Critical Points for MATH 139
Exam Relevance for MATH 139
Foundation for optimization. Find where f prime equals 0 or undefined. Check endpoints on closed intervals.
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What is a Critical Point?
A critical point of a function $f(x)$ is an $x$-value where:
- $f'(x) = 0$ (the derivative equals zero), OR
- $f'(x)$ does not exist (the derivative is undefined)
Why do we care? Critical points are the only places where a function can have a local maximum or minimum. They're the "candidates" for peaks and valleys.
🎯 This is a cornerstone concept. Finding critical points is one of the most important skills in calculus. Nearly every optimization problem, curve sketching question, and applied max/min problem relies on your ability to identify critical points quickly and accurately. Master this, and you'll be well-prepared for a large portion of your exams.
In this graph, notice how the tangent line is perfectly horizontal at each critical point — this is what $f'(x) = 0$ looks like visually.
The Two Types of Critical Points
Type 1: Where $f'(x) = 0$
At these points, the tangent line is horizontal (slope = 0).
Example: For $f(x) = x^2$, we have $f'(x) = 2x = 0$ when $x = 0$.
Type 2: Where $f'(x)$ Does Not Exist
This happens at:
- Corners/cusps (sharp points)
- Vertical tangents
- Discontinuities in the derivative
Example: For $f(x) = |x|$, the derivative doesn't exist at $x = 0$ (there's a sharp corner).
In this graph, notice the sharp "V" shape at the origin. The slope coming from the left is $-1$, and the slope coming from the right is $+1$. Since these don't match, the derivative doesn't exist at $x = 0$ — but it's still a critical point!
How to Find Critical Points
Step 1: Find $f'(x)$
Step 2: Set $f'(x) = 0$ and solve for $x$
Step 3: Find where $f'(x)$ is undefined (but $f(x)$ IS defined)
Step 4: List all values from Steps 2 and 3 — these are your critical points
Problem: Find all critical points of $f(x) = x^3 - 3x^2 - 9x + 5$
Step 1: Find the derivative: $$f'(x) = 3x^2 - 6x - 9$$
Step 2: Set $f'(x) = 0$ and solve: $$3x^2 - 6x - 9 = 0$$ $$x^2 - 2x - 3 = 0$$ $$(x-3)(x+1) = 0$$ $$x = 3 \quad \text{or} \quad x = -1$$
Step 3: Check where $f'(x)$ is undefined:
Since $f'(x) = 3x^2 - 6x - 9$ is a polynomial, it's defined everywhere. No additional critical points.
$$\boxed{x = -1 \text{ and } x = 3}$$
Problem: Find all critical points of $f(x) = \frac{x^2}{x-1}$
Step 1: Find the derivative using the quotient rule: $$f'(x) = \frac{2x(x-1) - x^2(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}$$
Step 2: Set $f'(x) = 0$:
A fraction equals zero when the numerator equals zero (and denominator ≠ 0): $$x^2 - 2x = 0$$ $$x(x-2) = 0$$ $$x = 0 \quad \text{or} \quad x = 2$$
Step 3: Check where $f'(x)$ is undefined:
$f'(x)$ is undefined when $(x-1)^2 = 0$, i.e., at $x = 1$.
But wait — is $f(1)$ defined? No! $f(1) = \frac{1}{0}$ is undefined.
Since $x = 1$ is not in the domain of $f$, it's not a critical point.
$$\boxed{x = 0 \text{ and } x = 2}$$
Problem: Find all critical points of $f(x) = x^{2/3}$
Step 1: Find the derivative: $$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3x^{1/3}}$$
Step 2: Set $f'(x) = 0$:
The numerator is $2$, which is never zero. So $f'(x) \neq 0$ for any $x$.
Step 3: Check where $f'(x)$ is undefined:
$f'(x) = \frac{2}{3x^{1/3}}$ is undefined when $x = 0$ (division by zero).
Is $f(0)$ defined? Yes! $f(0) = 0^{2/3} = 0$.
Since $x = 0$ is in the domain of $f$ but $f'(0)$ doesn't exist, $x = 0$ is a critical point.
$$\boxed{x = 0}$$
Problem: Find all critical points of $f(x) = \sin(x) + \cos(x)$ on $[0, 2\pi]$
Step 1: Find the derivative: $$f'(x) = \cos(x) - \sin(x)$$
Step 2: Set $f'(x) = 0$: $$\cos(x) - \sin(x) = 0$$ $$\cos(x) = \sin(x)$$ $$\tan(x) = 1$$ $$x = \frac{\pi}{4}, \frac{5\pi}{4}$$
Step 3: Check where $f'(x)$ is undefined:
$f'(x) = \cos(x) - \sin(x)$ is defined for all real $x$. No additional critical points.
$$\boxed{x = \frac{\pi}{4} \text{ and } x = \frac{5\pi}{4}}$$
Problem: Find all critical points of $f(x) = x^4 - 4x^3$
Step 1: Find the derivative: $$f'(x) = 4x^3 - 12x^2$$
Step 2: Set $f'(x) = 0$ and factor: $$4x^3 - 12x^2 = 0$$ $$4x^2(x - 3) = 0$$ $$x = 0 \quad \text{or} \quad x = 3$$
Step 3: $f'(x)$ is a polynomial — defined everywhere.
$$\boxed{x = 0 \text{ and } x = 3}$$
Important Distinction
⚠️ Critical points are $x$-values, not points!
When asked for critical points, give the $x$-coordinates: $x = -1, x = 3$
When asked for the critical point as an ordered pair, calculate $f(x)$ at each critical point:
- At $x = -1$: $f(-1) = ...$, so the point is $(-1, f(-1))$
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to check where $f'(x)$ is undefined
Wrong: Only solving $f'(x) = 0$ and missing critical points where the derivative doesn't exist.
Example: For $f(x) = |x|$, solving $f'(x) = 0$ gives no solutions, but $x = 0$ IS a critical point because $f'(0)$ doesn't exist.
Correct: Always check BOTH conditions: $f'(x) = 0$ AND where $f'(x)$ is undefined.
❌ Mistake: Including points not in the domain
Wrong: Saying $x = 1$ is a critical point of $f(x) = \frac{1}{x-1}$.
Why it's wrong: $f(1)$ is undefined, so $x = 1$ is not in the domain of $f$. A critical point must be in the domain.
Correct: Only include $x$-values where $f(x)$ is defined.
❌ Mistake: Confusing critical points with extrema
Wrong: "Every critical point is a maximum or minimum."
Why it's wrong: Some critical points are neither — they can be inflection points or saddle points.
Example: For $f(x) = x^3$, $x = 0$ is a critical point (since $f'(0) = 0$), but it's neither a max nor a min — it's an inflection point.
Correct: Critical points are candidates for extrema. Use the first or second derivative test to determine what type they are.
Critical Point Definition
A critical point is an x-value in the domain of f where the derivative is zero or undefined
Variables:
- $c$:
- x-value in domain of f
- $f'(c)$:
- derivative at c
Finding Critical Points
The three-step process to locate all critical points of a function
Variables:
- $f(x)$:
- original function
- $f'(x)$:
- derivative
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