ) \cup [ - 5 + \sqrt { 23 } , 1 ) \cup ( 4 , \infty )$
B) $- 4 - \sqrt { 23 } , - 5 , - 4 + \sqrt { 23 } , 1,5 ; [ - 4 - \sqrt { 23 } , - 5 ) \cup [ - 4 + \sqrt { 23 } , 1 ) \cup ( 5 , \infty )$
C) No critical values; $\varnothing$
D) $- 5 - \sqrt { 33 } , - 4 , - 5 + \sqrt { 33 } , 1,4 ; [ - 5 - \sqrt { 33 } , - 4 ) \cup [ - 5 + \sqrt { 33 } , 1 ) \cup ( 4 , \infty )$

Question

Quizplus · Accepted Answer

To solve the inequality, first identify the critical values by setting each denominator and the numerator (when combined into a single fraction) equal to zero. The denominators are $x^2 + 3x - 4$, $x^2 - 16$, and $x^2 - 5x + 4$. Factoring gives us the roots $x = -4, 1$ from $x^2 + 3x - 4$, $x = \pm 4$ from $x^2 - 16$, and $x = 1, 4$ from $x^2 - 5x + 4$. Combining the inequality into a single fraction and setting the numerator equal to zero gives a quadratic equation. Solving this quadratic equation yields two more critical values, which are $-5 \pm \sqrt{23}$. These are the points where the function could change sign. Testing intervals around these critical values determines where the function is less than or equal to zero, leading to the solution set provided in option A.

List the Critical Values of the Related Function xx2+3x−4+2x2−16≤2xx2−5x+4\frac { x } { x ^ { 2 } + 3 x - 4 } + \frac { 2 } { x ^ { 2 } - 16 } \leq \frac { 2 x } { x ^ { 2 } - 5 x + 4 }x2+3x−4x​+x2−162​≤x2−5x+42x​

List the Critical Values of the Related Function $\frac { x } { x ^ { 2 } + 3 x - 4 } + \frac { 2 } { x ^ { 2 } - 16 } \leq \frac { 2 x } { x ^ { 2 } - 5 x + 4 }$