Draw the graph of $f(x)=\min\{\lvert x \rvert-1,\lvert x-1\rvert -1,\vert x-2\rvert -1\}$

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I have had drawn this answer myself but I want clarification. What I do is:

\begin{cases} y = x-2 & \text{for } x\ge 1\\ \space \space = -x & \text{for }x<1 \end{cases}

\begin{cases} y = x-1 & \text{for } x\ge 0\\ \space \space = -x-1 & \text{for }x<0 \end{cases}

\begin{cases} y = x-3 & \text{for } x\ge 2\\ \space \space = -x+1 & \text{for }x<2 \end{cases}

Now we see that for $\space x<0, \;y=-x-1$ is the equation.

For $\; x\ge 0$ and $x\le \dfrac{1}{2},$ $y=x-1$ is the equation

For $\, x\ge \dfrac{1}{2},$ but $x<1$, we use $y=-x. $

For $\space x\ge 1 \space \text{and} x<\dfrac{3}{2}$ we have $y=x-2$.

For $\space x>\dfrac{3}{2} \space \text{and} x\le 2,$ we have $y=-x+1$

For $x>2$, we have $y=x-3$

Can anyone draw the graph and verify whether I am on the right track?