\(x=-1\) \(\frac{1}{1} \cdot-1\)
Since the denominator \((x)\) of the inequality may not always be positive, it will be wrong to directly cross multiply the inequality but rather we multiply through by the square of the denominator i.e. \(x^{2}\) which will always give a positive value whatever the value of \(x\) consider the inequality \(\frac{1}{x+1}<-1\). we cannot also cross multiply this because for all values of \(x\). the denommator is not always positive. Here we multiply through by the square of \((x+1)\) i.e. \((x+1)^{\text {? }}\) which will always give a positive value. Also consider this \(\frac{1}{x^{2}+1}<-1\). for this inequality. we are good to go when we cross multiply because whatever value of \(x\) (be it positive or negative). the denominator of \(\frac{1}{x+1}\) will always give a positive value. Going back to the problem at hand. \(\frac{1}{x}<-1\), we multiply through by \(x^{2}\)
\(\frac{1}{x} \times x^{2}<-1 \times x^{2}\)
\(x<-x^{2}\)
or \(-x>x^{2}: x^{2}+x<0\)
\(x(x+1)<0\)
set. \(x=0, x+1=0, x=-1\)
\begin{array}{c|c|c|c}
\multicolumn{5}{c}{\(x<-1\)} & \(-1-1<x<0 \quad 0 \quad x>0\) \\
\hline\(x\) & \(-\) ve & -ve & +ve \\
\hline\(x+1\) & \(-\) ve & +ve & \(+\) ve \\
\hline\(x(x+1)\) & +ve & -ve & \(+\) ve
\end{array}
Since \(-\) ve \(<0\) thus, \(-1<x<0\).