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2. Which recursively defined function has a first term equal to 10 and a common difference of 4? 1. f(1) 10 f(2) = f(2-1) +4 2. f(1) 4 f(2) – f(z - 1) + 10 f(1) 10 3. f(2) = 4f(2- 1) 4. S(1) 4 f(2) 10f(z- 1)

2. Which recursively defined function has a first term equal to 10 and a common difference of 4? 1. f(1) 10 f(2) = f(2-1) +4 2. f(1) 4 f(2) – f(z - 1) + 10 f(1) 10 3. f(2) = 4f(2- 1) 4. S(1) 4 f(2) 10f(z- 1)

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Question Which recursively defined function has a first term equal to 10 and a common difference of 4? ### Options 1. \[f(1) = 10\] \[f(x) = f(x-1) + 4\] 2. \[f(1) = 4\] \[f(x) = f(x-1) + 10\] 3. \[f(1) = 10\] \[f(x) = 4f(x-1)\] 4. \[f(1) = 4\] \[f(x) = 10f(x-1)\] ### Explanation To determine which of these recursively defined functions has a first term equal to 10 and a common difference of 4, consider the recursive formulas and their initial conditions: - **Option 1:** - Initial Condition: \(f(1) = 10\) - Recursive Formula: \(f(x) = f(x-1) + 4\) - This implies the sequence starts at 10 and increases by 4 each step, thus having a common difference of 4. - **Option 2:** - Initial Condition: \(f(1) = 4\) - Recursive Formula: \(f(x) = f(x-1) + 10\) - This sequence starts at 4 and increases by 10, which does not match the required first term or common difference. - **Option 3:** - Initial Condition: \(f(1) = 10\) - Recursive Formula: \(f(x) = 4f(x-1)\) - This generates a geometric sequence, not an arithmetic one with a common difference of 4. - **Option 4:** - Initial Condition: \(f(1) = 4\) - Recursive Formula: \(f(x) = 10f(x-1)\) - This also generates a geometric sequence, not matching the conditions. ### Correct Answer **Option 1:** \[f(1) = 10\] \[f(x) = f(x-1) + 4\] This option meets the conditions with the first term being 10 and a common difference of 4.
Transcribed Image Text:### Question Which recursively defined function has a first term equal to 10 and a common difference of 4? ### Options 1. \[f(1) = 10\] \[f(x) = f(x-1) + 4\] 2. \[f(1) = 4\] \[f(x) = f(x-1) + 10\] 3. \[f(1) = 10\] \[f(x) = 4f(x-1)\] 4. \[f(1) = 4\] \[f(x) = 10f(x-1)\] ### Explanation To determine which of these recursively defined functions has a first term equal to 10 and a common difference of 4, consider the recursive formulas and their initial conditions: - **Option 1:** - Initial Condition: \(f(1) = 10\) - Recursive Formula: \(f(x) = f(x-1) + 4\) - This implies the sequence starts at 10 and increases by 4 each step, thus having a common difference of 4. - **Option 2:** - Initial Condition: \(f(1) = 4\) - Recursive Formula: \(f(x) = f(x-1) + 10\) - This sequence starts at 4 and increases by 10, which does not match the required first term or common difference. - **Option 3:** - Initial Condition: \(f(1) = 10\) - Recursive Formula: \(f(x) = 4f(x-1)\) - This generates a geometric sequence, not an arithmetic one with a common difference of 4. - **Option 4:** - Initial Condition: \(f(1) = 4\) - Recursive Formula: \(f(x) = 10f(x-1)\) - This also generates a geometric sequence, not matching the conditions. ### Correct Answer **Option 1:** \[f(1) = 10\] \[f(x) = f(x-1) + 4\] This option meets the conditions with the first term being 10 and a common difference of 4.
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