Piecewise Functions - 4

For $f^{-1}$ to be a function, $f$ must be bijective. It must pass the horizontal line test, and its range must be equal to codomain. The range of $f$ must be $\mathbb{R}^+$ .

If $x \not \in \mathbb{Z}, ~ f(x) = x \lfloor x \rfloor$ .

For any non-zero integer $n$ , we can define this as

$f(x) = nx, \text{ if } n < x < (n+ 1)$

---

For any general $n$ , range of $f$ is

If $n > 0$ , $f$ is an increasing function,

$\implies f(n^+) < f(x) < f((n+1)^-)$

If $n < 0$ , $f$ is a decreasing function,

$\implies f((n+1)^-) < f(x) < f(n^+)$

The limiting values can be found by substitution.

$\displaystyle\lim_{x \rightarrow n^{+}} f(x) = n^2$ and $\displaystyle\lim_{x \rightarrow (n+1)^{-}} f(x) = n(n + 1)$

Since $n + 1$ is an integer as well, we can replace $n+1$ by $n$

$\displaystyle\lim_{x \rightarrow n^{-}} f(x) = (n-1)n =n^2 -n$

---

On observation, for some positive integer $k$ ,

$\begin{array}{l|l} n = -k & k(k-1) < f(x) < k^2 \\ n = k & k^2 < f(x) < k(k+1) \\ n = -(k + 1) & k(k+1) < f(x) < (k+1)^2 \\ n = k +1 & (k+1)^2 < f(x) < (k+1)(k+2)\\ \end{array}$

We see that if we extend this table for all $k \geq 1$ , then the range of $f$ is all positive real numbers except the limiting values.

If $g(x)$ and $h(x)$ were $x^2$ and $x^2 -x$ , or $x^2 + x$ and $x^2$ , then range of $f$ will become $\mathbb{R}^+$ , and it will become bijective.

So the functions are $g(x) = x^2$ and $h(x) = x^2 - x$ , or $x^2 + x$ and $x^2$ .

In both cases, we get $g(x) - h(x) = x$ .

We get the result $5 \times \lfloor g(3.5) - h(3.5) \rfloor = 5 \lfloor 3.5 \rfloor = 5 \times 3 = \boxed{15}$