Maxima/Minima Based Problems - Global Maxima or Minima

Understanding the concepts of global maxima and minima is crucial for solving optimization problems in calculus. These concepts help us determine the highest or lowest values of a function within a given domain.

Definitions

• Global Maximum: A function $f(x)$ has a global maximum at $x = c$ within a domain $D$ if $f(c) \geq f(x)$ for all $x$ in $D$.
• Global Minimum: A function $f(x)$ has a global minimum at $x = c$ within a domain $D$ if $f(c) \leq f(x)$ for all $x$ in $D$.

Critical Points and Endpoints

To find global maxima or minima, we need to consider critical points and endpoints of the domain.

• Critical Points: Points in the domain of $f(x)$ where $f'(x) = 0$ or $f'(x)$ does not exist.
• Endpoints: Points at the boundaries of the domain of $f(x)$.

Conditions for Global Maxima/Minima

1. First Derivative Test: If $f'(x)$ changes sign from positive to negative at $x = c$, then $f(c)$ is a local maximum. If it changes from negative to positive, then $f(c)$ is a local minimum. To be a global maximum or minimum, these points must also satisfy the global definitions above.
2. Second Derivative Test: If $f''(c) > 0$, then $f(c)$ is a local minimum. If $f''(c) < 0$, then $f(c)$ is a local maximum. Again, to be global, they must satisfy the global definitions.

Steps to Find Global Maxima/Minima

1. Find the domain of the function $f(x)$.
2. Calculate the derivative $f'(x)$.
3. Find the critical points by solving $f'(x) = 0$.
4. Evaluate the second derivative $f''(x)$ at the critical points (if applicable).
5. Evaluate $f(x)$ at the critical points and endpoints.
6. Compare the values to determine the global maxima and minima.

Table: Differences and Important Points

Feature	Global Maximum	Global Minimum
Definition	Highest value of $f(x)$ in domain $D$	Lowest value of $f(x)$ in domain $D$
First Derivative Test	$f'(x)$ changes from positive to negative	$f'(x)$ changes from negative to positive
Second Derivative Test	$f''(x) < 0$	$f''(x) > 0$
Critical Points	Necessary to check	Necessary to check
Endpoints	Must be evaluated	Must be evaluated

Examples

Example 1: Quadratic Function

Consider the function $f(x) = -2x^2 + 4x + 1$.

1. The domain of $f(x)$ is all real numbers.
2. The derivative $f'(x) = -4x + 4$.
3. Critical points are found by solving $f'(x) = 0 \Rightarrow x = 1$.
4. The second derivative $f''(x) = -4$, which is always negative, indicating a maximum.
5. Evaluate $f(x)$ at $x = 1$: $f(1) = -2(1)^2 + 4(1) + 1 = 3$.
6. Since the domain is all real numbers, there are no endpoints, and $f(1) = 3$ is the global maximum.

Example 2: Cubic Function with Restricted Domain

Consider $f(x) = x^3 - 3x^2 + 2$ with the domain $[0, 3]$.

1. The domain is $[0, 3]$.
2. The derivative $f'(x) = 3x^2 - 6x$.
3. Critical points: $f'(x) = 0 \Rightarrow x^2 - 2x = 0 \Rightarrow x = 0, 2$.
4. Second derivative $f''(x) = 6x - 6$.
   • At $x = 0$: $f''(0) = -6$, indicating a local maximum.
   • At $x = 2$: $f''(2) = 6$, indicating a local minimum.
5. Evaluate $f(x)$ at critical points and endpoints:
   • $f(0) = 2$ (endpoint)
   • $f(2) = -2$ (critical point)
   • $f(3) = 2$ (endpoint)
6. Comparing values, $f(2) = -2$ is the global minimum, and $f(0) = f(3) = 2$ are both global maxima within the domain.

By understanding these concepts and following the steps outlined, you can effectively solve problems involving global maxima and minima in calculus.