Second Derivative and Turning Points

Defining the Second Derivative

• Second Derivative: The second derivative, represented by $f''(x)$ or $\frac{d^2y}{dx^2}$, is the derivative of the first derivative.
  • It describes the rate at which the slope of the graph changes, analogous to the concept of acceleration in physics.

Significance in Graph Curvature and Concavity

• Curvature Analysis:
  • Concave Up: If $f''(x) > 0$, the graph "smiles," indicating positive curvature.
  • Concave Down: If $f''(x) < 0$, the graph "frowns," indicating negative curvature.
• Example with Quadratic Function: Consider the function $y = x^2$:
  • The second derivative is positive ($f''(x) = 2$), showing the graph consistently smiles or remains concave up.

Introduction to Stationary Points

Importance: Stationary points are essential for analysing functions to identify changes in direction or constant regions. They are crucial in areas like physics, engineering, and economics, where they signify optimal points, such as peak demand or lowest energy states.

• Local Maximum: A point at which the function shifts from increasing to decreasing.
• Local Minimum: A point at which the function shifts from decreasing to increasing.
• Point of Inflection: A point where the curve changes concavity without reaching a true extremum.

Second Derivative Test

• Purpose and Process: The second derivative test aids in determining if stationary points are maxima, minima, or require further investigation when $f''(x) = 0$.
• Test Breakdown:
  • If $f''(x) > 0$: The point is a local minimum (concave up).
  • If $f''(x) < 0$: The point is a local maximum (concave down).
  • If $f''(x) = 0$: Additional analysis is required.

Common Student Errors

• Typical Mistakes: Students may incorrectly apply the test or omit further investigation when $f''(x) = 0$.

Points of Inflection

Definition and Significance

• Point of Inflection: Point on a graph where the curve changes concavity.
• Mathematical Context:
  • Signalled when the second derivative changes sign, transitioning from positive to negative or vice versa.

Example 1: Polynomial Function

Consider $f(x) = x^3$.

• Calculate $f''(x) = 6x$.
• At $f''(0) = 0$, verify a sign change occurs.
• $x = 0$ is an inflection point due to the sign change.

Worked Examples

Example 1: Polynomial Function

For $f(x) = x^3 - 3x^2 + 2$:

1. Compute $f'(x) = 3x^2 - 6x$.
2. Set $f'(x) = 0$ to find $x = 0$ or $x = 2$.
3. Calculate $f''(x) = 6x - 6$.
   • At $x = 0$: $f''(0) = -6$ (negative), so this is a local maximum.
   • At $x = 2$: $f''(2) = 6$ (positive), so this is a local minimum.

Example 2: Trigonometric Function

For $g(x) = \sin(x)$:

1. Calculate $g'(x) = \cos(x)$.
2. Solve $g'(x) = 0$ at $x = \frac{\pi}{2} + n\pi$ ($n \in \mathbb{Z}$).
3. Calculate $g''(x) = -\sin(x)$.
   • When $x = \frac{\pi}{2}$: $g''(\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1$ (negative), so this is a local maximum.
   • When $x = \frac{3\pi}{2}$: $g''(\frac{3\pi}{2}) = -\sin(\frac{3\pi}{2}) = 1$ (positive), so this is a local minimum.

Graph Sketching Using the Second Derivative

Step-by-Step Guide to Graph Sketching

1. Identify Stationary Points

• Stationary Points: Occur where $f'(x) = 0$.

2. Determine Concavity

• Concave Up: $f''(x) > 0$.
• Concave Down: $f''(x) < 0$.
• Assess the behaviour as $x$ approaches infinity ($x \to \infty$) or negative infinity ($x \to -\infty$).

Practical Application Exercises

• Problem 1: Classify the stationary points of $h(x) = x^4 - 4x^2$.

  Solution:

  1. Find $h'(x) = 4x^3 - 8x$
  2. Set $h'(x) = 0$: $4x^3 - 8x = 0$
  3. Factor: $4x(x^2 - 2) = 0$
  4. Stationary points occur at $x = 0$, $x = \sqrt{2}$, and $x = -\sqrt{2}$
  5. Calculate $h''(x) = 12x^2 - 8$
  6. At $x = 0$: $h''(0) = -8$ (negative), so this is a local maximum
  7. At $x = \sqrt{2}$: $h''(\sqrt{2}) = 12(\sqrt{2})^2 - 8 = 24 - 8 = 16$ (positive), so this is a local minimum
  8. At $x = -\sqrt{2}$: $h''(-\sqrt{2}) = 16$ (positive), so this is a local minimum

• Problem 2: Evaluate stationary points for $p(x) = \tan(x)$ within $[0, 2\pi]$.

  Solution:

  1. Find $p'(x) = \sec^2(x)$
  2. Since $\sec^2(x) > 0$ for all values in the domain, $p'(x)$ is never zero
  3. Therefore, $p(x) = \tan(x)$ has no stationary points within $[0, 2\pi]$
  4. Note that $\tan(x)$ has vertical asymptotes at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ within this interval