Solve the inequality and graph the solution.

Question

v
–
1
+1≤3
Plot the endpoints. Select an endpoint to change it from closed to open. Select the middle of the segment, ray, or line to delete it.

Answer 1

To solve the inequality \( \sqrt{v} - 1 + 1 \leq 3 \), we first simplify it:

\[ \sqrt{v} \leq 3. \]

Next, we square both sides to eliminate the square root, but we must consider that both sides are non-negative in this context:

\[ v \leq 9. \]

Now, we also need to consider the domain of the square root. The expression \( \sqrt{v} \) is defined for \( v \geq 0 \). Therefore, we combine this with the previous inequality:

\[ 0 \leq v \leq 9. \]

This gives us the solution set of the inequality.

Solution in interval notation:

\[ [0, 9]. \]

Graphing the Solution

1. Plot the endpoints:

   • The endpoint \( 0 \) will be a closed dot (since it is included).
   • The endpoint \( 9 \) will also be a closed dot (since it is included).

2. Change an endpoint from closed to open:

   • For example, we could change \( 9 \) to an open dot if the inequality were strict \( (<) \), but since we have \( \leq \), we keep both as closed dots.

3. Deleting part of the segment:

   • Since the inequality includes both endpoints, we will not delete any part.

Final Graph Representation:

• The interval is represented on a number line from 0 to 9 with closed dots at both ends:

\[ \text{[----------]} \quad (0 \text{ closed dot}) \quad (9 \text{ closed dot}) \]

This shows that the solution includes all values of \( v \) from 0 to 9. If you were instructed to change one endpoint to open, we could consider it a modification of your instruction, but the solution to your specific inequality keeps both endpoints closed because of the \( \leq \) symbol.