To draw the graph of $|y|=|\log(1-|x|)|$, I see that $x=0, y=0$ satisfies it. Also, the asymptotes would be $x=1, x=-1$. I have made a graph which has sort of eight curves in it but I don't know how to show that here. Can anyone draw a graph here so that I could verify if what I have drawn is correct or not. Thanks.
$\endgroup$ 02 Answers
$\begingroup$You can enter implicit graphs in Desmos:
To figure out what this looks like by hand, experiment in Desmos with adding and removing the absolute value signs to see how each affects the graph $y = \log(1-x)$.
$\endgroup$ $\begingroup$We can write this as $y= log(1- |x|)$ and $y= -log(1- |x|)$.
Now take the absolute value off x:
$y= log(1- x)$ for $x\ge 0$Of course, log(1- x) is only defined for $1- x> 0$ so $0\le x< 1$.
$y= log(1+ x)$ for $x< 0$Of course, log(1+ x) is only defined for $1+ x> 0$ so $-1< x\le 0$.
$y= -log(1- x)$ for $xge 0$Of course, log(1- x) is only defined for $1- x> 0$ so $0\le x< 1$.
$y= -log(1+ x)$ for $x< 0$. Of course, log(1+ x) is only defined for $1+ x> 0$ so $-1< x\le 0$.
