6. (8) Let f(x) = 4 x ³ - x 4 3 a) Find the domain D (f) and the Y-intercept b) Find f'(x) the intervals on which f(x)is increasing, the intervals on which f(x) is decreasing, and any local max
6. (8) Let f(x) = 4 x ³ - x 4 3 a) Find the domain D (f) and the Y-intercept b) Find f'(x) the intervals on which f(x)is increasing, the intervals on which f(x) is decreasing, and any local max
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.3: Zeros Of Polynomials
Problem 3E
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![### Calculus Problem on Polynomial Functions
**Problem 6:**
Given the function \( f(x) = 4x^3 - x + 4 \):
1. **(a) Find the domain \(D(f)\) and the y-intercept.**
* **Domain:** The domain of \( f(x) \) is the set of all real numbers, since it is a polynomial function.
* **Y-intercept:** The y-intercept is found by evaluating \( f(x) \) at \( x = 0 \). Hence, \( f(0) = 4(0)^3 - 0 + 4 = 4 \). Therefore, the y-intercept is \( (0, 4) \).
2. **(b) Find \( f'(x) \), the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and any local maxima or minima.**
* **First Derivative:** \( f'(x) = \frac{d}{dx}(4x^3 - x + 4) = 12x^2 - 1 \).
* **Critical Points:** To find the critical points, set \( f'(x) \) to zero:
\[
12x^2 - 1 = 0 \implies x^2 = \frac{1}{12} \implies x = \pm \frac{1}{2\sqrt{3}}
\]
* **Intervals of Increase and Decrease:**
- For intervals of increase and decrease, test the sign of \( f'(x) \) in the intervals determined by the critical points \( x = \frac{1}{2\sqrt{3}} \) and \( x = -\frac{1}{2\sqrt{3}} \).
Rather than providing a purely theoretical explanation, students are advised to compute the values in intervals:
- \( (-\infty, -\frac{1}{2\sqrt{3}}) \)
- \( (-\frac{1}{2\sqrt{3}}, \frac{1}{2\sqrt{3}}) \)
- \( (\frac{1}{2\sqrt{3}}, \infty) \)
* **Local Maxima and Minima:**
- By evaluating the second derivative](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2fc8d61-50af-4e4e-bb29-e3320480694b%2F55ce607c-1816-4655-ad6c-8b131709091a%2Fkptgek2d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem on Polynomial Functions
**Problem 6:**
Given the function \( f(x) = 4x^3 - x + 4 \):
1. **(a) Find the domain \(D(f)\) and the y-intercept.**
* **Domain:** The domain of \( f(x) \) is the set of all real numbers, since it is a polynomial function.
* **Y-intercept:** The y-intercept is found by evaluating \( f(x) \) at \( x = 0 \). Hence, \( f(0) = 4(0)^3 - 0 + 4 = 4 \). Therefore, the y-intercept is \( (0, 4) \).
2. **(b) Find \( f'(x) \), the intervals on which \( f(x) \) is increasing, the intervals on which \( f(x) \) is decreasing, and any local maxima or minima.**
* **First Derivative:** \( f'(x) = \frac{d}{dx}(4x^3 - x + 4) = 12x^2 - 1 \).
* **Critical Points:** To find the critical points, set \( f'(x) \) to zero:
\[
12x^2 - 1 = 0 \implies x^2 = \frac{1}{12} \implies x = \pm \frac{1}{2\sqrt{3}}
\]
* **Intervals of Increase and Decrease:**
- For intervals of increase and decrease, test the sign of \( f'(x) \) in the intervals determined by the critical points \( x = \frac{1}{2\sqrt{3}} \) and \( x = -\frac{1}{2\sqrt{3}} \).
Rather than providing a purely theoretical explanation, students are advised to compute the values in intervals:
- \( (-\infty, -\frac{1}{2\sqrt{3}}) \)
- \( (-\frac{1}{2\sqrt{3}}, \frac{1}{2\sqrt{3}}) \)
- \( (\frac{1}{2\sqrt{3}}, \infty) \)
* **Local Maxima and Minima:**
- By evaluating the second derivative
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