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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![The mathematical exercise provided requires solving the given function taking into account an initial condition.
We start with the second derivative of a function:
\[ f''(x) = 2x + 2 \]
We are also provided with initial conditions:
\[ f'(-2) = -1 \]
\[ f(-2) = 4 \]
**Steps to solve:**
1. **Finding the first derivative, \( f'(x) \):**
Integrate \( f''(x) \):
\[ f'(x) = \int (2x + 2) \, dx \]
\[ f'(x) = x^2 + 2x + C_1 \]
Use the initial condition \( f'(-2) = -1 \):
\[ -1 = (-2)^2 + 2(-2) + C_1 \]
\[ -1 = 4 - 4 + C_1 \]
\[ C_1 = -1 \]
So, \( f'(x) = x^2 + 2x - 1 \).
2. **Finding the function, \( f(x) \):**
Integrate \( f'(x) \):
\[ f(x) = \int (x^2 + 2x - 1) \, dx \]
\[ f(x) = \frac{x^2}{2} + x^2 - x + C_2 \]
Combine terms:
\[ f(x) = \frac{1}{3}x^3 + x^2 - x + C_2 \]
Use the initial condition \( f(-2) = 4 \):
\[ 4 = \frac{-8}{3} + 4 + 2 + C_2 \]
\[ 4 = -\frac{8}{3} + \frac{12}{3} + \frac{6}{3} + C_2 \]
\[ 4 = \frac{10}{3} + C_2 \]
\[ C_2 = \frac{12}{3} - \frac{10}{3} \]
\[ C_2 = \frac{2}{3} \]
So, \( f(x) = \frac{1}{3}x^3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F686df84f-2572-4ea8-99e4-a6aa8d03e207%2Fa7cba020-856e-41b4-bb83-f16809e1bda1%2F85dttsx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The mathematical exercise provided requires solving the given function taking into account an initial condition.
We start with the second derivative of a function:
\[ f''(x) = 2x + 2 \]
We are also provided with initial conditions:
\[ f'(-2) = -1 \]
\[ f(-2) = 4 \]
**Steps to solve:**
1. **Finding the first derivative, \( f'(x) \):**
Integrate \( f''(x) \):
\[ f'(x) = \int (2x + 2) \, dx \]
\[ f'(x) = x^2 + 2x + C_1 \]
Use the initial condition \( f'(-2) = -1 \):
\[ -1 = (-2)^2 + 2(-2) + C_1 \]
\[ -1 = 4 - 4 + C_1 \]
\[ C_1 = -1 \]
So, \( f'(x) = x^2 + 2x - 1 \).
2. **Finding the function, \( f(x) \):**
Integrate \( f'(x) \):
\[ f(x) = \int (x^2 + 2x - 1) \, dx \]
\[ f(x) = \frac{x^2}{2} + x^2 - x + C_2 \]
Combine terms:
\[ f(x) = \frac{1}{3}x^3 + x^2 - x + C_2 \]
Use the initial condition \( f(-2) = 4 \):
\[ 4 = \frac{-8}{3} + 4 + 2 + C_2 \]
\[ 4 = -\frac{8}{3} + \frac{12}{3} + \frac{6}{3} + C_2 \]
\[ 4 = \frac{10}{3} + C_2 \]
\[ C_2 = \frac{12}{3} - \frac{10}{3} \]
\[ C_2 = \frac{2}{3} \]
So, \( f(x) = \frac{1}{3}x^3
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