MHB Number Line & Intervals (Part 2)

  • Thread starter Thread starter mathdad
  • Start date Start date
  • Tags Tags
    intervals Line
AI Thread Summary
The discussion focuses on solving inequalities involving absolute values and representing the solutions on a number line. For the inequality |x - 4| < 4, the solution is the interval (0, 8), indicating all real numbers between 0 and 8. For the inequality |x + 5| >= 2, the solution consists of two intervals: (-∞, -7] and [-3, ∞), representing all numbers whose distance from -5 is greater than or equal to 2. Participants confirm the accuracy of these intervals and provide visual representations on a number line. The conversation emphasizes understanding absolute value inequalities and their graphical interpretations.
mathdad
Messages
1,280
Reaction score
0
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?
 
Mathematics news on Phys.org
a) $$|x-4|<4$$

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) $$|x+5|\ge2$$

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}
 
MarkFL said:
a) $$|x-4|<4$$

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) $$|x+5|\ge2$$

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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top