Finding Absolute Extrema for MATH 139
Exam Relevance for MATH 139
On closed intervals: evaluate f at critical points AND endpoints. Compare all values for absolute max/min.
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What Are Absolute Extrema?
The absolute maximum is the single highest value a function achieves on its entire domain.
The absolute minimum is the single lowest value a function achieves on its entire domain.
Unlike local extrema (peaks and valleys), absolute extrema are the overall highest and lowest points.
Important: Absolute extrema don't always exist! It depends on the function and the interval.
The Closed Interval Method
On a closed interval $[a, b]$, finding absolute extrema is straightforward:
Absolute extrema can only occur at:
- Critical points (where $f'(x) = 0$ or $f'(x)$ doesn't exist)
- Endpoints ($x = a$ and $x = b$)
The Method:
- Find all critical points in $(a, b)$
- Evaluate $f$ at each critical point
- Evaluate $f$ at both endpoints $a$ and $b$
- Compare all values — the largest is the absolute max, the smallest is the absolute min
Find the absolute maximum and minimum of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.
Step 1: Find critical points
$$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$$
Set $f'(x) = 0$: $$x = -1 \text{ or } x = 1$$
Both are in $[-2, 2]$ ✓
Step 2: Evaluate $f$ at critical points and endpoints
| $x$ | $f(x) = x^3 - 3x + 1$ |
|---|---|
| $-2$ | $(-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$ |
| $-1$ | $(-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$ |
| $1$ | $(1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1$ |
| $2$ | $(2)^3 - 3(2) + 1 = 8 - 6 + 1 = 3$ |
Step 3: Compare values
$$\boxed{\text{Absolute max} = 3 \text{ at } x = -1 \text{ and } x = 2}$$ $$\boxed{\text{Absolute min} = -1 \text{ at } x = -2 \text{ and } x = 1}$$
Find the absolute extrema of $f(x) = x^{2/3}$ on $[-1, 8]$.
Step 1: Find critical points
$$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}}$$
$f'(x) = 0$: Never (the numerator is always $2/3$)
$f'(x)$ undefined: At $x = 0$ (division by zero)
So $x = 0$ is a critical point.
Step 2: Evaluate at critical point and endpoints
| $x$ | $f(x) = x^{2/3}$ |
|---|---|
| $-1$ | $(-1)^{2/3} = 1$ |
| $0$ | $0^{2/3} = 0$ |
| $8$ | $8^{2/3} = (\sqrt[3]{8})^2 = 4$ |
Step 3: Compare
$$\boxed{\text{Absolute min} = 0 \text{ at } x = 0}$$ $$\boxed{\text{Absolute max} = 4 \text{ at } x = 8}$$
Open Intervals: When Extrema Don't Exist
On open intervals $(a, b)$ or unbounded domains like $(-\infty, \infty)$, absolute extrema might not exist.
Find the absolute extrema of $f(x) = \frac{1}{x}$ on $(0, 1)$.
Step 1: Analyze the behavior
As $x \to 0^+$: $f(x) = \frac{1}{x} \to +\infty$
As $x \to 1^-$: $f(x) = \frac{1}{x} \to 1$
Step 2: Look for critical points
$$f'(x) = -\frac{1}{x^2}$$
$f'(x) = 0$: Never (always negative)
No critical points in $(0, 1)$.
Step 3: Conclusion
- The function gets arbitrarily large as $x \to 0^+$, but never reaches a maximum value.
- The function approaches $1$ as $x \to 1^-$, but $x = 1$ is not in the interval, so it never achieves $1$.
$$\boxed{\text{No absolute maximum}}$$ $$\boxed{\text{No absolute minimum}}$$
Find the absolute extrema of $f(x) = x^2$ on $(-\infty, \infty)$.
Step 1: Find critical points
$$f'(x) = 2x = 0$$ $$x = 0$$
Step 2: Analyze behavior
At $x = 0$: $f(0) = 0$
As $x \to \pm\infty$: $f(x) = x^2 \to +\infty$
Step 3: Conclusion
- $f(0) = 0$ is the lowest value (check: $x^2 \geq 0$ for all $x$)
- The function grows without bound in both directions
$$\boxed{\text{Absolute min} = 0 \text{ at } x = 0}$$ $$\boxed{\text{No absolute maximum (function unbounded)}}$$
Find the absolute extrema of $f(x) = \begin{cases} x & 0 \leq x < 2 \\ 1 & x = 2 \end{cases}$ on $[0, 2]$.
Step 1: Analyze the function
For $0 \leq x < 2$: $f(x) = x$, which increases toward $2$
At $x = 2$: $f(2) = 1$
Step 2: Look for extrema
- Minimum: $f(0) = 0$ ✓
- Maximum: As $x \to 2^-$, $f(x) \to 2$, but $f(2) = 1$. The function never actually achieves the value $2$.
$$\boxed{\text{Absolute min} = 0 \text{ at } x = 0}$$ $$\boxed{\text{No absolute maximum (supremum is 2, but never achieved)}}$$
Key insight: Even on a closed interval, a discontinuous function might not have an absolute maximum or minimum!
Summary: When Do Absolute Extrema Exist?
| Situation | Absolute Extrema? |
|---|---|
| Continuous on closed interval $[a, b]$ | Guaranteed to exist |
| Open interval $(a, b)$ | May or may not exist |
| Unbounded domain $(-\infty, \infty)$ | May or may not exist |
| Discontinuous function | May or may not exist |
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to check endpoints
Wrong: "The only critical point is $x = 1$, so the absolute max is there."
Why it's wrong: On a closed interval, absolute extrema can occur at endpoints even if there's no critical point there.
Correct: Always evaluate $f$ at both endpoints in addition to critical points.
❌ Mistake: Assuming extrema always exist
Wrong: "Every function has an absolute max and min."
Why it's wrong: On open intervals or unbounded domains, the function might approach a value without reaching it, or grow without bound.
Correct: Check if the interval is closed and the function is continuous. If not, extrema might not exist.
❌ Mistake: Only checking where $f'(x) = 0$
Wrong: "$f'(x)$ is never zero, so there are no critical points."
Why it's wrong: Critical points also occur where $f'(x)$ doesn't exist.
Example: $f(x) = |x|$ has a critical point at $x = 0$ even though $f'(0)$ doesn't exist.
Closed Interval Method
On a closed interval [a, b], absolute extrema can only occur at: (1) critical points where f'(x) = 0, (2) critical points where f'(x) doesn't exist, or (3) the endpoints x = a and x = b. Evaluate f at all these candidates and compare.
Variables:
- $a$:
- Left endpoint of the interval
- $b$:
- Right endpoint of the interval
- $f'(x) = 0$:
- Where the derivative equals zero (horizontal tangent)
- $f'(x) \text{ DNE}$:
- Where the derivative doesn't exist (corners, cusps)
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