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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![### Example 4
Find \( f'(x) \) given that \( f(x) = \frac{2x^2 + 3}{x^2 - 5x + 2} \).
**Solution:**
To find the derivative of \( f(x) \), we can use the quotient rule, which states:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
where \( u = 2x^2 + 3 \) and \( v = x^2 - 5x + 2 \).
First, we find \( u' \) and \( v' \):
\[ u = 2x^2 + 3 \implies u' = 4x \]
\[ v = x^2 - 5x + 2 \implies v' = 2x - 5 \]
Now, we apply the quotient rule:
\[ f'(x) = \left( \frac{u}{v} \right)' = \frac{(4x)(x^2 - 5x + 2) - (2x^2 + 3)(2x - 5)}{(x^2 - 5x + 2)^2} \]
Next, we simplify the numerator:
\[ (4x)(x^2 - 5x + 2) = 4x^3 - 20x^2 + 8x \]
\[ (2x^2 + 3)(2x - 5) = (2x^2)(2x) + (2x^2)(-5) + (3)(2x) + (3)(-5) \]
\[ = 4x^3 - 10x^2 + 6x - 15 \]
Therefore,
\[ f'(x) = \frac{4x^3 - 20x^2 + 8x - (4x^3 - 10x^2 + 6x - 15)}{(x^2 - 5x + 2)^2} \]
\[ = \frac{4x^3 - 20x^2 + 8x - 4x^3 + 10x^2 - 6x + 15}{(x^2 - 5x + 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16f66f84-778a-42af-b26b-121b7bbe9a0a%2F7e035cb8-b282-4c59-b76f-a3cec40e019a%2Fzxpdv4t_processed.png&w=3840&q=75)
Transcribed Image Text:### Example 4
Find \( f'(x) \) given that \( f(x) = \frac{2x^2 + 3}{x^2 - 5x + 2} \).
**Solution:**
To find the derivative of \( f(x) \), we can use the quotient rule, which states:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
where \( u = 2x^2 + 3 \) and \( v = x^2 - 5x + 2 \).
First, we find \( u' \) and \( v' \):
\[ u = 2x^2 + 3 \implies u' = 4x \]
\[ v = x^2 - 5x + 2 \implies v' = 2x - 5 \]
Now, we apply the quotient rule:
\[ f'(x) = \left( \frac{u}{v} \right)' = \frac{(4x)(x^2 - 5x + 2) - (2x^2 + 3)(2x - 5)}{(x^2 - 5x + 2)^2} \]
Next, we simplify the numerator:
\[ (4x)(x^2 - 5x + 2) = 4x^3 - 20x^2 + 8x \]
\[ (2x^2 + 3)(2x - 5) = (2x^2)(2x) + (2x^2)(-5) + (3)(2x) + (3)(-5) \]
\[ = 4x^3 - 10x^2 + 6x - 15 \]
Therefore,
\[ f'(x) = \frac{4x^3 - 20x^2 + 8x - (4x^3 - 10x^2 + 6x - 15)}{(x^2 - 5x + 2)^2} \]
\[ = \frac{4x^3 - 20x^2 + 8x - 4x^3 + 10x^2 - 6x + 15}{(x^2 - 5x + 2
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