Applications of Differentiation: Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

Applications of Differentiation

Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

This section explores how the second derivative of a function can be used to determine concavity, locate relative extrema (maximum and minimum points), and sketch the graph of a function. These concepts are fundamental in Business Calculus for analyzing and optimizing functions that model real-world scenarios.

Concavity of Functions

• Definition: The concavity of a function describes the direction in which the graph bends. If a function f is twice differentiable on an interval I:

  • Concave Up: f is concave up on I if its graph bends upwards, resembling a cup ($f''(x) > 0$).

  • Concave Down: f is concave down on I if its graph bends downwards, resembling a cap ($f''(x) < 0$).

• Visual Representation: Concave up graphs have slopes that increase, while concave down graphs have slopes that decrease.

Theorem 4: A Test for Concavity

• If $f''(x) > 0$ on an interval I, the graph of f is concave up on I.

• If $f''(x) < 0$ on an interval I, the graph of f is concave down on I.

• Application: This test helps identify regions where a function is increasing at an increasing rate (concave up) or increasing at a decreasing rate (concave down).

Theorem 5: The Second Derivative Test for Relative Extrema

• Suppose f is differentiable for every x in an open interval $(a, b)$ and there is a critical value $c$ in $(a, b)$ where $f'(c) = 0$.

• Relative Minimum: $f(c)$ is a relative minimum if $f''(c) > 0$.

• Relative Maximum: $f(c)$ is a relative maximum if $f''(c) < 0$.

• If $f''(c) = 0$, use the First Derivative Test to determine whether $f(c)$ is a relative extremum.

Example 1: Finding Relative Extrema

Given the function $f(x) = x^3 + 3x^2 - 9x - 13$, find its relative extrema.

• Step 1: Find the first and second derivatives.

  • $f'(x) = 3x^2 + 6x - 9$

  • $f''(x) = 6x + 6$

• Step 2: Find critical values by solving $f'(x) = 0$.

  • $3x^2 + 6x - 9 = 0$

  • $x^2 + 2x - 3 = 0$

  • $(x + 3)(x - 1) = 0$

  • Critical values: $x = -3$, $x = 1$

• Step 3: Apply the Second Derivative Test.

  • $f''(-3) = 6(-3) + 6 = -12$ (less than 0): Relative maximum at $x = -3$

  • $f''(1) = 6(1) + 6 = 12$ (greater than 0): Relative minimum at $x = 1$

• Step 4: Find the function values at the critical points.

  • $f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) - 13 = 14$

  • $f(1) = (1)^3 + 3(1)^2 - 9(1) - 13 = -18$

  • Conclusion: Relative maximum at $(-3, 14)$ and relative minimum at $(1, -18)$

Summary Table: Second Derivative Test

Condition	Conclusion
$f''(c) > 0$	Relative minimum at $x = c$
$f''(c) < 0$	Relative maximum at $x = c$
$f''(c) = 0$	Use First Derivative Test

Key Terms

• Critical Value: A point $x = c$ where $f'(c) = 0$ or $f'(c)$ does not exist.

• Relative Maximum: The highest point in a local region of the graph.

• Relative Minimum: The lowest point in a local region of the graph.

• Concavity: The direction in which the graph bends (upward or downward).

Example Application

• Business Context: In economics and business, finding maximum and minimum values helps identify optimal production levels, profit maximization, and cost minimization.

• Graphical Analysis: Sketching the graph using critical points and concavity provides insight into the behavior of functions modeling business scenarios.

Additional info: These notes are based on textbook slides for Chapter 3 of "Calculus and Its Applications," focusing on the use of second derivatives in curve sketching and optimization, which are core topics in Business Calculus.