A function $k$ whose domain is the set of positive integers is defined as $k(1)=4$ and $k(n)=k(n-1)-2$ . $\newline$ Function $k$ was evaluated for several numbers. Which of the following are true? $\newline$ Select each correct answer. $\newline$ (A) $k(-1)=-4$ $\newline$ (B) $k(0)=-2$ $\newline$ (C) $k(2)=2$ $\newline$ (D) $k(3)=0$ $\newline$ (E) $k(6)=3$

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Algebra 1

Negative exponents

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Q. A function

k

whose domain is the set of positive integers is defined as

k(1)=4

and

k(n)=k(n-1)-2

\newline

Function

k

was evaluated for several numbers. Which of the following are true?

\newline

Select each correct answer.

\newline

(A)

k(-1)=-4

\newline

(B)

k(0)=-2

\newline

(C)

k(2)=2

\newline

(D)

k(3)=0

\newline

(E)

k(6)=3

Understand Function Definition: Understand the function definition and domain. $\newline$ The function $k$ is defined for positive integers, with $k(1)=4$ and $k(n)=k(n-1)-2$ for $n > 1$ . This means that for each step $n$ , the function value decreases by $2$ from its previous value.
Evaluate Statement A: Evaluate the truth of statement A. $\newline$ Since the domain of $k$ is the set of positive integers, $k(-1)$ is not defined. Therefore, statement A is false.
Evaluate Statement B: Evaluate the truth of statement B. Similarly, $k(0)$ is not defined because $0$ is not a positive integer. Therefore, statement B is false.
Evaluate Statement C: Evaluate the truth of statement C. $\newline$ Using the definition of $k$ , we calculate $k(2)$ as follows: $\newline$ $k(2) = k(2-1) - 2$ $\newline$ $k(2) = k(1) - 2$ $\newline$ $k(2) = 4 - 2$ $\newline$ $k(2) = 2$ $\newline$ Therefore, statement C is true.
Evaluate Statement D: Evaluate the truth of statement D. $\newline$ Using the definition of $k$ , we calculate $k(3)$ as follows: $\newline$ $k(3) = k(3-1) - 2$ $\newline$ $k(3) = k(2) - 2$ $\newline$ $k(3) = 2 - 2$ (from the previous step) $\newline$ $k(3) = 0$ $\newline$ Therefore, statement D is true.
Evaluate Statement E: Evaluate the truth of statement E. $\newline$ Using the definition of $k$ , we calculate $k(6)$ as follows: $\newline$ $k(6) = k(6-1) - 2$ $\newline$ $k(6) = k(5) - 2$ $\newline$ We need to find $k(5)$ first, which requires finding $k(4)$ : $\newline$ $k(4) = k(4-1) - 2$ $\newline$ $k(4) = k(3) - 2$ $\newline$ $k(4) = 0 - 2$ (from step $5$ ) $\newline$ $k(4) = -2$ $\newline$ Now we find $k(5)$ : $\newline$ $k(6)$ $1$ $\newline$ $k(6)$ $2$ $\newline$ $k(6)$ $3$ $\newline$ $k(6)$ $4$ $\newline$ Finally, we find $k(6)$ : $\newline$ $k(6) = k(5) - 2$ $\newline$ $k(6)$ $7$ $\newline$ $k(6)$ $8$ $\newline$ Therefore, statement E is false.