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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.
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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.
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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.
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.
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).
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.
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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)
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).
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.
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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)
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:
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).
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.
Learn essential terms and foundational ideas that form the basis of the topic.
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.
See how the concepts apply in real-world scenarios to understand their practical implications.
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.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
To find a min or max, take the first for a fax; when it's zero, check the sign to find success in kind!
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.
M&M for Maxima & Minima: When doing calculus, remember M&M (Maxima & Minima)! Always check your first derivative!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Critical Point
Definition:
A point where the first derivative of a function is either zero or undefined.
Term: Local Maximum
Definition:
A point where a function reaches a high value relative to its immediate surroundings.
Term: Local Minimum
Definition:
A point where a function reaches a low value relative to its immediate surroundings.
Term: First Derivative Test
Definition:
A method to determine the nature of critical points by checking the sign change of the first derivative.
Term: Second Derivative Test
Definition:
A method to classify critical points based on the concavity of the function using the second derivative.