Question Number 86894 by M±th+et£s last updated on 01/Apr/20
$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\lceil{x}−\mathrm{1}\rceil} \\ $$$${find}\:{the}\:{domain}\:{and}\:{f}\:'\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:{if}\:{exist}. \\ $$$$ \\ $$$$\lceil…\rceil\:{ceiling}\:{function} \\ $$
Commented by mathmax by abdo last updated on 01/Apr/20
![if [..] mean integr part we have f(x) =((1−x^2 )/([x]−1)) let [x]=p ⇒p≤x<p+1 ⇒f(x)=((1−x^2 )/(p−1)) and for p≠1 we get f^′ (x) =(1/(p−1))(−2x) =((−2x)/(p−1)) x=−(1/2) ⇒[−(1/2)]=−1 ⇒f^′ (−(1/2)) =(((−2)×(−(1/2)))/(−2)) =−(1/2) domain of f [x]−1=0 ⇔[x]=1 ⇒1≤x<2 ⇒D_f =R−[1,2[](https://www.tinkutara.com/question/Q86905.png)
$${if}\:\left[..\right]\:{mean}\:{integr}\:{part}\:\:{we}\:{have}\:{f}\left({x}\right)\:=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left[{x}\right]−\mathrm{1}} \\ $$$${let}\:\left[{x}\right]={p}\:\Rightarrow{p}\leqslant{x}<{p}+\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{p}−\mathrm{1}}\:\:{and}\:{for}\:{p}\neq\mathrm{1}\:{we}\:{get} \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{p}−\mathrm{1}}\left(−\mathrm{2}{x}\right)\:=\frac{−\mathrm{2}{x}}{{p}−\mathrm{1}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\left[−\frac{\mathrm{1}}{\mathrm{2}}\right]=−\mathrm{1}\:\Rightarrow{f}^{'} \left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\left(−\mathrm{2}\right)×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)}{−\mathrm{2}}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${domain}\:{of}\:{f} \\ $$$$\left[{x}\right]−\mathrm{1}=\mathrm{0}\:\Leftrightarrow\left[{x}\right]=\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{x}<\mathrm{2}\:\Rightarrow{D}_{{f}} ={R}−\left[\mathrm{1},\mathrm{2}\left[\right.\right. \\ $$
Commented by M±th+et£s last updated on 01/Apr/20
$$\lceil…\rceil\:{smallest}\:{integer}\:{function}\: \\ $$