Maxima minima of $f(x)=\frac{x-\lfloor x\rfloor}{\sqrt{x}}$

Question

Let $f(x)=\dfrac{x-\lfloor x\rfloor}{\sqrt{x}}$

$$\forall x\in {]}0,+\infty{[},\quad 0\leq f(x)<1$$

• How can I show that $0$ is minimum value for $f$ and $1$ isn't maximum value for $f$

Answer 1

Your function $f$ is non-negative because the numerator and the denominator are non-negative.

$f(1)=0$ so the minimum is zero.

For $x > 1$ the numerator is less than one and $\sqrt{x} > 1$ so $f$ is strictly less than one. For $x \in]0,1[$, $f(x)=\sqrt{x}$ which is strictly less than one.