The question is, "Find the domain and the range of the real function $f$ defined by $f(x) = |x-1|$. I have no idea what $|x-1|$ means, someone help me?
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$\begingroup$The function is defined this way: $$f\left(x\right)=\left|x-1\right|=\begin{cases} x-1 & \text{if }\:x-1\geq0\\ -x+1 & \text{if }\:x-1<0 \end{cases}=\begin{cases} x-1 & \text{if }\:x\geq1\\ -x+1 & \text{if }\:x<1 \end{cases} $$
$\endgroup$ $\begingroup$It means absolute value, the distance from $0$. First calculate the number inside of the $||$, then if it's negative make it positive and if it's positive, keep it positive. For example: $$|10-1|=|9|=9$$ and: $$|-23-1|=|-24|=24$$
$\endgroup$ 2 $\begingroup$$|x|$ is the absolute value of $x$. Suppose the value of $x$ is negative, then after plugging it to the absolute function($| |$), it will return the value with positive sign. For example, $|-3|=3$. If the value of $x$ is already positive, then it will return as it is. In your question, you are supposed to find the absolute value of $x-1$ for all values that $x$ can take, namely Real Numbers (Domain). The trick to find the range is to break the expression into two parts - one where the expression is positive for a given range of values of $x$ and other for the range where the value of the expression is negative. Take the union of the result of those two and that's your range.
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