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**---#### Problem1. 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:

We an find the point if the given function where the slope of the tangent line is 1.

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...

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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:

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