Maxima and Minima: Fresh Take

Define and Identify the highest and lowest points of a function on a graph, both overall and within specific sections
Locate points on a function within a specific range where the slope is zero or undefined (critical points)
Explain how to use critical points to find the highest or lowest values of a function within a limited range

Extrema and Critical Points

The Main Idea

Absolute Extrema:
- Absolute Maximum:
  - $f(c) \ge f(x)$ for all $x$ in the interval
- Absolute Minimum:
  - $f(c) \le f(x)$ for all $x$ in the interval
- Can be positive, negative, or zero
Extreme Value Theorem:
- For a continuous function f on a closed interval [a,b]:
  - There exists a point where $f$ has an absolute maximum
  - There exists a point where $f$ has an absolute minimum
Conditions for Extrema:
- Continuity is crucial for the existence of extrema
- Closed, bounded intervals guarantee extrema for continuous functions
Types of Domains:
- Infinite intervals may lack extrema
- Open intervals may lack extrema even for continuous functions
- Discontinuities can prevent extrema from existing

Analyze the function $f(x) = x^3 - 3x$ on the interval $[-2, 2]$ for absolute extrema.

Check if the Extreme Value Theorem applies:
- $f(x)$ is a polynomial, so it’s continuous
- The interval $[-2, 2]$ is closed and bounded
- Therefore, absolute extrema must exist on this interval
Find critical points in [−2,2]:
- $f'(x) = 3x^2 - 3 = 3(x^2 - 1)$
- Set $f'(x) = 0$ : $x^2 = 1$ , so $x = \pm 1$
- Both critical points are in the interval
Evaluate f(x) at critical points and endpoints:
- $f(-2) = -8 + 6 = -2$
- $f(-1) = -1 + 3 = 2$
- $f(0) = 0$
- $f(1) = 1 - 3 = -2$
- $f(2) = 8 - 6 = 2$
Identify extrema:
- Absolute maximum: $2$ , occurs at $x = -1$ and $x = 2$
- Absolute minimum: $-2$ , occurs at $x = -2$ and $x = 1$

The Main Idea

Local Extrema:
- Local Maximum:
  - $f(c) \ge f(x)$ for all $x$ in some open interval containing $c$
- Local Minimum:
  - $f(c) \le f(x)$ for all $x$ in some open interval containing $c$
Critical Points:
- Interior point $c$ where $f'(c) = 0$ or $f'(c)$ is undefined
- Candidates for local extrema
Fermat’s Theorem:
- If $f$ has a local extremum at $c$ and $f$ is differentiable at $c$ , then $f'(c) = 0$
Key Relationships:
- Local extrema occur only at critical points
- Not all critical points yield local extrema
- Endpoint extrema are not considered local extrema
Types of Critical Points:
- Horizontal tangent line: $f'(c) = 0$
- Vertical tangent line or corner point: $f'(c)$ undefined

Find all critical points for $f(x)=x^3-\frac{1}{2}x^2-2x+1$ .

Calculate $f^{\prime}(x)$ .

$x=-\frac{2}{3}$ , $x=1$

Find the critical points of $f(x) = x^{2/3}(x-2)$ and determine if they correspond to local extrema.

Find $f'(x)$ using the product rule:

$f'(x) = \frac{2}{3}x^{-1/3}(x-2) + x^{2/3}$
$f'(x) = \frac{2}{3}x^{-1/3}(x-2) + x^{2/3} \cdot 1$

Simplify:

$f'(x) = \frac{2x^{2/3} - 4x^{-1/3} + 3x^{2/3}}{3x^{1/3}}$
$f'(x) = \frac{5x^{2/3} - 4x^{-1/3}}{3x^{1/3}}$

Find critical points:

a) Where $f'(x) = 0$ :

$5x^{2/3} - 4x^{-1/3} = 0$
$5x - 4 = 0$
$x = \frac{4}{5}$

b) Where $f'(x)$ is undefined:

$x = 0$ (due to division by $x^{1/3}$ )

Evaluate behavior at critical points:

At $x = 0$ : function has a vertical tangent line (cusp)
At $x = \frac{4}{5}$ : function has a horizontal tangent line

Conclusion:

$x = 0$ is a critical point but not a local extremum (inflection point)
$x = \frac{4}{5}$ is a critical point and a local minimum

The Main Idea

Extreme Value Theorem:
- A continuous function on a closed, bounded interval has both an absolute maximum and minimum
Location of Absolute Extrema:
- Occur at endpoints of the interval or at critical points within the interval
Critical Points:
- Points where $f'(x) = 0$ or $f'(x)$ is undefined
Procedure for Finding Absolute Extrema:
- Evaluate function at endpoints
- Find and evaluate function at critical points within the interval
- Compare all values to determine absolute maximum and minimum
Interior vs. Endpoint Extrema:
- Absolute extrema at interior points are also local extrema
- Endpoint extrema are not considered local extrema

Find the absolute maximum and absolute minimum of $f(x)=x^2-4x+3$ over the interval $[1,4]$ .

Look for critical points. Evaluate $f$ at all critical points and at the endpoints.

The absolute maximum is 3 and it occurs at $x=4$ . The absolute minimum is -1 and it occurs at $x=2$ .

Find the absolute extrema of $f(x) = x^3 - 6x^2 + 9x$ on the interval $[0,4]$ .

Evaluate at endpoints:

$f(0) = 0$
$f(4) = 64 - 96 + 36 = 4$

Find critical points:

$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$

Critical points:

$x = 1$ and $x = 3$ (both within $[0,4]$ )

Evaluate at critical points:

$f(1) = 1 - 6 + 9 = 4$
$f(3) = 27 - 54 + 27 = 0$

Compare all values:

$\begin{array}{|c|c|} \hline x & f(x) \\ \hline 0 & 0 \\ 1 & 4 \\ 3 & 0 \\ 4 & 4 \\ \hline \end{array}$

Conclusion:

Absolute maximum: $4$ , occurs at $x = 1$ and $x = 4$
Absolute minimum: $0$ , occurs at $x = 0$ and $x = 3$