First Derivative and Turning Points

Introduction

Understanding the behaviour of functions is vital in various disciplines, ranging from physics to finance. The first derivative plays a key role in providing insights into the rate at which functions change at specific points, aiding in optimisation and decision-making processes.

Example Analogy:

• First Derivative: Comparable to a car's speedometer, showing how speed changes as you accelerate or slow down.

Key Definitions

The Derivative as a Rate of Change

• First Derivative $f'(x)$: Indicates the slope of the tangent line and the rate of change at a specific point.
• Real-world Analogy: Comparable to a plane taking off, representing initial velocity.

Notation for Derivatives

Notation	Context
$\frac{dy}{dx}$	Physics, engineering
$f'(x)$	Mathematics

Increasing and Decreasing Functions

• Increasing: A function is increasing if $f(a) < f(b)$ for any $a < b$. In this case, $f'(x) > 0$.
• Decreasing: A function is decreasing if $f(a) > f(b)$ for any $a < b$. In this instance, $f'(x) < 0$.
• Problem Exercise: Determine intervals of increase and decrease for $f(x) = e^x$.

Solution: Since $f'(x) = e^x > 0$ for all values of $x$, the function $f(x) = e^x$ is always increasing.

Stationary Points

Stationary points occur where a function's derivative equals zero, indicating horizontal tangents.

• Local Maxima: Transition of $f'(x)$ from positive to negative.
• Local Minima: Transition of $f'(x)$ from negative to positive.
• Point of Inflection: Identified where $f''(x) = 0$, signalling a change in concavity.

Finding and Classifying Stationary Points

Step-by-Step Procedure:

• Step 1: Differentiate the function to obtain $f'(x)$.
• Step 2: Solve $f'(x) = 0$ to find potential stationary points.
• Step 3: Examine changes in $f'(x)$ around these points.

Example Problem: Find and classify stationary points of $f(x) = x^3 - 3x + 2$.

Solution:

• Step 1: Differentiate $f(x)$ to get $f'(x) = 3x^2 - 3$.
• Step 2: Solve $3x^2 - 3 = 0$ $3x^2 = 3$ $x^2 = 1$ $x = 1$ or $x = -1$
• Step 3: Classify using the first derivative test:
  • For $x < -1$: $f'(x) > 0$, function increasing
  • For $-1 < x < 1$: $f'(x) < 0$, function decreasing
  • For $x > 1$: $f'(x) > 0$, function increasing
  • Therefore, $x = -1$ is a local maximum and $x = 1$ is a local minimum

Points of Inflection

Points of inflection are critical for understanding how curves of functions change direction.

• Misconception: They are distinct from stationary points; focus on changes in concavity.

Real-world Applications of Maxima and Minima

• Economics: Identifying peaks in sales data.
• Engineering: Optimising material usage.

Worked Examples

Example: Physics Application

• Function: $s(t) = t^3 - 3t^2 + 2t$.
• Purpose: Describes an object's position over time.

Solution:

• Find the velocity function (first derivative): $s'(t) = 3t^2 - 6t + 2$
• To find critical points, solve $s'(t) = 0$: $3t^2 - 6t + 2 = 0$ Using the quadratic formula: $t = \frac{6 \pm \sqrt{36-24}}{6} = \frac{6 \pm \sqrt{12}}{6}$ $t ≈ 0.42$ or $t ≈ 1.58$
• Analyse intervals:
  • For $t < 0.42$: $s'(t) > 0$, position increasing
  • For $0.42 < t < 1.58$: $s'(t) < 0$, position decreasing
  • For $t > 1.58$: $s'(t) > 0$, position increasing
• Therefore, at $t ≈ 0.42$ there is a local maximum, and at $t ≈ 1.58$ there is a local minimum.

Example: Financial Application

• Function: $R(x) = \ln(x) - \frac{x}{5}$.
• Purpose: Models investment returns.

Solution:

• Find the derivative: $R'(x) = \frac{1}{x} - \frac{1}{5}$
• Solve for critical points: $\frac{1}{x} - \frac{1}{5} = 0$ $\frac{1}{x} = \frac{1}{5}$ $x = 5$
• Check the second derivative: $R''(x) = -\frac{1}{x^2} < 0$ for all $x > 0$
• Since $R''(5) < 0$, the point $x = 5$ is a local maximum
• This means maximum returns occur when $x = 5$