Could somebody show me their method for finding the domain of this function? $\ln(1-(1/x))$
With the usual method I get that $1-(1/x)>0$ which results in $x>1$. however, the actual answer is a union of $x<0$ and $x>1$.
Could someone explain how to get to the actual (complete) answer?
thanks in advance!
$\endgroup$4 Answers
$\begingroup$You just forgot that $$\frac1x<1\iff x>1\quad\textbf{or}\quad x<0. $$ Another way to see this consists in rewriting the inequation as $$\frac{x-1}x>0\iff x(x-1)>0 \enspace\text{and}\enspace x\ne 0.$$
$\endgroup$ $\begingroup$When you get to $$\frac{1}{x}<1$$ The natural thing is to multiply everything with $x$ and get $$x>1$$ But this isn't valid for all $x$ it's valid only if $x>0$ because if $x<0$ you have to flip the signs $>$ turns into $<$ and $<$ into $>$, so for $x<0$ we get $\frac 1x\cdot x>1\cdot x$ after we multiply the inequality by $x$ which is just $x<1$.
To avoid having to deal with cases you can either leave everything on one side or multiply by something that is always positive (in this case it can be multiplied by $x^2$ as suggested by A. Goodier).
$\endgroup$ $\begingroup$The domain for the $\;\log\quad$ function is $\;(0,\infty).\quad$ For $\;\ln\left(1-\frac{1}{x}\right),\quad$ we require $\;1-\frac{1}{x}>0.\quad$ Multiplying both sides of the inequality by $\;x^2\quad$ gives $\;x^2-x>0.\quad$ The solution to this inequality is $\;x>1\quad$ or $\;x<0.$
$\endgroup$ $\begingroup$Domain of $\ln(y) : y >0$, real.
$y:= 1-1/x;$ $y =(1-1/x) >0;$
$\star) $ $1> 1/x;$
1) $x>0$; then: $x>1$.
(Multiply both sides of $\star)$ by $x$ (positive).)
2) $x<0$;
Then the RHS of $\star)$ is negative, I.e.
$1>1/x$ is true.
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