Answer
$y$ is
increasing on $(-\infty, 0)$ and $(1, \infty)$ , and
decreasing on $(0,1/2)$ and $(1/2, 1).$
Work Step by Step
From the graph, $y$ is
increasing on $(-\infty, 0)$ and $(1, \infty)$ , and
decreasing on $(0,1/2)$ and $(1/2, 1).$
Analytically,
$y=\displaystyle \frac{x^{2}}{2x-1},$ defined everywhere, except for x=1/2
$y^{\prime}=\displaystyle \frac{(2x-1)2x-x^{2}(2)}{(2x-1)^{2}}$
$=\displaystyle \frac{2x^{2}-2x}{(2x-1)^{2}}$
$=\displaystyle \frac{2x(x-1)}{(2x-1)^{2}}$, differentiable everywhere except for x=1/2
Critical numbers: $x=0,1$
Discontinuity: $x=1/2$
$\left[\begin{array}{lllll}
Interval & (-\infty,0) & (0,1/2) & (1/2,1) & (1,\infty)\\
\text{test point} & -1 & 0.1 & 0.6 & 2\\
f^{\prime}(\text{test point}) & 0.44444 & -0.28125 & -12 & 0.44444\\
\text{sign} & + & - & - & +\\
& \nearrow & \searrow & \searrow & \nearrow
\end{array}\right]$
