**Content for Educational Website**Title: Graphing a Function and a Secant LineDescription: In this lesson, we will explore the graph of the function $ f(x) = x + \frac{5}{x} $ and the secant line that intersects the function at specific points. **Objective:**- To understand the behavior of the function $ f(x) = x + \frac{5}{x} $.- To learn how to graph a secant line passing through given points.**Instructions:**1. **Function to Graph:** - The function is given by: \[ f(x) = x + \frac{5}{x} \]2. **Secant Line:** - The secant line passes through the points $(1, 6)$ and $(10, 10.5)$.3. **Viewing Rectangle:** - The viewing rectangle for graphing these is defined by the intervals $[0, 12]$ along both the x-axis and y-axis.**Graph Explanation:**- Plot the function $ f(x) $ over the domain $[0, 12]$. Ensure the calculation and plotting consider both the linear component $ x $ and the rational component $ \frac{5}{x} $.- Plot the points $(1, 6)$ and $(10, 10.5)$ on the same graph.- Draw the secant line that connects these two points. This line provides a linear approximation over the interval between these points.**Further Exploration:**- Investigate how changing the points through which the secant line passes affects its slope and position.- Calculate the slope of the secant line using the formula: \[ \text{slope} = \frac{10.5 - 6}{10 - 1} \]**Conclusion:**By graphing the function and the secant line, students can visually interpret the function's behavior and the concept of secant lines in calculus. Understanding the intersection of different mathematical elements graphically enhances comprehension and problem-solving skills. **Transcription for Educational Website**---**Instruction:**Find the number $ c $ that satisfies the conclusion of the **Mean Value Theorem** for this function $ f $ and the interval $[1, 10]$.**Explanation:**This prompt involves applying the Mean Value Theorem, which is a fundamental concept in calculus. According to the theorem, if a function $ f $ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one number $ c $ in $(a, b)$ such that:\[f'(c) = \frac{f(b) - f(a)}{b - a}\]In this context, you need to find the value of $ c $ that satisfies this condition for the given interval from 1 to 10.

The given function is and the secant line that passes through the points (1, 6) and (10,…

Answered: Find the number c that satisfies the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

**Content for Educational Website**

Title: Graphing a Function and a Secant Line

Description: In this lesson, we will explore the graph of the function $ f(x) = x + \frac{5}{x} $ and the secant line that intersects the function at specific points.

**Objective:**
- To understand the behavior of the function $ f(x) = x + \frac{5}{x} $.
- To learn how to graph a secant line passing through given points.

**Instructions:**

1. **Function to Graph:**
- The function is given by:
\[
f(x) = x + \frac{5}{x}
\]

2. **Secant Line:**
- The secant line passes through the points $(1, 6)$ and $(10, 10.5)$.

3. **Viewing Rectangle:**
- The viewing rectangle for graphing these is defined by the intervals $[0, 12]$ along both the x-axis and y-axis.

**Graph Explanation:**

- Plot the function $ f(x) $ over the domain $[0, 12]$. Ensure the calculation and plotting consider both the linear component $ x $ and the rational component $ \frac{5}{x} $.
- Plot the points $(1, 6)$ and $(10, 10.5)$ on the same graph.
- Draw the secant line that connects these two points. This line provides a linear approximation over the interval between these points.

**Further Exploration:**

- Investigate how changing the points through which the secant line passes affects its slope and position.
- Calculate the slope of the secant line using the formula:
\[
\text{slope} = \frac{10.5 - 6}{10 - 1}
\]

**Conclusion:**
By graphing the function and the secant line, students can visually interpret the function's behavior and the concept of secant lines in calculus. Understanding the intersection of different mathematical elements graphically enhances comprehension and problem-solving skills.

**Transcription for Educational Website**

---

**Instruction:**

Find the number $ c $ that satisfies the conclusion of the **Mean Value Theorem** for this function $ f $ and the interval $[1, 10]$.

**Explanation:**

This prompt involves applying the Mean Value Theorem, which is a fundamental concept in calculus. According to the theorem, if a function $ f $ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one number $ c $ in $(a, b)$ such that:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

In this context, you need to find the value of $ c $ that satisfies this condition for the given interval from 1 to 10.

Expert Solution

Step 1: Finding the secant line:

The given function is $f open parentheses x close parentheses equals x plus 5 over x$ and the secant line that passes through the points (1, 6) and (10, 10.5).

Obtain the secant line passes through the points (1, 6) and (10, 10.5).

Substitute $open parentheses x subscript 1 comma y subscript 1 close parentheses equals open parentheses 1 comma 6 close parentheses space a n d space open parentheses x subscript 2 comma y subscript 2 close parentheses equals open parentheses 10 comma 10.5 close parentheses$ in $fraction numerator y minus y subscript 1 over denominator y subscript 2 minus y subscript 1 end fraction equals fraction numerator x minus x subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction$ .

$table row cell fraction numerator y minus 6 over denominator 10.5 minus 6 end fraction end cell equals cell fraction numerator x minus 1 over denominator 10 minus 1 end fraction end cell row cell fraction numerator y minus 6 over denominator 4.5 end fraction end cell equals cell fraction numerator x minus 1 over denominator 9 end fraction end cell row cell 9 y minus 54 end cell equals cell 4.5 x minus 4.5 end cell row y equals cell 0.5 x plus 5.5 end cell end table$