## University Calculus: Early Transcendentals (3rd Edition)

$(-\infty,-5)\cup (-1,1)$ $\:$ 
$\mathrm{See\:the\:graph\:above.}$ $a.\:\:\:$ By observing the graph, we come to know that the point of intersection is $\:x=-5.\:$ Based on the graph, the function $\:y=\frac{3}{x-1}\:$ (represented by red curve) is below the function $\:\frac{2}{x+1}\:$ (represented by blue curve), when: $x < -5\:\:$ and $\:\:-1< x <1\:\:$ (vertical asymptotes are at $\:x=1\:\:\:\mathrm{and}\:\:x=-1\:$ because for those numbers the denominators of the functions are zero.) $b.$ $\frac{3}{x-1} -5$ $x^2-1 < 0\:\:\rightarrow x^2 < 1\:\:$ when $\:\:-1 < x < 1$ Solution of this case is $\:x\in(-1,1).$ $\mathrm{Case\:II:}$ $x+5 < 0,\:\:$ when $\:\:x < -5$ $x^2-1 > 0\:\:\rightarrow\:\: x^2 > 1\:\:$ when $\:\:-1 > x > 1$ Solution of this case is $\:x\in(-\infty,-5).$ Combined solution from both cases: $\:(-\infty,-5)\cup (-1,1).$