Question

given, ( f(x) = -(x - 5)(x + 1)^2(x - 3)^3 ), create a sign chart to:
a. determine the intervals when the function value is greater than zero and when the function value is less than zero.
b. sketch the graph of ( f(x) ).

Explanation:

Part (a) - Sign Chart and Intervals

Step 1: Find Critical Points

The critical points are the roots of the function, found by setting each factor to zero:

• \( x - 5 = 0 \implies x = 5 \)
• \( x + 1 = 0 \implies x = -1 \) (multiplicity 2)
• \( x - 3 = 0 \implies x = 3 \) (multiplicity 3)

These divide the number line into intervals: \( (-\infty, -1) \), \( (-1, 3) \), \( (3, 5) \), \( (5, \infty) \).

Step 2: Analyze Sign of Each Factor

• For \( x < -1 \) (e.g., \( x = -2 \)):

\( -(x-5) = -(-7) = 7 \) (positive), \( (x+1)^2 = (-1)^2 = 1 \) (positive), \( (x-3)^3 = (-5)^3 = -125 \) (negative).
Product: \( + \cdot + \cdot - = - \).

• For \( -1 < x < 3 \) (e.g., \( x = 0 \)):

\( -(x-5) = 5 \) (positive), \( (x+1)^2 = 1 \) (positive), \( (x-3)^3 = (-3)^3 = -27 \) (negative).
Product: \( + \cdot + \cdot - = - \).

• For \( 3 < x < 5 \) (e.g., \( x = 4 \)):

\( -(x-5) = 1 \) (positive), \( (x+1)^2 = 25 \) (positive), \( (x-3)^3 = 1^3 = 1 \) (positive).
Product: \( + \cdot + \cdot + = + \).

• For \( x > 5 \) (e.g., \( x = 6 \)):

\( -(x-5) = -1 \) (negative), \( (x+1)^2 = 49 \) (positive), \( (x-3)^3 = 27 \) (positive).
Product: \( - \cdot + \cdot + = - \).

• At \( x = -1 \): \( (x+1)^2 = 0 \implies f(x) = 0 \).
• At \( x = 3 \): \( (x-3)^3 = 0 \implies f(x) = 0 \).
• At \( x = 5 \): \( (x-5) = 0 \implies f(x) = 0 \).

Sign Chart Summary:

Interval	\( -\infty < x < -1 \)	\( -1 < x < 3 \)	\( 3 < x < 5 \)	\( 5 < x < \infty \)
Zero at	No	\( x = -1 \)	\( x = 3 \)	\( x = 5 \)

Part (b) - Sketching the Graph

Key Features:

1. Roots (x-intercepts):

• \( x = -1 \) (multiplicity 2: graph touches the x-axis and turns around).
• \( x = 3 \) (multiplicity 3: graph crosses the x-axis with a "flatter" slope).
• \( x = 5 \) (multiplicity 1: graph crosses the x-axis).

1. End Behavior:

The leading term (highest degree) of \( f(x) \) is \( -x^{1+2+3} = -x^6 \). For even degree (\( 6 \)) and negative leading coefficient, as \( x \to \pm\infty \), \( f(x) \to -\infty \).

1. Intervals of Increase/Decrease (Simplified):

• For \( x < -1 \): \( f(x) < 0 \) (decreasing, since end behavior is \( -\infty \) and \( f(-1) = 0 \)).
• For \( -1 < x < 3 \): \( f(x) < 0 \) (decreasing, as \( f(3) = 0 \)).
• For \( 3 < x < 5 \): \( f(x) > 0 \) (increasing, since \( f(3) = 0 \) and \( f(5) = 0 \)).
• For \( x > 5 \): \( f(x) < 0 \) (decreasing, as \( x \to \infty \), \( f(x) \to -\infty \)).

Sketch Steps:

1. Plot the x-intercepts: \( (-1, 0) \) (touch and turn), \( (3, 0) \) (cross with flatness), \( (5, 0) \).
2. Draw the end behavior: both ends go down (\( -\infty \)).
3. Connect the points, ensuring:

• At \( x = -1 \), the graph touches the axis and turns.
• At \( x = 3 \), the graph crosses the axis with a flatter slope.
• In \( (3, 5) \), the graph is above the x-axis; in other intervals, below (except at roots).

Final Answers (Part a)

• \( f(x) > 0 \) when \( \boldsymbol{3 < x < 5} \).
• \( f(x) < 0 \) when \( \boldsymbol{(-\infty, -1) \cup (-1, 3) \cup (5, \infty)} \).

(For the graph, follow the sketching steps above. The sign chart and intervals are the primary results for part (a).)

Answer:

Part (a) - Sign Chart and Intervals

Step 1: Find Critical Points

The critical points are the roots of the function, found by setting each factor to zero:

• \( x - 5 = 0 \implies x = 5 \)
• \( x + 1 = 0 \implies x = -1 \) (multiplicity 2)
• \( x - 3 = 0 \implies x = 3 \) (multiplicity 3)

These divide the number line into intervals: \( (-\infty, -1) \), \( (-1, 3) \), \( (3, 5) \), \( (5, \infty) \).

Step 2: Analyze Sign of Each Factor

• For \( x < -1 \) (e.g., \( x = -2 \)):

\( -(x-5) = -(-7) = 7 \) (positive), \( (x+1)^2 = (-1)^2 = 1 \) (positive), \( (x-3)^3 = (-5)^3 = -125 \) (negative).
Product: \( + \cdot + \cdot - = - \).

• For \( -1 < x < 3 \) (e.g., \( x = 0 \)):

\( -(x-5) = 5 \) (positive), \( (x+1)^2 = 1 \) (positive), \( (x-3)^3 = (-3)^3 = -27 \) (negative).
Product: \( + \cdot + \cdot - = - \).

• For \( 3 < x < 5 \) (e.g., \( x = 4 \)):

\( -(x-5) = 1 \) (positive), \( (x+1)^2 = 25 \) (positive), \( (x-3)^3 = 1^3 = 1 \) (positive).
Product: \( + \cdot + \cdot + = + \).

• For \( x > 5 \) (e.g., \( x = 6 \)):

\( -(x-5) = -1 \) (negative), \( (x+1)^2 = 49 \) (positive), \( (x-3)^3 = 27 \) (positive).
Product: \( - \cdot + \cdot + = - \).

• At \( x = -1 \): \( (x+1)^2 = 0 \implies f(x) = 0 \).
• At \( x = 3 \): \( (x-3)^3 = 0 \implies f(x) = 0 \).
• At \( x = 5 \): \( (x-5) = 0 \implies f(x) = 0 \).

Sign Chart Summary:

Interval	\( -\infty < x < -1 \)	\( -1 < x < 3 \)	\( 3 < x < 5 \)	\( 5 < x < \infty \)
Zero at	No	\( x = -1 \)	\( x = 3 \)	\( x = 5 \)

Part (b) - Sketching the Graph

Key Features:

1. Roots (x-intercepts):

• \( x = -1 \) (multiplicity 2: graph touches the x-axis and turns around).
• \( x = 3 \) (multiplicity 3: graph crosses the x-axis with a "flatter" slope).
• \( x = 5 \) (multiplicity 1: graph crosses the x-axis).

1. End Behavior:

The leading term (highest degree) of \( f(x) \) is \( -x^{1+2+3} = -x^6 \). For even degree (\( 6 \)) and negative leading coefficient, as \( x \to \pm\infty \), \( f(x) \to -\infty \).

1. Intervals of Increase/Decrease (Simplified):

• For \( x < -1 \): \( f(x) < 0 \) (decreasing, since end behavior is \( -\infty \) and \( f(-1) = 0 \)).
• For \( -1 < x < 3 \): \( f(x) < 0 \) (decreasing, as \( f(3) = 0 \)).
• For \( 3 < x < 5 \): \( f(x) > 0 \) (increasing, since \( f(3) = 0 \) and \( f(5) = 0 \)).
• For \( x > 5 \): \( f(x) < 0 \) (decreasing, as \( x \to \infty \), \( f(x) \to -\infty \)).

Sketch Steps:

1. Plot the x-intercepts: \( (-1, 0) \) (touch and turn), \( (3, 0) \) (cross with flatness), \( (5, 0) \).
2. Draw the end behavior: both ends go down (\( -\infty \)).
3. Connect the points, ensuring:

• At \( x = -1 \), the graph touches the axis and turns.
• At \( x = 3 \), the graph crosses the axis with a flatter slope.
• In \( (3, 5) \), the graph is above the x-axis; in other intervals, below (except at roots).

Final Answers (Part a)

• \( f(x) > 0 \) when \( \boldsymbol{3 < x < 5} \).
• \( f(x) < 0 \) when \( \boldsymbol{(-\infty, -1) \cup (-1, 3) \cup (5, \infty)} \).

(For the graph, follow the sketching steps above. The sign chart and intervals are the primary results for part (a).)