Let's solve the inequality $(x - 7)(x - 1) < 0$ step by step to understand how it should be graphed on the number line.

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Step 1: Find the Critical Points

The product $(x - 7)(x - 1)$ changes sign at the values that make either factor equal to zero:

• $x - 7 = 0 \Rightarrow x = 7$
• $x - 1 = 0 \Rightarrow x = 1$

These values, $x = 1$ and $x = 7$, divide the number line into three intervals:

• $(-\infty, 1)$
• $(1, 7)$
• $(7, \infty)$

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Step 2: Test the Sign in Each Interval

We need to determine where the product $(x - 7)(x - 1)$ is negative.

• For $x \in (-\infty, 1)$ (e.g., $x = 0$): $(0 - 7)(0 - 1) = (-7)(-1) = 7$ (positive)

• For $x \in (1, 7)$ (e.g., $x = 2$): $(2 - 7)(2 - 1) = (-5)(1) = -5$ (negative)

• For $x \in (7, \infty)$ (e.g., $x = 8$): $(8 - 7)(8 - 1) = (1)(7) = 7$ (positive)

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Step 3: Determine the Solution

We are looking for where the product is negative, which happens in the interval:

• $(1, 7)$

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Step 4: Graph the Solution

• At $x = 1$ and $x = 7$, the expression is equal to 0, so these points are not included in the solution. Use open circles at $x = 1$ and $x = 7$.

• Draw a line segment between $1$ and $7$ (not including the endpoints).

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Final Answer

On the number line:

1. Place open circles at $x = 1$ and $x = 7$.
2. Draw a line segment between $1$ and $7$.

Let me know if you need further clarifications!

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Related Questions:

1. How do we solve inequalities with more than two factors?
2. What changes if the inequality were $(x - 7)(x - 1) \leq 0$?
3. How can this method be extended to quadratic inequalities?
4. What is the importance of testing values between critical points?
5. How do open and closed circles affect solutions on a number line?

Tip: Always test points from different regions to verify the behavior of expressions!