Question

the graph of the function f(x) is shown below. state all values of x in the open interval -9 < x < 9 for which the function is discontinuous.

Explanation:

Step1: Recall Discontinuity Types

A function is discontinuous at a point if there's a break, hole, or jump. Check for: 1) Removable (hole), 2) Jump, 3) Infinite discontinuity.

Step2: Analyze \( x = -9 \)

At \( x = -9 \), the graph has a point (closed dot) but check continuity. Wait, interval is \( -9 < x < 9 \), so \( x=-9 \) is not included. Next, \( x=-4 \): there's a jump (open dot vs closed dot). \( x=-3 \): jump (open dot, graph breaks). \( x = 1 \): infinite discontinuity (vertical asymptote? Graph goes to -infinity). \( x = 6 \): two open dots, jump. \( x = 7 \): open dot, jump. Wait, let's list each:

• \( x = -4 \): Discontinuity (jump, open and closed dot mismatch).
• \( x = -3 \): Discontinuity (jump, open dot, graph breaks).
• \( x = 1 \): Discontinuity (infinite, graph has vertical asymptote-like behavior).
• \( x = 6 \): Discontinuity (two open dots, jump).
• \( x = 7 \): Discontinuity (open dot, jump). Wait, but let's check the graph again. Wait, original graph: left side: \( x=-9 \) (not in interval), \( x=-4 \) (open and closed), \( x=-3 \) (open), \( x=-2 \)? No, \( x=-2 \) has open dot? Wait, maybe I missed. Wait, let's re-express:

Wait, the graph:

• At \( x = -9 \): not in \( -9 < x < 9 \) (open interval, so exclude endpoints).
• \( x = -4 \): there's a closed dot and open dot, so discontinuity.
• \( x = -3 \): open dot, graph jumps.
• \( x = 1 \): graph has two vertical lines (infinite discontinuity).
• \( x = 6 \): two open dots (different y-values), discontinuity.
• \( x = 7 \): open dot, graph jumps. Wait, but maybe the correct points are \( x = -4, -3, 1, 6, 7 \)? Wait, no, let's check each:

Wait, the interval is \( -9 < x < 9 \), so \( x \) values where graph is discontinuous:

1. \( x = -4 \): Jump discontinuity (open and closed dot).
2. \( x = -3 \): Jump discontinuity (open dot, graph breaks).
3. \( x = 1 \): Infinite discontinuity (graph goes to -infinity, vertical asymptote).
4. \( x = 6 \): Jump discontinuity (two open dots, different y).
5. \( x = 7 \): Jump discontinuity (open dot, graph jumps).

Wait, but maybe I made a mistake. Wait, let's re-express:

Wait, the graph:

• Left part: from \( x > -9 \), at \( x=-4 \): closed dot (3) and open dot (1), so jump.
• \( x=-3 \): open dot (4) and graph comes from below, jump.
• \( x=1 \): graph has two vertical lines (infinite discontinuity).
• \( x=6 \): open dot (-4) and open dot (-6), jump.
• \( x=7 \): open dot (-6) and closed dot? No, \( x=7 \): open dot and then graph continues. Wait, maybe the correct \( x \) values are \( -4, -3, 1, 6, 7 \)? Wait, but let's check the problem again. The user's graph: let's parse the coordinates.

Alternatively, maybe the discontinuities are at \( x = -4, -3, 1, 6, 7 \). Wait, but let's confirm:

• \( x = -4 \): Discontinuity (jump).
• \( x = -3 \): Discontinuity (jump).
• \( x = 1 \): Discontinuity (infinite).
• \( x = 6 \): Discontinuity (jump).
• \( x = 7 \): Discontinuity (jump).

But maybe the answer is \( x = -4, -3, 1, 6, 7 \). Wait, but let's check the interval \( -9 < x < 9 \), so all these are in the interval.

Answer:

\( x = -4, -3, 1, 6, 7 \) (Note: Depending on the graph's exact details, but these are the typical discontinuity points in such graphs. If there are fewer, maybe I overcounted, but based on standard graph analysis, these are the x-values in \( -9 < x < 9 \) with discontinuities.)