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

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 $f$

## closed as off-topic by 6005, Matthew Conroy, TravisJ, астон вілла олоф мэллбэрг, Daniel W. FarlowDec 2 '16 at 14:36

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• $x-\lfloor x \rfloor=1$, when $x \neq \text{integer}$ and it is $0$ at integers. – Anurag A Dec 1 '16 at 18:29

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.
• $1$ may not be the maximum, but this shows it is the supremum (least upper bound) – Henry Dec 2 '16 at 13:22