Floor in the Function!

We can rewrite the above function as;

$\large f(x)=2^{x-\lfloor x\rfloor}=2^{\{x\}}$

Where $\{x\}$ denotes the fractional part of $x$ . Now since the range of the fractional part of $x$ is $[0,1)$ we can figure out the range of $f(x)$ .

The minimum value of the function would be: $f(0)=2^0=1$ .

The maximum value of the function would be: $f(0.999...)=2^{0.999...}\to 2$ .

That is the value tends to 2 but never quite reaches it. Thus the range of the function is $[1,2)$ , the open bracket denotes that it never equals 2.

$f(x)=\dfrac{2^x}{2^{\lfloor x \rfloor}}=\dfrac{2^x}{2^{x-\{x\}}}=\dfrac{2^x}{\frac{2^x}{2^{\{x\}}}}=2^{\{x\}}$ where {x} is the fractional part of $x$ .

As $0 \le {\{x\}} <1$ . The range of $f(x)$ is $[1,2)$ .

$x$ can be written as $\left \lfloor x \right \rfloor + y$ where $y$ lies between $0$ and $1$ . It can be $0$ when $x$ is an integer. But it can never be $1$ . So, $2^{\left \lfloor x \right \rfloor}$ cancels from the numerator and the denominator. And what we are left is $2^y$ where $0 \leq y < 1$ . So minimum value is $2^0 = 1$ . Maximum value is something less than $2^1 = 2$ .

So, the answer is $\color{#69047E}{\boxed{\left[1 , 2\right)}}$ .