Introduction: Finding the "Sweet Spot"

In the world of Mathematics: Applications and Interpretation, calculus isn't just about abstract formulas. It’s a powerful tool for finding the "best" possible outcome. Whether a company wants to maximize its profit, a designer wants to minimize the material used for a box, or an athlete wants to find the peak height of a ball, they are all looking for stationary points.

In this chapter, we will learn how to identify these "turning points" on a graph and use them to solve real-world optimisation problems. Don't worry if it sounds complex—it's essentially like finding the peaks and valleys on a roller coaster map!


1. Increasing and Decreasing Functions

Before we find the stationary points, we need to understand how a function behaves as it moves along the \(x\)-axis. We look at the gradient (the slope) to tell us what is happening.

  • Increasing Functions: A function is increasing if the graph is going "uphill" from left to right. In calculus terms, this happens when the derivative is positive: \(f'(x) > 0\).
  • Decreasing Functions: A function is decreasing if the graph is going "downhill." This happens when the derivative is negative: \(f'(x) < 0\).

Analogy: Imagine you are walking along the curve. If you are climbing up, the gradient is positive. If you are sliding down, the gradient is negative. If you are standing on perfectly flat ground at the very top or very bottom, the gradient is zero.


2. What is a Stationary Point?

A stationary point occurs at any point on a graph where the gradient is zero. This is where the function "stops" increasing or decreasing for a brief moment.

To find these points mathematically, we simply solve the equation:
\(f'(x) = 0\)

There are two main types of stationary points you will focus on:

  1. Local Maximum: The "peak" of a hill. The graph stops increasing and starts decreasing.
  2. Local Minimum: The "bottom" of a valley. The graph stops decreasing and starts increasing.

Note: These are called "local" because they are the highest or lowest points in a specific neighborhood, not necessarily for the entire function.


3. Step-by-Step: Finding Stationary Points

When you are asked to find the stationary points of a function, follow these steps:

Step 1: Differentiate. Find the derivative of the function, \(f'(x)\).

Step 2: Set to zero. Set your derivative equal to zero: \(f'(x) = 0\).

Step 3: Solve for \(x\). Solve the equation to find the \(x\)-coordinates where the gradient is zero.

Step 4: Find the \(y\)-coordinates. Plug your \(x\)-values back into the original function \(f(x)\) to find the corresponding \(y\)-values.

Common Mistake: Many students stop after finding \(x\). Remember, a point needs both an \(x\) and a \(y\)! Always plug \(x\) back into the original equation, not the derivative.


4. Determining the Nature of Stationary Points

Once you've found a stationary point, you need to know if it's a maximum or a minimum. There are two ways to do this:

Method A: The First Derivative Test (Sign Table)

Check the gradient slightly to the left and slightly to the right of your stationary point.

  • If the gradient goes from Positive \(\to\) Zero \(\to\) Negative, you have found a Local Maximum.
  • If the gradient goes from Negative \(\to\) Zero \(\to\) Positive, you have found a Local Minimum.

Method B: The Second Derivative Test (HL Only)

For Higher Level students, you can use the second derivative, written as \(f''(x)\), which measures how the gradient is changing.

  • If \(f''(x) > 0\) at the stationary point, the curve is "happy" (concave up), so it’s a Local Minimum.
  • If \(f''(x) < 0\) at the stationary point, the curve is "sad" (concave down), so it’s a Local Maximum.

Memory Aid: A positive second derivative is like a positive smile (\(\cup\)) \(\to\) Minimum. A negative second derivative is like a negative frown (\(\cap\)) \(\to\) Maximum.


5. Optimisation Problems in Context

Optimisation is just a fancy word for finding the "best" value. In exams, these are usually word problems. You might be asked to find the dimensions of a fence that give the maximum area, or the production level that results in the minimum cost.

How to approach Optimisation:
  1. Identify the variables: What are you trying to maximize or minimize? (e.g., Area \(A\), Volume \(V\), Cost \(C\)).
  2. Write an equation: Express that variable in terms of one other variable (e.g., \(A\) in terms of \(x\)). You might need to use a second piece of information from the question to substitute and simplify.
  3. Differentiate: Find the derivative (e.g., \(\frac{dA}{dx}\)).
  4. Solve: Set the derivative to zero and solve for the variable.
  5. Answer the question: Read carefully—do they want the dimensions (\(x\)) or the actual maximum value (\(A\))?

Did you know? Using your Graphic Display Calculator (GDC) is vital here. In many Paper 2 questions, you can graph the function and use the "G-Solve" or "Analyze Graph" feature to find the Maximum or Minimum directly!


Summary Checklist

Key Takeaways:

  • A stationary point happens where the gradient is zero (\(f'(x) = 0\)).
  • \(f'(x) > 0\) means the function is increasing; \(f'(x) < 0\) means it is decreasing.
  • Use the first or second derivative test to decide if a point is a maximum or minimum.
  • In optimisation, always translate the words into an equation first, then differentiate and set to zero.
  • Don't forget to use your GDC to verify your answers or find points of intersection and extrema!

Don't worry if this seems tricky at first! Optimisation is often the most challenging part of Topic 5. Practice setting up the equations from the word problems, and the calculus steps will become second nature.