Question - Absolute Value Inequality and Interval Notation
Solution:
The given inequality is $$ |x| \leq \frac{11}{3} $$, and we are asked to express this set in interval notation and graph it on a number line.First, let's solve the absolute value inequality:An absolute value inequality like $$ |x| \leq \frac{11}{3} $$ means that the distance between $$ x $$ and 0 on the number line is less than or equal to $$ \frac{11}{3} $$. We break this into two separate inequalities:1. $$ x \leq \frac{11}{3} $$, because $$ x $$ can be to the right of 0 and still be within $$ \frac{11}{3} $$ units of 0.2. $$ x \geq -\frac{11}{3} $$, because $$ x $$ can also be to the left of 0 and be within $$ \frac{11}{3} $$ units of 0.The solution to the inequality is the set of all $$ x $$ that satisfy both conditions, meaning $$ x $$ is between $$ -\frac{11}{3} $$ and $$ \frac{11}{3} $$, inclusive of both endpoints.The interval notation for this inequality is $$[- \frac{11}{3}, \frac{11}{3}] $$.In terms of graphing this set on a number line, we would mark and shade the interval from $$ -\frac{11}{3} $$ to $$ \frac{11}{3} $$, including the endpoints, which are marked with a closed dot to indicate that they are part of the solution set.The grouping symbols for this interval notation would be ([]) which indicates that the set includes its endpoints.The endpoints entered into the boxes as shown in the image should be:- Left endpoint: $$ -\frac{11}{3} $$- Right endpoint: $$ \frac{11}{3} $$For the number line provided in part (b), the shaded region would start at $$ -\frac{11}{3} $$ and end at $$ \frac{11}{3} $$. You would ensure that there is a closed dot or a solid circle at each of the endpoints $$ -\frac{11}{3} $$ and $$ \frac{11}{3} $$ to depict that these points are included in the interval.
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