## Thomas' Calculus 13th Edition

$x\in(0,1)$
Think of the equation as $f(x) \lt 0,\quad f(x)=x^{2}-x=x(x-1)$ A geometric interpretation is to find the values of $x$ for which the graph of $f(x)$ is below the x-axis. The graph of $f$ is a parabola that opens up, because the leading coefficient is $+1$, a positive number. $f(x)=x(x-1)$ so it has zeros (crosses the x-axis) at $x=0$ and $x=1$. Since it opens up, the graph will be below the x-axis in the interval between the two zeros, $x\in(0,1)$. --- Solving algebraically, we analyze $x(x-1) \lt 0$ asking "when is the product of two numbers on the LHS negative?" Setting up a table where we test the signs of the product: $\left[\begin{array}{ccccccc} & -\infty & & 0 & & 1 & & \infty\\ \text{test point } t & & -1 & & 0.5 & & 2 & \\ f(t) & & +2 & & -0.25 & & +2 & \\ \text{sign of }f & & + & ( & - & ) & + & \end{array}\right]$ We see that the inequality is true for $x\in(0,1)$