Question

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Explanation:

Part A (Function Machines)

Machine A: \( x \to 2x \to f(x) \)

• For \( x = 5 \):

Step1: Substitute \( x = 5 \) into \( 2x \)

\( 2 \times 5 = 10 \)

• For \( x = -3 \):

Step2: Substitute \( x = -3 \) into \( 2x \)

\( 2 \times (-3) = -6 \)

Machine B: \( x \to x^2 + 1 \to f(x) \)

• For \( x = 5 \):

Step1: Substitute \( x = 5 \) into \( x^2 + 1 \)

\( 5^2 + 1 = 25 + 1 = 26 \)

• For \( x = -3 \):

Step2: Substitute \( x = -3 \) into \( x^2 + 1 \)

\( (-3)^2 + 1 = 9 + 1 = 10 \)

Function Notation

• Machine A: \( f(x) = 2x \)
• Machine B: \( f(x) = x^2 + 1 \)

Evaluating \( f(x) = 3x + 9 \) for \( x = 5 \)

Step1: Substitute \( x = 5 \) into \( 3x + 9 \)

\( f(5) = 3(5) + 9 \)

Step2: Calculate the result

\( 3(5) + 9 = 15 + 9 = 24 \)

Evaluating the Given Functions

1. \( f(x) = 6(x + 9) \)

• a. \( f(4) \):

Step1: Substitute \( x = 4 \) into \( 6(x + 9) \)

\( f(4) = 6(4 + 9) \)

Step2: Simplify inside the parentheses

\( 4 + 9 = 13 \)

Step3: Multiply by 6

\( 6 \times 13 = 78 \)

• b. \( f(-6) \):

Step1: Substitute \( x = -6 \) into \( 6(x + 9) \)

\( f(-6) = 6(-6 + 9) \)

Step2: Simplify inside the parentheses

\( -6 + 9 = 3 \)

Step3: Multiply by 6

\( 6 \times 3 = 18 \)

2. \( g(x) = x^2 - 12 \)

• a. \( g(1) \):

Step1: Substitute \( x = 1 \) into \( x^2 - 12 \)

\( g(1) = 1^2 - 12 \)

Step2: Calculate \( 1^2 \)

\( 1^2 = 1 \)

Step3: Subtract 12

\( 1 - 12 = -11 \)

• b. \( g(-3) \):

Step1: Substitute \( x = -3 \) into \( x^2 - 12 \)

\( g(-3) = (-3)^2 - 12 \)

Step2: Calculate \( (-3)^2 \)

\( (-3)^2 = 9 \)

Step3: Subtract 12

\( 9 - 12 = -3 \)

3. \( p(x) = \frac{8 - x}{x} \)

• a. \( p(-8) \):

Step1: Substitute \( x = -8 \) into \( \frac{8 - x}{x} \)

\( p(-8) = \frac{8 - (-8)}{-8} \)

Step2: Simplify the numerator

\( 8 - (-8) = 8 + 8 = 16 \)

Step3: Divide by -8

\( \frac{16}{-8} = -2 \)

• b. \( p(8) \):

Step1: Substitute \( x = 8 \) into \( \frac{8 - x}{x} \)

\( p(8) = \frac{8 - 8}{8} \)

Step2: Simplify the numerator

\( 8 - 8 = 0 \)

Step3: Divide by 8

\( \frac{0}{8} = 0 \)

Function Evaluation Restriction for \( p(x) = \frac{8 - x}{x} \)

A function with a denominator cannot have a denominator of zero (since division by zero is undefined). For \( p(x) = \frac{8 - x}{x} \), the denominator is \( x \). Setting \( x = 0 \) makes the denominator zero, so \( p(x) \) is undefined at \( x = 0 \).

Answer:

Yes, \( x = 0 \) makes the function undefined (division by zero is not allowed).

Final Answers (Key Parts)

• Machine A Outputs: \( 10, -6 \)
• Machine B Outputs: \( 26, 10 \)
• Function Notation: \( \boldsymbol{f(x) = 2x} \) (A), \( \boldsymbol{f(x) = x^2 + 1} \) (B)
• \( f(5) = 24 \)
• 1a. \( \boldsymbol{78} \); 1b. \( \boldsymbol{18} \)
• 2a. \( \boldsymbol{-11} \); 2b. \( \boldsymbol{-3} \)
• 3a. \( \boldsymbol{-2} \); 3b. \( \boldsymbol{0} \)
• Restriction: \( x = 0 \) (function undefined)