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Find a function f(x) such that f'(x)% = - 3a? + 8 and f(0) = 1 f(z) = %3D

Find a function f(x) such that f'(x)% = - 3a? + 8 and f(0) = 1 f(z) = %3D

Calculus: Early Transcendentals
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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**Problem Statement:** Find a function \( f(x) \) such that \( f'(x) = -3x^2 + 8 \) and \( f(0) = 1 \). **Solution:** To find \( f(x) \), integrate the derivative \( f'(x) = -3x^2 + 8 \). 1. Integrate \( -3x^2 \) to get \( -x^3 \). 2. Integrate \( 8 \) to get \( 8x \). Thus, \( f(x) = -x^3 + 8x + C \), where \( C \) is the constant of integration. **Apply the Initial Condition:** Use the given condition \( f(0) = 1 \) to find \( C \): 1. Substitute \( x = 0 \) into \( f(x) = -x^3 + 8x + C \). 2. \( f(0) = 0 + 0 + C = 1 \). Therefore, \( C = 1 \). **Final Expression:** The function is \( f(x) = -x^3 + 8x + 1 \).
Transcribed Image Text:**Problem Statement:** Find a function \( f(x) \) such that \( f'(x) = -3x^2 + 8 \) and \( f(0) = 1 \). **Solution:** To find \( f(x) \), integrate the derivative \( f'(x) = -3x^2 + 8 \). 1. Integrate \( -3x^2 \) to get \( -x^3 \). 2. Integrate \( 8 \) to get \( 8x \). Thus, \( f(x) = -x^3 + 8x + C \), where \( C \) is the constant of integration. **Apply the Initial Condition:** Use the given condition \( f(0) = 1 \) to find \( C \): 1. Substitute \( x = 0 \) into \( f(x) = -x^3 + 8x + C \). 2. \( f(0) = 0 + 0 + C = 1 \). Therefore, \( C = 1 \). **Final Expression:** The function is \( f(x) = -x^3 + 8x + 1 \).
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