Understanding Maxima/Minima Based Problems: Local Maxima or Minima

Maxima and minima are important concepts in calculus, particularly in the study of functions and their applications. They refer to the highest and lowest points in a particular region of a function's graph. When we talk about local maxima or minima, we are referring to points where the function reaches a peak or a trough, but only in a small neighborhood around that point.

Definitions

• Local Maximum: A function $f(x)$ has a local maximum at $x = a$ if $f(a) \geq f(x)$ for all $x$ in some open interval around $a$.
• Local Minimum: A function $f(x)$ has a local minimum at $x = a$ if $f(a) \leq f(x)$ for all $x$ in some open interval around $a$.

Conditions for Local Maxima or Minima

To determine whether a function has a local maximum or minimum at a point, we use the first and second derivative tests.

First Derivative Test

1. Find the derivative $f'(x)$ of the function $f(x)$.
2. Identify the critical points by solving $f'(x) = 0$.
3. Analyze the sign of $f'(x)$ around each critical point.

Condition	Conclusion
$f'(x)$ changes from positive to negative at $x = a$	Local maximum at $x = a$
$f'(x)$ changes from negative to positive at $x = a$	Local minimum at $x = a$
$f'(x)$ does not change sign at $x = a$	Neither a local maximum nor a local minimum

Second Derivative Test

1. Find the second derivative $f''(x)$ of the function $f(x)$.
2. Evaluate $f''(x)$ at each critical point.

Condition	Conclusion
$f''(a) > 0$	Local minimum at $x = a$
$f''(a) < 0$	Local maximum at $x = a$
$f''(a) = 0$	Test is inconclusive

Examples

Let's go through some examples to illustrate how to find local maxima and minima.

Example 1: Quadratic Function

Consider the function $f(x) = -x^2 + 4x - 3$.

1. Find the first derivative: $f'(x) = -2x + 4$.
2. Solve for critical points: $-2x + 4 = 0 \Rightarrow x = 2$.
3. Use the second derivative test: $f''(x) = -2$.

Since $f''(2) < 0$, there is a local maximum at $x = 2$.

Example 2: Cubic Function

Consider the function $f(x) = x^3 - 3x^2 - 9x + 5$.

1. Find the first derivative: $f'(x) = 3x^2 - 6x - 9$.
2. Solve for critical points: $3x^2 - 6x - 9 = 0$.
3. Factor and find the roots: $(3x + 3)(x - 3) = 0 \Rightarrow x = -1, x = 3$.
4. Use the first derivative test to determine the nature of each critical point.

For $x = -1$:

• $f'(-2) > 0$ and $f'(0) < 0$ (sign change from positive to negative), so there is a local maximum at $x = -1$.

For $x = 3$:

• $f'(2) < 0$ and $f'(4) > 0$ (sign change from negative to positive), so there is a local minimum at $x = 3$.

Important Points to Remember

• Critical points are where $f'(x) = 0$ or $f'(x)$ does not exist.
• The first derivative test requires analyzing the sign of $f'(x)$ around critical points.
• The second derivative test uses the value of $f''(x)$ at the critical points.
• If the second derivative test is inconclusive, revert to the first derivative test or other methods.
• Local maxima and minima are not necessarily the absolute highest or lowest points on the graph of the function.

Understanding and applying these concepts is crucial for solving maxima/minima based problems in calculus. Practice with various functions will help solidify these methods and prepare you for exams.