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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Question
A) find the inflection point ( x and y value) of the graph of f
![### Differentiation Using the Product Rule
#### Problem Statement
Given the function:
\[ f(x) = 3x \cdot e^{-2x} \]
We aim to find its derivative.
#### Solution
To differentiate \( f(x) \), we will use the **Product Rule** for differentiation. The Product Rule states:
\[ \dfrac{d}{dx} [f(x) \cdot g(x)] = f(x) \cdot \dfrac{d}{dx} [g(x)] + g(x) \cdot \dfrac{d}{dx} [f(x)] \]
Applying the Product Rule to our function:
\[ f(x) = 3x \]
\[ g(x) = e^{-2x} \]
The solution is:
\[ 3x \cdot \dfrac{d}{dx} [e^{-2x}] + e^{-2x} \cdot \dfrac{d}{dx} [3x] \]
\[ 3x \cdot (-2e^{-2x}) + e^{-2x} \cdot 3 \]
\[ -6xe^{-2x} + 3e^{-2x} \]
Hence, the derivative of the function \( f(x) = 3x \cdot e^{-2x} \) is:
\[ \boxed{-6xe^{-2x} + 3e^{-2x}} \]
To review the step-by-step solution, please click on the provided link.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77922221-9e32-493b-ab71-d15b7753ec9e%2F98b128f1-94bb-4284-85de-1dacf6e3400b%2Festtbfmt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Differentiation Using the Product Rule
#### Problem Statement
Given the function:
\[ f(x) = 3x \cdot e^{-2x} \]
We aim to find its derivative.
#### Solution
To differentiate \( f(x) \), we will use the **Product Rule** for differentiation. The Product Rule states:
\[ \dfrac{d}{dx} [f(x) \cdot g(x)] = f(x) \cdot \dfrac{d}{dx} [g(x)] + g(x) \cdot \dfrac{d}{dx} [f(x)] \]
Applying the Product Rule to our function:
\[ f(x) = 3x \]
\[ g(x) = e^{-2x} \]
The solution is:
\[ 3x \cdot \dfrac{d}{dx} [e^{-2x}] + e^{-2x} \cdot \dfrac{d}{dx} [3x] \]
\[ 3x \cdot (-2e^{-2x}) + e^{-2x} \cdot 3 \]
\[ -6xe^{-2x} + 3e^{-2x} \]
Hence, the derivative of the function \( f(x) = 3x \cdot e^{-2x} \) is:
\[ \boxed{-6xe^{-2x} + 3e^{-2x}} \]
To review the step-by-step solution, please click on the provided link.
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