Q.If $f(x)=2\{x\}+5x$, where $\{x\}$ is the fractional part function of $x$, then find $f(-1)$.

The greatest integer function (floor function) $[x]$ gives the greatest integer $\le x$ — always rounding down, never toward zero (so $[-6.3]=-7$, not $-6$). It is piecewise-constant: $f(x)=n$ for every $x\in[n,n+1)$, giving a staircase graph with domain $R$ and range the integers $I$. Key identities: $x-1\le[x]<x$ always (sandwiching); $[x+n]=[x]+n$ for any integer $n$ (shifting by a whole number just carries straight through the floor); and $[x]+[-x]$ equals $0$ if $x$ is itself an integer, or $-1$ otherwise.

The fractional part function $\{x\}=x-[x]$ captures exactly what the floor throws away, always landing in $[0,1)$ — so $\{x\}=0$ precisely characterises the integers, and $x=[x]+\{x\}$ splits any real number into its integer and fractional halves. Its graph is a repeating sawtooth, resetting to $0$ at every integer and climbing back up to just under $1$. Key identities mirror the floor's: $\{x\pm n\}=\{x\}$ for integer $n$, and $\{x\}+\{-x\}$ equals $0$ if $x\in I$, or $1$ otherwise. …