1
$\begingroup$

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 $f$
$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 6005, Matthew Conroy, TravisJ, астон вілла олоф мэллбэрг, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $x-\lfloor x \rfloor=1$, when $x \neq \text{integer}$ and it is $0$ at integers. $\endgroup$ – Anurag A Dec 1 '16 at 18:29
4
$\begingroup$

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.