Examples
Enroll to start learning
You’ve not yet enrolled in this course. Please enroll for free to listen to audio lessons, classroom podcasts and take practice test.
Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Finding Local Minimum
🔒 Unlock Audio Lesson
Sign up and enroll to listen to this audio lesson
Today we will discuss how we find the local minimum of the function. Let's take the function f(x) = x^2 - 4x + 3. What do we do first?
We need to find the first derivative of the function, right?
Correct! So, what does the first derivative tell us?
It tells us where the slope is zero, which helps us find critical points.
Exactly! Let's compute that. The first derivative is f'(x) = 2x - 4. Now, what’s the next step?
We set f'(x) to zero to find critical points.
Right! Setting it to zero, we find x = 2. How do we classify this point?
Using the second derivative!
Absolutely! The second derivative, f''(x) = 2, is greater than zero, which indicates a local minimum. Therefore, we have a local minimum at (2, -1).
Can we summarize this whole process?
Sure! The steps are: Find the first derivative, set it to zero to find critical points, use the second derivative to classify them, and finally, substitute back to find the function values.
Finding Turning Points
🔒 Unlock Audio Lesson
Sign up and enroll to listen to this audio lesson
Now let's take another example to understand turning points. Consider the function f(x) = -x^3 + 3x^2 + 9. Who can tell me how we can start?
First, we need to find the first derivative.
Correct! The first derivative would be f'(x) = -3x^2 + 6x. What do we do next?
Set f'(x) to zero to find critical points!
Exactly! Setting -3x^2 + 6x = 0 gives us two critical points: x = 0 and x = 2. How do we classify these points?
We use the second derivative to check concavity!
Right again! The second derivative, f''(x) = -6x + 6, yields f''(0) = 6 and f''(2) = -6. So, what can we say about x = 0 and x = 2?
At x = 0, we have a local minimum and at x = 2, a local maximum!
Exactly! And the values calculated were (0, 9) as a local minimum and (2, 13) as a local maximum. Always remember, derivatives help us analyze the behavior of functions!
Can we go over how these concepts relate to optimization?
Of course! In real-life scenarios, understanding maxima and minima can help in optimizing resources effectively.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In this section, several detailed examples demonstrate the processes of using the first and second derivatives to identify local maxima and minima of functions. These examples are critical for understanding optimization in various contexts.
Detailed
Examples in Finding Maxima and Minima
In this section, we illustrate the techniques for finding local maxima and minima through illustrative examples using both the first and second derivative tests. These examples provide hands-on applications of theoretical concepts, solidifying the learners' understanding and ability to apply calculus to real-world problems.
Example 1
Finding Local Maximum or Minimum
Given the function:
\[ f(x) = x^2 - 4x + 3 \]
Steps to Solve:
1. Calculate the First Derivative:
\[ f'(x) = 2x - 4 \]
2. Set the First Derivative to Zero:
\[ 2x - 4 = 0 \]
\[ x = 2 \]
3. Second Derivative Test:
\[ f''(x) = 2 > 0 \]
Indicating that there is a local minimum at x = 2.
4. Find the Function Value:
\[ f(2) = (2)^2 - 4(2) + 3 = -1 \]
Thus, the local minimum occurs at the point (2, -1).
Example 2
Finding Turning Points
Given the function:
\[ f(x) = -x^3 + 3x^2 + 9 \]
Steps to Solve:
1. Calculate the First Derivative:
\[ f'(x) = -3x^2 + 6x \]
2. Set the First Derivative to Zero:
\[ -3x^2 + 6x = 0 \]
\[ x(−3x + 6) = 0 \]
Points at x = 0 and x = 2.
3. Second Derivative Test:
\[ f''(x) = −6x + 6 \]
- At x = 0:
\[ f''(0) = 6 > 0 \]
Local minimum at (0, 9).
- At x = 2:
\[ f''(2) = -6 < 0 \]
Local maximum at (2, 13).
4. Values Calculated Are:
Local minimum at (0, 9) and local maximum at (2, 13).
Understanding these examples emphasizes the significance of using derivatives to identify and classify critical points, thereby reinforcing the optimization strategies in calculus.
Audio Book
Dive deep into the subject with an immersive audiobook experience.
Example 1: Finding Local Extrema
Chapter 1 of 2
🔒 Unlock Audio Chapter
Sign up and enroll to access the full audio experience
Chapter Content
Example 1:
Find the local maximum or minimum of:
𝑓(𝑥) = 𝑥² − 4𝑥 + 3
Solution:
1. 𝑓′(𝑥) = 2𝑥 − 4
2. Set 𝑓′(𝑥) = 0:
2𝑥 − 4 = 0 ⇒ 𝑥 = 2
3. 𝑓″(𝑥) = 2 > 0 ⇒ Local minimum at 𝑥 = 2.
4. Find the value:
𝑓(2) = (2)² − 4(2) + 3 = 4 − 8 + 3 = −1
Answer: Local minimum at (2, −1)
Detailed Explanation
In this example, we want to find the local maximum or minimum of the function f(x) = x² − 4x + 3. We start by finding the first derivative, which allows us to determine critical points. The first derivative is f'(x) = 2x − 4. We set this equal to zero to find critical points. Solving for x gives us x = 2.
Next, we use the second derivative test to classify this critical point. The second derivative is f''(x) = 2, which is greater than zero, indicating that the graph is concave up at x = 2. This tells us there is a local minimum at this point.
Finally, we compute the value of the function at this point: f(2) = (2)² − 4(2) + 3 = −1. Thus, the local minimum occurs at the point (2, −1).
Examples & Analogies
Imagine you're looking for the lowest point in a valley. The function represents the shape of the valley. By finding where the slope of the valley becomes flat (the critical point), you're able to determine that the bottom of the valley is at (2, −1), which is the lowest point you could stand.
Example 2: Identifying Turning Points
Chapter 2 of 2
🔒 Unlock Audio Chapter
Sign up and enroll to access the full audio experience
Chapter Content
Example 2:
Find the turning points of:
𝑓(𝑥) = −𝑥³ + 3𝑥² + 9
Solution:
1. 𝑓′(𝑥) = −3𝑥² + 6𝑥
2. Set 𝑓′(𝑥) = 0:
−3𝑥² + 6𝑥 = 0 ⇒ 𝑥(−3𝑥 + 6) = 0 ⇒ 𝑥 = 0, 2
3. Use second derivative:
𝑓″(𝑥) = −6𝑥 + 6
o At 𝑥 = 0: 𝑓″(0) = 6 > 0 ⇒ Local minimum
o At 𝑥 = 2: 𝑓″(2) = −6 < 0 ⇒ Local maximum
4. Values:
o 𝑓(0) = −0 + 0 + 9 = 9
o 𝑓(2) = −8 + 12 + 9 = 13
Answer: Local minimum at (0, 9), local maximum at (2, 13)
Detailed Explanation
In this example, we analyze the function f(x) = −x³ + 3x² + 9 to find its turning points. We first calculate the first derivative, which is f'(x) = −3x² + 6x. Setting this equal to zero gives us the critical points: x = 0 and x = 2.
Next, we apply the second derivative test to determine whether each critical point is a maximum or minimum. The second derivative is f''(x) = −6x + 6. Plugging in our critical points:
- For x = 0, we get f''(0) = 6 (which is greater than zero), indicating a local minimum.
- For x = 2, we get f''(2) = −6 (less than zero), indicating a local maximum.
Finally, we find the function values at these points: f(0) = 9 (local minimum) and f(2) = 13 (local maximum). Thus, we conclude that there is a local minimum at (0, 9) and a local maximum at (2, 13).
Examples & Analogies
Think of this function as a roller coaster ride. The turning points are where the coaster goes from going up to down (like at the peak) and from down to up (like at the valley). At the peak (2, 13), you have your highest thrill before it starts descending again, and at the valley (0, 9), you have your lowest drop before it goes back up.
Key Concepts
-
Critical Points: Points where the first derivative is zero or undefined, indicative of potential maxima or minima.
-
Local Maximum: The highest point surrounding by lower points on the graph of the function.
-
Local Minimum: The lowest point surrounding by higher points on the graph.
-
First Derivative Test: Assesses whether a critical point is a maximum or minimum based on the sign change.
-
Second Derivative Test: Determines the concavity of the graph at the critical point to classify it as a maximum or minimum.
Examples & Applications
Example 1: Finding local maximum or minimum with f(x) = x^2 - 4x + 3.
Example 2: Finding turning points with f(x) = -x^3 + 3x^2 + 9.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
To find a min or max, take the first for a fax; when it's zero, check the sign to find success in kind!
Stories
Imagine hiking up and down a mountain. Every peak is a local maximum, and every valley is a local minimum. You check your elevation continuously, adjusting your path based on what you observe, like using derivatives for maxima and minima.
Memory Tools
M&M for Maxima & Minima: When doing calculus, remember M&M (Maxima & Minima)! Always check your first derivative!
Acronyms
D2C for Derivatives check to classify
Differentiate
Derive
Classify for maxima or minima!
Flash Cards
Glossary
- Critical Point
A point where the first derivative of a function is either zero or undefined.
- Local Maximum
A point where a function reaches a high value relative to its immediate surroundings.
- Local Minimum
A point where a function reaches a low value relative to its immediate surroundings.
- First Derivative Test
A method to determine the nature of critical points by checking the sign change of the first derivative.
- Second Derivative Test
A method to classify critical points based on the concavity of the function using the second derivative.
Reference links
Supplementary resources to enhance your learning experience.