Find the nth-derivative, f(n), of f(x) e2x+4. 1. f(n)(x) = 2. f(n)(x) = 3. f(n) (x) = 2ne2x+4 1 - n! = 1 n! 2ne2x+4 e2x+4 4. f(n) (x) = e 5. f(n)(x) = n! 2ne²x+4 6. f(n) (x) e2x+4 = n! e²x+4
Find the nth-derivative, f(n), of f(x) e2x+4. 1. f(n)(x) = 2. f(n)(x) = 3. f(n) (x) = 2ne2x+4 1 - n! = 1 n! 2ne2x+4 e2x+4 4. f(n) (x) = e 5. f(n)(x) = n! 2ne²x+4 6. f(n) (x) e2x+4 = n! e²x+4
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![## Part 1 of 4
### Problem:
Find the \( n^{th} \)-derivative, \( f^{(n)} \), of
\[ f(x) = e^{2x+4}. \]
### Solutions:
1. \( f^{(n)}(x) = 2^n e^{2x+4} \)
2. \( f^{(n)}(x) = \frac{1}{n!} 2^n e^{2x+4} \)
3. \( f^{(n)}(x) = \frac{1}{n!} e^{2x+4} \)
4. \( f^{(n)}(x) = e^{2x+4} \)
5. \( f^{(n)}(x) = n! 2^n e^{2x+4} \)
6. \( f^{(n)}(x) = n! e^{2x+4} \)
---
## Part 2 of 4
### Problem:
Find the degree three Taylor polynomial \( T_3 \) centered at \( x = 0 \) for \( f \) when
\[ f(x) = \ln(3 - 4x). \]
### Solutions:
1. \( T_3(x) = \ln 3 - \frac{4}{3}x - \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
2. \( T_3(x) = \frac{4}{3}x - \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
3. \( T_3(x) = \ln 3 + \frac{4}{3}x - \frac{8}{9}x^2 + \frac{32}{81}x^3 \)
4. \( T_3(x) = \frac{4}{3}x + \frac{8}{9}x^2 + \frac{64}{81}x^3 \)
5. \( T_3(x) = \ln 3 - \frac{4}{3}x + \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
6. \( T_3(x) = \frac{4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc88a61a8-930a-4496-81c3-aab24a173799%2F3b44815f-1ae7-4f1c-bbb3-9ea196cd14af%2Fn5f8h17_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Part 1 of 4
### Problem:
Find the \( n^{th} \)-derivative, \( f^{(n)} \), of
\[ f(x) = e^{2x+4}. \]
### Solutions:
1. \( f^{(n)}(x) = 2^n e^{2x+4} \)
2. \( f^{(n)}(x) = \frac{1}{n!} 2^n e^{2x+4} \)
3. \( f^{(n)}(x) = \frac{1}{n!} e^{2x+4} \)
4. \( f^{(n)}(x) = e^{2x+4} \)
5. \( f^{(n)}(x) = n! 2^n e^{2x+4} \)
6. \( f^{(n)}(x) = n! e^{2x+4} \)
---
## Part 2 of 4
### Problem:
Find the degree three Taylor polynomial \( T_3 \) centered at \( x = 0 \) for \( f \) when
\[ f(x) = \ln(3 - 4x). \]
### Solutions:
1. \( T_3(x) = \ln 3 - \frac{4}{3}x - \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
2. \( T_3(x) = \frac{4}{3}x - \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
3. \( T_3(x) = \ln 3 + \frac{4}{3}x - \frac{8}{9}x^2 + \frac{32}{81}x^3 \)
4. \( T_3(x) = \frac{4}{3}x + \frac{8}{9}x^2 + \frac{64}{81}x^3 \)
5. \( T_3(x) = \ln 3 - \frac{4}{3}x + \frac{8}{9}x^2 - \frac{64}{81}x^3 \)
6. \( T_3(x) = \frac{4
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