Question

part 2 of 5 (b) the solution set in interval notation for |h - 1| > 6 is (-∞, -5) ∪ (7, ∞). part: 2 / 5 part 3 of 5 graph the solution set for |h - 1| > 6. ← -13 -12 -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 →

Explanation:

Step1: Recall the solution set

From part (b), the solution set of \(|h - 1|>6\) is \((-\infty,- 5)\cup(7,\infty)\). This means we need to graph all real numbers less than - 5 or greater than 7.

Step2: Graph the first interval \((-\infty,-5)\)

For the interval \((-\infty,-5)\), we use an open circle at \(h=-5\) (because - 5 is not included in the solution set, since the inequality is strict, \(>\) not \(\geq\)) and draw an arrow pointing to the left (towards negative infinity) to represent all numbers less than - 5.

Step3: Graph the second interval \((7,\infty)\)

For the interval \((7,\infty)\), we use an open circle at \(h = 7\) (because 7 is not included in the solution set) and draw an arrow pointing to the right (towards positive infinity) to represent all numbers greater than 7.

To graph it on the given number line:

• At \(h=-5\), place an open circle (hollow dot) and draw a line with an arrow going to the left (passing through - 6, - 7, etc.).
• At \(h = 7\), place an open circle (hollow dot) and draw a line with an arrow going to the right (passing through 8, 9, etc.).

(Note: Since we can't draw the actual graph here, the description above tells how to construct it. If we were to represent it textually on the given number line: mark an open circle at - 5 and draw an arrow left, mark an open circle at 7 and draw an arrow right.)

Answer:

To graph the solution set \((-\infty, - 5)\cup(7,\infty)\) on the number line:

1. At \(h=-5\), draw an open circle (hollow dot) and draw a ray (line with an arrow) extending to the left (towards \(-\infty\)).
2. At \(h = 7\), draw an open circle (hollow dot) and draw a ray (line with an arrow) extending to the right (towards \(\infty\)).