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Understanding Global Maximums and Minimums

Expert reviewed • 21 July 2024 • 7 minute read

HSC Maths Advanced Syllabus

  • use the first derivative to investigate the shape of the graph of a function
    • determine the greatest or least value of a function over a given domain (if the domain is not given, the natural domain of the function is assumed) and distinguish between local and global minima and maxima
  • use any of the functions covered in the scope of this syllabus and their derivatives to solve practical and abstract problems AAM
    • evaluate solutions and their reasonableness given the constraints of the domain and formulate appropriate conclusions to optimisation problems

Note:

Video coming soon!

What is a Global Maximum and Minimum?

A global maximum is the highest point over the entire range of a given function or within a specific region. A global minimum is the lowest point over the entire range of a function or within a specified region.

Both of these can present themselves in many forms on a graph, ranging from an end point of a curve, to a turning point. Additionally, a graph does not always contain a global max and min. So how do we find them?

The simplest way to determine the global maximum and minimum of a curve is to sketch it and visually examine the graph. This can be done using the graphing steps, outlined in the previous chapter. Let’s look at some examples.

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How we display our answer is highly important. If a curve f(x)f(x) has a global maximum at point (2,36)(2,36), we can display our answer in point form, or say; a global maximum of 36 occurs when x=2x=2.

Practice Question 1

Determine the global maximum and minimum of the following graph f(x)=x3+3x2f(x)=x^3+3x^2, with the limit −2.5<x<0.5-2.5<x<0.5. The graph is provided below.

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Looking at the graph we can see that it has both a global maximum and minimum which are turning points. Thus, we need to determine the turning points of the graph.

Deriving the given function:

f(x)=x3+3x2f‘(x)=3x2+6xf(x)=x^3+3x^2\\f^`(x)=3x^2+6x

Now to find the turning points, we must equate the derived function with zero.

0=3x2+6x=x2+2x=x(x+2)x=0,x=−20=3x^2+6x\\=x^2+2x\\=x(x+2)\\x=0\quad , \quad x=-2

Substituting the x-values into f(x)f(x):

f(0)=(0)3+3(0)2=0f(−2)=(−2)3+3(−2)2=−8+12=4f(0)=(0)^3+3(0)^2\\=0\\f(-2)=(-2)^3+3(-2)^2\\=-8+12\\=4

∴\therefore The turning points of the graph are (0,0)(0,0) and (−2,4)(-2,4)

Thus, inspecting the graph, we can see that there is a global minimum at the origin (0,0)(0,0), with the global maximum of 4 occurring when x=2x=2.

Maximisation and Minimisation Problems

The concept of global maximum and minimums aren’t only used in a cartesian plane or in geometric problems. We frequently see questions in HSC exams, relating to maximisation and minimisation. The process to solve these questions is essentially the same as solving the questions provided above, however, they can be slightly more complicated. See the video for a worked HSC example!

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How to Sketch a Graph using the Derivative

Return to Module 2: Curve Sketching Using the Derivative