Find the equation of the tangent line to y = x + tan (x) where x = TI.

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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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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### Determining the Equation of the Tangent Line

**Problem Statement:**
Find the equation of the tangent line to the function \( y = x + \tan(x) \) at the point where \( x = \pi \).

**Solution Outline:**
To determine the equation of the tangent line to the given function at a specific point, we need to follow these steps:

1. **Find the value of \( y \) at \( x = \pi \).**
2. **Calculate the derivative of the function to determine the slope of the tangent line at \( x = \pi \).**
3. **Use the point-slope form of the line equation to write the equation of the tangent line.**

**Step-by-Step Solution:**

1. **Evaluate \( y \) at \( x = \pi \):**
   Given the function \( y = x + \tan(x) \):
   \[
   y(\pi) = \pi + \tan(\pi)
   \]
   Since \(\tan(\pi) = 0\), we have:
   \[
   y(\pi) = \pi + 0 = \pi
   \]
   Therefore, the point of tangency is \((\pi, \pi)\).

2. **Calculate the derivative of \( y \):**
   The derivative \( y' \) of the function \( y = x + \tan(x) \) is:
   \[
   \frac{dy}{dx} = 1 + \sec^2(x)
   \]
   At \( x = \pi \), we have:
   \[
   \frac{dy}{dx} \bigg|_{x = \pi} = 1 + \sec^2(\pi)
   \]
   Since \(\sec(\pi) = -1\), \(\sec^2(\pi) = 1\), thus:
   \[
   \frac{dy}{dx} \bigg|_{x = \pi} = 1 + 1 = 2
   \]
   Therefore, the slope of the tangent line at \( x = \pi \) is \( 2 \).

3. **Write the equation of the tangent line:**
   Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_
Transcribed Image Text:### Determining the Equation of the Tangent Line **Problem Statement:** Find the equation of the tangent line to the function \( y = x + \tan(x) \) at the point where \( x = \pi \). **Solution Outline:** To determine the equation of the tangent line to the given function at a specific point, we need to follow these steps: 1. **Find the value of \( y \) at \( x = \pi \).** 2. **Calculate the derivative of the function to determine the slope of the tangent line at \( x = \pi \).** 3. **Use the point-slope form of the line equation to write the equation of the tangent line.** **Step-by-Step Solution:** 1. **Evaluate \( y \) at \( x = \pi \):** Given the function \( y = x + \tan(x) \): \[ y(\pi) = \pi + \tan(\pi) \] Since \(\tan(\pi) = 0\), we have: \[ y(\pi) = \pi + 0 = \pi \] Therefore, the point of tangency is \((\pi, \pi)\). 2. **Calculate the derivative of \( y \):** The derivative \( y' \) of the function \( y = x + \tan(x) \) is: \[ \frac{dy}{dx} = 1 + \sec^2(x) \] At \( x = \pi \), we have: \[ \frac{dy}{dx} \bigg|_{x = \pi} = 1 + \sec^2(\pi) \] Since \(\sec(\pi) = -1\), \(\sec^2(\pi) = 1\), thus: \[ \frac{dy}{dx} \bigg|_{x = \pi} = 1 + 1 = 2 \] Therefore, the slope of the tangent line at \( x = \pi \) is \( 2 \). 3. **Write the equation of the tangent line:** Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_
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