0
$\begingroup$

It goes like : find domain of $f(x)= \frac{1}{\sqrt{(\{x+1\}-x^{2}+2x)}}$ where {.} denotes fractional part of x. I went on with the usual methodology putting denominator greater than 0,

$\{x+1\}-(x)^{2}+2x>0$

I changed fractional part to (x+1)-[x+1] where [.] denoted greatest integer function.

solving further,
$3x-(x)^{2}+1>[x+1]$
$3x-(x)^{2}>[x]$
after this i tried making graph of this but to no avail as i couldn't specify the exact points satisfying the condition. Algebraically I'm not able to make out the next step. Any clues will be helpful,thanks in advance!

$\endgroup$
  • $\begingroup$ This can help, perhaps: For one thing you need to have $(x^2 - 2x<1)$ to make sure that the denominator is not negative (otherwise it is negative and the square root is not defined). Then you can break into cases, try considering one case: $x^2-2x<-1$ and then separately another case: $-1<x^2-2x<1$ $\endgroup$ – them Aug 20 '19 at 9:56
0
$\begingroup$

Hint:

Recall that $x-1<[x]$ for all real $x.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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