Answer
$(-1,1- \frac {\sqrt 2}{2}) U (1+ \frac {\sqrt 2}{2}, ∞)$
Work Step by Step
$\frac{x+1}{2x^2 - 4x + 1} \leq 0$
Find the zeros of the expressions in the numerator AND the denominator
$x+1 = 0$
$ x = -1$
$2x^2 - 4x = -1$
$2(x^2 - 2x + 1) = -1 + 2$
$(x-1)^2 = 1/2$
$x = 1 + \frac {\sqrt 2}{2}, 1- \frac {\sqrt 2}{2}$
$2x^2 - 4x + 1$ = $(x - 1 - \frac {\sqrt 2}{2}) (x - 1+ \frac {\sqrt 2}{2})$
Test numbers in between those zero values to determine if the function is negative or positive
$(-∞, -1)$ $\frac{(-)}{(-)(-)} = (-)$
$(-1,1- \frac {\sqrt 2}{2})$ $\frac{(+)}{(-)(-)} = (+)$
$(1- \frac {\sqrt 2}{2}, 1+ \frac {\sqrt 2}{2})$ $\frac{(+)}{(+)(-)} = (-)$
$(1+ \frac {\sqrt 2}{2}, ∞)$ $\frac{(+)}{(+)(+)} = (+)$
Thus the solution is $(-1,1- \frac {\sqrt 2}{2}) U (1+ \frac {\sqrt 2}{2}, ∞)$
