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Edexcel GCSE Maths (Higher) Study Notes

2.10.4 Discriminating Maxima and Minima

In calculus, understanding how to discriminate between maxima and minima is crucial for analyzing the behavior of functions. This involves techniques such as examining the first and second derivatives of a function, as well as gradient inspection. Through step-by-step examples, we'll demonstrate these techniques with detailed calculations.

Techniques for Discriminating Maxima and Minima

First Derivative Test

The first derivative of a function, f(x)f'(x), indicates the gradient or slope at any point on the function. Setting f(x)=0f'(x) = 0 helps identify potential maxima or minima, where the function's slope changes.

Second Derivative Test

The second derivative, f(x)f''(x), helps determine the nature (maximum or minimum) of these points. If f''(x) > 0, the point is a minimum; if f''(x) < 0, it's a maximum.

Gradient Inspection

Inspecting the gradient around critical points (where f(x)=0f'(x) = 0) can also help determine their nature. A change from positive to negative indicates a maximum, while a change from negative to positive suggests a minimum.

Worked Examples

Example 1: Identifying Maxima and Minima

Consider the function f(x)=x36x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1.

Solution:

1. First Derivative: f(x)=3x212x+9f'(x) = 3x^2 - 12x + 9

2. Critical Points: Solving f(x)=0f'(x) = 0 yields x=1x = 1 and x=3x = 3.

3. Second Derivative: f(x)=6x12f''(x) = 6x - 12

  • At x=1x = 1, f(1)=6f''(1) = -6 (negative), indicating a maximum.
  • At x=3x = 3, f(3)=6f''(3) = 6 (positive), indicating a minimum.

Example 2: Using Gradient Inspection

Consider the function g(x)=x44x2g(x) = x^4 - 4x^2.

Solution:

  1. First Derivative: g(x)=4x38xg'(x) = 4x^3 - 8x
  2. Critical Points: Solving g(x)=0g'(x) = 0 yields x=0x = 0, x=2x = -\sqrt{2}, and x=2x = \sqrt{2}.
  3. Second Derivative: g(x)=12x28g''(x) = 12x^2 - 8
    • At x=0x = 0, g(0)=8g''(0) = -8 (negative), indicating a maximum.
    • At x=2x = -\sqrt{2} and (x=2),(g(2)=g(2)=16(x = \sqrt{2}), (g''(-\sqrt{2}) = g''(\sqrt{2}) = 16(positive), both indicating minima.

Practical Application

By applying these techniques, we can accurately identify and distinguish between the maxima and minima of a function. This approach is not only crucial for mathematical analysis but also has applications in various fields such as economics, engineering, and physics. Practice with different functions to gain proficiency in these methods.

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