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IB DP Maths AA SL Study Notes

5.3.1 Finding Local Extrema

Critical Points

A critical point occurs on a function when its derivative is either zero or undefined. These points are potential candidates for local extrema.

  • Definition: A point x = c is a critical point of a function f(x) if f'(c) = 0 or f'(c) does not exist.
  • Finding Critical Points: To find the critical points of a function f(x):
    • 1. Differentiate the function to get f'(x).
    • 2. Solve for x when f'(x) = 0 or f'(x) is undefined. You can learn more about the product and quotient rules used in differentiation here.

Example: Consider the function f(x) = x3 - 3x2.

To find its critical points:

1. Differentiate: f'(x) = 3x2 - 6x

2. Set f'(x) = 0: 3x2 - 6x = 0 Solving for x, we get x = 0 and x = 2. These are the critical points.

First Derivative Test

The first derivative test helps determine whether a critical point is a local maximum, local minimum, or neither. For complex problems, understanding the chain rule can be essential.

  • If f'(x) changes from positive to negative as x increases through the critical point, then f(x) has a local maximum at that point.
  • If f'(x) changes from negative to positive as x increases through the critical point, then f(x) has a local minimum at that point.
  • If f'(x) does not change sign, then f(x) has no local extrema at that point.

Example: Using the function from above, f'(x) is positive for x < 0, negative between 0 < x < 2, and positive for x > 2. Thus, x = 0 is a local minimum and x = 2 is a local maximum.

Second Derivative Test

The second derivative test provides another method to determine the nature of a critical point. Sometimes, knowing the equation of a tangent line is useful in these tests.

  • If f''(x) > 0 at a critical point, then f(x) has a local minimum there.
  • If f''(x) < 0 at a critical point, then f(x) has a local maximum there.
  • If f''(x) = 0 at a critical point, the test is inconclusive.

Example: For the function f(x) = x3 - 3x2, the second derivative is f''(x) = 6x - 6. At x = 0, f''(0) = -6 indicating a local maximum and at x = 2, f''(2) = 6 indicating a local minimum.

Application in Real-world Problems

Local extrema are crucial in various fields such as economics, physics, and engineering. For instance, businesses might use calculus to find the optimal price to charge for a product to maximise profit. Similarly, engineers might use it to determine the optimal dimensions for a container that minimises material costs. The process often involves advanced concepts like definite integration.

Example: A company finds that the profit P from selling x units of a product is given by P(x) = -x2 + 40x - 200. To find the number of units they should produce to maximise profit, we can find the critical points and use the first or second derivative test.

In this case, the critical point that maximises profit is x = 20 units.

Understanding the Behaviour of Functions

The concept of local extrema is pivotal in understanding the overall behaviour of functions. By identifying where a function might have peaks or troughs, we can gain insights into its general shape and trend. This is particularly useful when modelling real-world scenarios where understanding the behaviour of a function can lead to better decision-making. Sometimes, the mathematical rigor required can be supported by techniques such as proof by mathematical induction.

For instance, in environmental science, understanding the local extrema of a function modelling the population of a species can help in conservation efforts. If the population is reaching a local maximum and is expected to decline, interventions can be made to prevent that decline.

Practice Questions

1. Question: Find the local extrema of the function g(x) = x4 - 4x3.

Solution:

  • 1.1. Differentiate the function: g'(x) = 4x3 - 12x2.
  • 1.2. Set g'(x) = 0 to find the critical points.
  • 1.3. Use the second derivative test to determine the nature of each critical point.

2. Question: A company's revenue from selling x items is given by R(x) = -2x2 + 50x. Determine the number of items they should sell to maximise their revenue.

Solution:

  • 2.1. Differentiate the revenue function.
  • 2.2. Find the critical points.
  • 2.3. Use the first or second derivative test to determine the nature of the critical point.

FAQ

No, not all critical points are guaranteed to be local maxima or minima. While a critical point is a point where the derivative is zero or undefined, it doesn't necessarily mean it's a peak or trough. It could be a point of inflection, where the function changes concavity but doesn't have a maximum or minimum. This is why tests like the first and second derivative tests are essential. They help classify the nature of the critical point, determining whether it's a local maximum, local minimum, or neither.

The first derivative test looks at the sign change of the first derivative before and after the critical point. If the derivative changes from positive to negative, it indicates a local maximum, and if it changes from negative to positive, it indicates a local minimum. The second derivative test, on the other hand, evaluates the second derivative at the critical point. A positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum. If the second derivative is zero, the test is inconclusive.

Yes, a function can have multiple local maxima or minima. Depending on the nature of the function and its degree, it might have several peaks and troughs. Each of these points represents a local maximum or minimum. However, among these, there might be a single point that is the highest (global maximum) or the lowest (global minimum). It's also possible for functions to have no global maxima or minima but still have local ones.

Points of inflection are points where the function changes concavity, i.e., from concave up to concave down or vice versa. While they are not local maxima or minima, they are essential in understanding the overall behaviour of the function. In the context of local extrema, a point of inflection can exist between two local maxima or two local minima. It signifies a change in the curvature of the graph but not necessarily a peak or trough. Recognising points of inflection can provide insights into the function's shape and potential turning points.

Critical points are essential in real-world applications because they often represent optimal solutions or turning points in various scenarios. For instance, in business, a critical point of a profit function might indicate the optimal price to charge for a product to maximise profit. In engineering, it might represent the optimal dimensions of a structure to maximise strength while minimising material costs. By identifying and analysing critical points, professionals in various fields can make informed decisions that lead to optimal outcomes.

Practice Questions

Given the function f(x) = x^4 - 6x^3 + 8x^2, determine the critical points and use the second derivative test to classify them as local maxima, local minima, or neither.

The derivative of the function is f'(x) = 16x - 18x2 + 4x3. Setting this equal to zero, we find the critical points to be x = 0, x = (9 - square root of 17)/4, and x = (9 + square root of 17)/4.

The second derivative is f''(x) = 16 - 36x + 12x2. Evaluating this at each critical point:

  • For x = 0, f''(0) = 16 which is positive, indicating a local minimum.
  • For x = (9 - square root of 17)/4, f''(x) is negative, indicating a local maximum.
  • For x = (9 + square root of 17)/4, f''(x) is positive, indicating a local minimum.
A company's profit function is modelled by P(x) = x^4 - 6x^3 + 8x^2 where x is the number of items produced. Determine the number of items the company should produce to maximise its profit.

To maximise the profit, we need to find the critical points of the profit function and determine which of these is a local maximum. As calculated in the previous question, the critical points are x = 0, x = (9 - square root of 17)/4, and x = (9 + square root of 17)/4.

Using the second derivative test, we found that x = (9 - square root of 17)/4 is a local maximum. Therefore, to maximise profit, the company should produce approximately (9 - square root of 17)/4 items.


Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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