Find the points on the graph of y = (2x²+1)(x-5) 4 where the tangent line Slope is 1. 0=y= (2x²+1) (x-1) (product Find the points=2001 = (n on the graph where thereS tangent Slope is 0.00014002010 +00. = (25) ab sal= (all +30 dy dx = 4x(x-1) + (2x²+1) Cx 0= 4x² - 4x + 2x² + 1 0 = 6x² - 4x + 1 X = -(-4) ± √(-4) ²-4 (6)(1)_ _4±-√16-24 216) 12 X=4± √√-8 12 * Ans X= NO Point on graph
Find the points on the graph of y = (2x²+1)(x-5) 4 where the tangent line Slope is 1. 0=y= (2x²+1) (x-1) (product Find the points=2001 = (n on the graph where thereS tangent Slope is 0.00014002010 +00. = (25) ab sal= (all +30 dy dx = 4x(x-1) + (2x²+1) Cx 0= 4x² - 4x + 2x² + 1 0 = 6x² - 4x + 1 X = -(-4) ± √(-4) ²-4 (6)(1)_ _4±-√16-24 216) 12 X=4± √√-8 12 * Ans X= NO Point on graph
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...
Related questions
Question
PLEASE REVIEW THE EXAMPLE PROBLEM
AND DO THE QUES 1 ACCORDINGLY!
![### Applied Calculus Class: Problem Solving Example
#### Expert Q&A
**Applied Calculus Class: Please see the completed review example for reference and solve the given question. (show all work). Thanks! Solve ques. 1**
---
#### Problem
1. Find the points on the graph of
\[ y = (2x^4 + 1)(x - 5) \]
where the tangent line slope is 1.
---
#### Solution Steps:
1. **Question Identification**
Find the points on the graph where the tangent slope is 1.
2. **Function given**
\[ y = (2x^4 + 1)(x - 5) \]
3. **Apply Product Rule**
- To find the points where the tangent slope is 1, first determine the derivative of the function using the product rule.
\[ \frac{dy}{dx} = \frac{d}{dx} [ (2x^4 + 1)(x - 5) ] \]
- Apply the product rule \(\frac{d}{dx} [uv] = u'v + uv'\)
\[ u = (2x^4 + 1), \quad v = (x - 5) \]
- Derivatives:
\[ u' = \frac{d}{dx} (2x^4 + 1) = 8x^3 \]
\[ v' = \frac{d}{dx} (x - 5) = 1 \]
- Substitute into the product rule:
\[ \frac{dy}{dx} = (2x^4 + 1)(1) + (x - 5)(8x^3) \]
- Simplify:
\[ \frac{dy}{dx} = 2x^4 + 1 + 8x^4 - 40x^3 \]
\[ \frac{dy}{dx} = 10x^4 - 40x^3 + 1 \]
4. **Equate to 1**
\[ 1 = 10x^4 - 40x^3 + 1 \]
- Simplify to solve for \(x\):
\[ 0 = 10x^4 - 40x^3 \]
- Factor out:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b8d1efa-c07b-4629-8963-670ddc0d1c0b%2Fa2f9fa38-723f-464c-ade5-28dacd8420c5%2F8a7i7o9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Applied Calculus Class: Problem Solving Example
#### Expert Q&A
**Applied Calculus Class: Please see the completed review example for reference and solve the given question. (show all work). Thanks! Solve ques. 1**
---
#### Problem
1. Find the points on the graph of
\[ y = (2x^4 + 1)(x - 5) \]
where the tangent line slope is 1.
---
#### Solution Steps:
1. **Question Identification**
Find the points on the graph where the tangent slope is 1.
2. **Function given**
\[ y = (2x^4 + 1)(x - 5) \]
3. **Apply Product Rule**
- To find the points where the tangent slope is 1, first determine the derivative of the function using the product rule.
\[ \frac{dy}{dx} = \frac{d}{dx} [ (2x^4 + 1)(x - 5) ] \]
- Apply the product rule \(\frac{d}{dx} [uv] = u'v + uv'\)
\[ u = (2x^4 + 1), \quad v = (x - 5) \]
- Derivatives:
\[ u' = \frac{d}{dx} (2x^4 + 1) = 8x^3 \]
\[ v' = \frac{d}{dx} (x - 5) = 1 \]
- Substitute into the product rule:
\[ \frac{dy}{dx} = (2x^4 + 1)(1) + (x - 5)(8x^3) \]
- Simplify:
\[ \frac{dy}{dx} = 2x^4 + 1 + 8x^4 - 40x^3 \]
\[ \frac{dy}{dx} = 10x^4 - 40x^3 + 1 \]
4. **Equate to 1**
\[ 1 = 10x^4 - 40x^3 + 1 \]
- Simplify to solve for \(x\):
\[ 0 = 10x^4 - 40x^3 \]
- Factor out:
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning