An example of Inequality Could someone please show me the step by step solution to the following problem?
$$
\frac{x^2+1}{x-1} \leq x
$$
the anwer should be $-1 \leq x <1$.
I'd like to know how to do this without the use of a table. I am aware that I must do two different solutions. One based on $x>1$ and one where $x<1$. However I can't seem to get it right so a step by step solution would be much appreciated
 A: Note that if $x\ne1$,
$$
\frac{x^2+1}{x-1}-x=\frac{x+1}{x-1}=\frac{x^2-1}{(x-1)^2}
$$
This is negative if and only if $x^2\lt1$. Therefore,
$$
\frac{x^2+1}{x-1}\lt x
$$
precisely when $x^2\lt1$ which is $-1\lt x\lt1$. When $x=1$, the fraction is not defined, but when $x=-1$ we have equality.
A: You have: $$\dfrac{x^2+1}{x-1}\leqslant x$$
You would like to multiply by $x-1$ both sides by you need to make sure that $x-1$ is positive or negative. So:


*

*If $x-1\gt0$:
$$x^2+1\leqslant x(x-1),\\\Leftrightarrow\\1\leqslant -x,\\\Leftrightarrow\\x\leqslant -1.$$
You start by $x-1\gt 0$ and you get $x\leqslant-1$, so $ x\leqslant-1$ and $x\gt1$. Hence there is no solution.

*If $x-1\lt0$:
$$x^2+1\geqslant x(x-1),\\\Leftrightarrow\\1\geqslant -x,\\\Leftrightarrow\\x\geqslant -1.$$
You start by $x-1\lt 0$ and you get $x\geqslant-1$, so $-1\leqslant x\lt1$.
A: Case 1: $x>1$
$x^2+1\leq x(x-1)=x^2-x\Rightarrow x\leq -1$ which is always false.
Case 2: $x<1$
$x^2+1\geq x(x-1)=x^2-x\Rightarrow x\geq -1$
Thus the answer is $-1\leq x< 1$ as desired.
$x\neq 1$ because then the denominator of the rational function will not be defined.
A: *

*Case 1. $x-1>0$. Multiplying both sides by $x-1$, we obtain $$x^2+1\leq x(x-1),$$ that is, $$1\leq -x,$$ or $$x\leq -1,$$ a contradiction.

*Case 2. $x-1<0$. Multiplying both sides by $x-1$, we obtain $$x^2+1\geq x(x-1),$$ that is, $$1\geq -x,$$ or $$x\geq-1.$$Thus the solution is $$-1\leq x<1.$$

A: Due to continuity $f(x)=\frac{x^2+1}{x-1}-x$ may change sign only at its zeroes (at $ x=-1$) or where it's not defined (at $x=1$).  Now plug in some appropriate numbers.
A: $$\frac{x^2+1}{x-1}-x\leq 0\\\frac{x^2+1-x(x-1)}{x-1}\leq0\\\frac{x^2+1-x^2+x}{x-1}\leq0\\\frac{x+1}{x-1}\leq 0$$
Now only one has to be negative,if $x<-1$ both $x-1$,$x+1$ are negative,if $x>1$ both are positive if they are between $x-1$ is negative and $x+1$ is positive
A: You can always multiply through by the non-negative number $(x-1)^2$ to clear fractions to obtain $$x^3-x^2+x-1\le x^3-2x^2+x$$ which reduces immediately to $$x^2\le 1$$
