(2) Draw the graph of a function that is continuous on [0, 4] where f(0) = 1 and ƒ(4) = 5 and that does NOT satisfy the conclusion of the Mean Value Theorem on [0, 4].
(2) Draw the graph of a function that is continuous on [0, 4] where f(0) = 1 and ƒ(4) = 5 and that does NOT satisfy the conclusion of the Mean Value Theorem on [0, 4].
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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![### Task (2)
**Problem Statement:**
Draw the graph of a function that is continuous on \([0, 4]\) where \( f(0) = 1 \) and \( f(4) = 5 \) and that does NOT satisfy the conclusion of the Mean Value Theorem on \([0, 4]\).
**Graph Description:**
- The figure includes a graph with a standard Cartesian coordinate system.
- The x-axis and y-axis intersect at the origin \((0, 0)\), with unit grid lines in both directions.
- There are no additional annotations, lines, or curves on the graph currently.
### Explanation for Students:
**The Mean Value Theorem (MVT):**
The Mean Value Theorem states that for a function \( f \) that is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one point \( c \) in the interval \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
### Conditions for the Problem:
- The function \( f \) is continuous on \([0, 4]\).
- \( f(0) = 1 \) and \( f(4) = 5 \).
### Hint for Drawing the Graph:
To ensure that the function does not satisfy the conclusion of the Mean Value Theorem, you need to construct a function that does not have a derivative at any point where the slope of the secant line from \((0,1)\) to \((4,5)\) would be matched by the derivative of the function.
One way to do this can be creating a function where it is continuous but not differentiable on the open interval \((0, 4)\) by including sharp points or vertical segments within the interval.
### Approach:
- You may create a piecewise function with sharp turns (such as an absolute value function) that is continuous but not differentiable at certain points.
- Ensure that for any segments connecting the endpoints, no local point \(c\) has the consistent derivative required by the Mean Value Theorem.
### Example:
One classic example solution is a piecewise-linear function with a sharp turn:
\[ f(x) =
\begin{cases}
3x + 1 & \text{for }](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2F607e6c79-7b18-4dfa-8750-348749df65e0%2F6m0d28_processed.png&w=3840&q=75)
Transcribed Image Text:### Task (2)
**Problem Statement:**
Draw the graph of a function that is continuous on \([0, 4]\) where \( f(0) = 1 \) and \( f(4) = 5 \) and that does NOT satisfy the conclusion of the Mean Value Theorem on \([0, 4]\).
**Graph Description:**
- The figure includes a graph with a standard Cartesian coordinate system.
- The x-axis and y-axis intersect at the origin \((0, 0)\), with unit grid lines in both directions.
- There are no additional annotations, lines, or curves on the graph currently.
### Explanation for Students:
**The Mean Value Theorem (MVT):**
The Mean Value Theorem states that for a function \( f \) that is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one point \( c \) in the interval \((a, b)\) such that:
\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]
### Conditions for the Problem:
- The function \( f \) is continuous on \([0, 4]\).
- \( f(0) = 1 \) and \( f(4) = 5 \).
### Hint for Drawing the Graph:
To ensure that the function does not satisfy the conclusion of the Mean Value Theorem, you need to construct a function that does not have a derivative at any point where the slope of the secant line from \((0,1)\) to \((4,5)\) would be matched by the derivative of the function.
One way to do this can be creating a function where it is continuous but not differentiable on the open interval \((0, 4)\) by including sharp points or vertical segments within the interval.
### Approach:
- You may create a piecewise function with sharp turns (such as an absolute value function) that is continuous but not differentiable at certain points.
- Ensure that for any segments connecting the endpoints, no local point \(c\) has the consistent derivative required by the Mean Value Theorem.
### Example:
One classic example solution is a piecewise-linear function with a sharp turn:
\[ f(x) =
\begin{cases}
3x + 1 & \text{for }
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