Interval for the solutions of $\{x+1\}
Find the interval(s) which contain solutions of $$\{x+1\}<x^2-2x$$ where $\{x\}$ is the fractional part of $x$.
I was told that one way of solving this would be graphically. However I generally don't use graphical methods as I find them a tad difficult. Could somebody please show me how to solve this using a graph, as well as through a non-graphical method?
Secondly, from plotting the curves for $\{x+1\}$ and $x^2-2x$, wouldn't I only get points of intersection and thus values of $x$ which satisfy both equations? How then could I use the method to generate an interval for which the given inequality holds?
Many thanks!
 A: Here's an algebraic way of doing it.
$\{x+1 \} = \{ x\}$ because the fractional part function has period $1$. Also, $0 \leq \{x \} < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x \in (0,2)$ and non-negative everywhere else. So, $\{x\} < x^2 - 2x$ can hold only outside the interval $[0,2]$.
Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x \geq 2$. Also, $x^2 - 2x$ and $\{x\}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = \{x\}$ at $x = 2$, then $x^2 - 2x > \{x\}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.
Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = \{x\}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = \{x\}$ coincides with the function $\tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
$$
x = \frac{3 \pm \sqrt{13}}{2}
$$
so the negative root is the one with the minus sign. Thus, $x^2 - 2x > \{ x\}$ for all $x$ less than this root.
Thus, there are two intervals which do the job: $(-\infty,(3 - \sqrt{13})/2)$ and $(2,\infty)$.

In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.

I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].
