Number Line & Intervals (Part 2)

[mathdad]

The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the intervals on a number line.

(A) |x - 4| < 4

(B) |x + 5| >= 2

For (A), I did the following:

-4 < x - 4 < 4

I now add 4 to each term.

0 < x < 8

On the number line, I would need to plot (0, 8). Is this correct?

For (B), we have the following:

|x + 5| >= 2

x + 5 < -2 or x + 5 >= 2

x =< -2 - 5 or x >= 2 - 5

x =< - 7 or x >= -3

I must plot [-infinity, -7] and [-3, infinity] on the number line. Is this right?

 

[MarkFL]

a) \(\displaystyle |x-4|<4\)

I would read this as a distance formula, that is, all real numbers $x$ that are less than 4 units from 4, which as you stated is the interval $(0,8)$.

\begin{tikzpicture}[scale=2.5]
\draw[very thick] (0,0) -- (8,0);
\path [draw=black, fill=white, thick] (0,0) circle (2pt);
\path [draw=black, fill=white, thick] (8,0) circle (2pt);
\draw[latex-latex] (-0.5,0) -- (8.5,0) ;
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

b) \(\displaystyle |x+5|\ge2\)

I would read this as all real numbers $x$ whose distance from $-5$ is greater than or equal to 2, which as you stated is the interval $(-\infty,-7]\,\cup\,[-3,\infty)$.

\begin{tikzpicture}[scale=2.5]
\path [draw=black, fill=black, thick] (-7,0) circle (2pt);
\path [draw=black, fill=black, thick] (-3,0) circle (2pt);
\draw[latex-latex] (-9.5,0) -- (-0.5,0) ;
\draw[->,thick] (-7,0) -- (-9.25,0);
\draw[->,thick] (-3,0) -- (-0.75,0);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

 

[mathdad]

a) \(\displaystyle |x-4|<4\)

I would read this as a distance formula, that is, all real numbers $x$ that are less than 4 units from 4, which as you stated is the interval $(0,8)$.

\begin{tikzpicture}[scale=2.5]
\draw[very thick] (0,0) -- (8,0);
\path [draw=black, fill=white, thick] (0,0) circle (2pt);
\path [draw=black, fill=white, thick] (8,0) circle (2pt);
\draw[latex-latex] (-0.5,0) -- (8.5,0) ;
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

b) \(\displaystyle |x+5|\ge2\)

I would read this as all real numbers $x$ whose distance from $-5$ is greater than or equal to 2, which as you stated is the interval $(-\infty,-7]\,\cup\,[-3,\infty)$.

\begin{tikzpicture}[scale=2.5]
\path [draw=black, fill=black, thick] (-7,0) circle (2pt);
\path [draw=black, fill=black, thick] (-3,0) circle (2pt);
\draw[latex-latex] (-9.5,0) -- (-0.5,0) ;
\draw[->,thick] (-7,0) -- (-9.25,0);
\draw[->,thick] (-3,0) -- (-0.75,0);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

Cool. Thanks.