Critical Points: How to Find Them Fast and Classify Them Correctly on the FE Exam

Every critical points problem hinges on one decision: did you check both conditions?

f'(x) = 0 catches the obvious peaks and valleys where the tangent line goes horizontal.

f'(x) doesn’t exist catches the corners, cusps, and fractional exponents that break at specific x-values.

Check both, and you find every critical point.

Miss the second condition, and you report an incomplete set.

The problem marks you wrong even though you took the derivative correctly and factored it perfectly.

That’s not a calculus failure. It’s a checklist failure.

On the FE, critical points show up in optimization problems, curve sketching, and any question asking where a function’s behavior changes from increasing to decreasing.

The setup looks clean—take the derivative, set it equal to zero, solve.

But between “set equal to zero” and “state your answer,” there’s a second gate most students forget to check: where does this derivative not exist?

That’s where points get lost.

Not in the differentiation. Not in the algebra.

In the systematic check of both conditions before calling the problem done.

This guide walks you through a repeatable process that handles any critical points problem the FE throws at you—from polynomial functions where factoring is straightforward to transcendental functions with exponentials or products. You’ll learn exactly how to identify critical points by checking both f'(x) = 0 and f'(x) DNE, how to factor derivatives efficiently without missing roots, and how to classify critical points as local minima or maxima using the second derivative test.

The goal is clean, systematic execution that finds every critical point and classifies them correctly every time.

Before we walk through it step by step, watch this short video. It shows you the full process from taking the derivative to factoring it completely to identifying all critical points and classifying them using the second derivative test.

You’ll see exactly where students typically get stuck on critical points problems—usually in the factoring step or in remembering that “doesn’t exist” also creates critical points—and how to avoid those traps completely.

What Is a Critical Point?

A critical point of a function f(x) is any x-value where the function could have a local maximum, local minimum, or an inflection point in its slope behavior.

Mathematically, x = c is a critical point if f(c) exists and one of two things happens: either the derivative f'(c) equals zero, or the derivative f'(c) doesn’t exist at that point.

The “equals zero” condition catches peaks and valleys where the tangent line is horizontal. The “doesn’t exist” condition catches sharp corners, cusps, or vertical tangents where the slope isn’t defined.

Think of critical points as the places where the function changes its mind about whether it’s increasing or decreasing. At these points, the function either pauses (horizontal tangent) or makes an abrupt turn (no derivative). Either way, these are the x-values you need to check if you’re optimizing something, sketching a curve, or analyzing function behavior.

On the FE Exam, critical points show up when you’re asked to find local extrema, optimize a cost or efficiency function, or analyze where a function’s rate of change transitions from positive to negative. The exam isn’t testing whether you understand what critical points are—it’s testing whether you can execute the derivative, factor or simplify it cleanly, solve for the critical points without missing any, and then classify critical points correctly using the second derivative test.

Critical points aren’t about memorizing definitions. They’re about recognizing that f'(x) = 0 and f'(x) = DNE are the only two gates you need to check, then handling the algebra and classification systematically.

Breaking Down Critical Points Step by Step

Finding critical points requires checking two gates, not one.

Most students solve f'(x) = 0, find the roots, and stop.

They forget that f'(x) doesn’t exist is the second condition—equally important but easier to overlook.

That’s where incomplete answers come from.

You factor a derivative perfectly, solve for three roots, report those three critical points, and miss the fourth point at x = 0 where the derivative has a fractional exponent and doesn’t exist.

The workflow below eliminates that.

It’s a four-step process that checks both gates systematically, factors derivatives completely before solving, and classifies critical points using the second derivative test if needed.

Once you internalize the structure, critical points problems become mechanical—no guessing about whether you’ve found them all, no uncertainty about which test to apply.

Let’s lay it out.

Step 1: Find the derivative f'(x)

The first thing you need to do is compute the derivative of the given function. This is non-negotiable—you can’t find critical points without it.

Use the power rule, product rule, quotient rule, or chain rule depending on what the function looks like. Write the derivative out completely and simplify it as much as you can. If you have a product of terms, factor out any common factors immediately—it’ll make Step 2 cleaner.

Step 2: Solve f'(x) = 0 to find where the derivative equals zero

Now that you have the derivative, set it equal to zero and solve for x. This is where most of the critical points will come from.

If the derivative is a polynomial, factor it completely. Pull out common factors first, then factor what’s left. If the derivative is a product of terms (especially if one involves an exponential), remember that exponentials are never zero, so ignore them and focus on the other factors.

Solve for all x-values that make f'(x) = 0. Each solution is a critical point.

Step 3: Check where f'(x) doesn’t exist

This step gets skipped more than any other, but it’s just as important as solving f'(x) = 0 when you’re finding critical points.

Look at your derivative. Are there any x-values where the derivative is undefined? This happens when you have a denominator that equals zero, or when you have a term like x^(1/2) or x^(2/3) that isn’t defined at x = 0, or when you have an absolute value or piecewise function that creates a sharp corner.

If you find any x-values where f'(x) doesn’t exist and f(x) does exist, those are also critical points.

Step 4: Classify the critical points using the second derivative test (if needed)

If the problem asks you to determine whether each critical point is a local minimum, local maximum, or neither, use the second derivative test.

Take the second derivative f”(x), then plug in each critical point. If f”(c) > 0, the function is concave up at that point, so x = c is a local minimum. If f”(c) < 0, the function is concave down, so x = c is a local maximum. If f”(c) = 0, the test is inconclusive.

Write down the classification next to each critical point.

Example Problem: Critical Points

The workflow handles any critical points problem cleanly once you trust the structure. Right now, we’re going to practice it on a real FE-style problem so you see exactly how the steps eliminate uncertainty and keep the algebra from spiraling.

The goal isn’t speed yet—it’s building confidence that the process works every time. Speed comes after a few reps.

With that laid out, let’s put these steps into practice.

This problem states:

Solution: Critical Points

When you see a polynomial function and you’re asked to find critical points, the question isn’t whether you can take the derivative—it’s whether you’ll factor it completely before solving, and whether you’ll remember to check where it doesn’t exist after finding where it equals zero.

Rush the factoring and you miss roots hiding inside a quadratic that factors further.

Forget the “doesn’t exist” check and you report an incomplete set even though your algebra was perfect.

This is exactly why we use a workflow.

It forces both gates—f'(x) = 0 and f'(x) DNE—so you don’t rely on memory or skip the second condition under pressure.

The workflow turns incomplete searches into systematic checks you can trust.

Let’s walk through it step by step.

Step 1: Find the derivative V'(t)

The first thing we need to do is take the derivative of V(t) = 2t³ – 15t² + 24t + 50.

Using the power rule on each term:

V'(t) = 6t² – 30t + 24

That’s the derivative. Before moving on, let’s factor out the common factor of 6:

V'(t) = 6(t² – 5t + 4)

Now we have a cleaner quadratic inside the parentheses, which will make finding the critical points easier.

Step 2: Solve V'(t) = 0 to find where the derivative equals zero

Next, we need to set the derivative equal to zero and solve for t to find the critical points.

6(t² – 5t + 4) = 0

Divide both sides by 6:

t² – 5t + 4 = 0

Now factor the quadratic:

(t – 1)(t – 4) = 0

So t = 1 and t = 4 are the two values where V'(t) = 0. These are critical points.

Step 3: Check where V'(t) doesn’t exist

Now we need to verify whether there are any critical points where the derivative doesn’t exist.

V'(t) = 6t² – 30t + 24 is a polynomial, and polynomials are defined everywhere. There are no fractions, no square roots, no absolute values. So there are no points where V'(t) doesn’t exist.

That means our critical points come entirely from Step 2, and we haven’t missed any by skipping the second gate.

Step 4: State the critical points

Bringing it together, the critical points are t = 1 and t = 4.

So the final answer to this problem is A) t = 1 and t = 4.

These are the times when the rate of volume change (the derivative) equals zero, meaning the tank’s volume is transitioning from increasing to decreasing or vice versa. At t = 1 and t = 4, the rate of change pauses before switching direction.

Common Mistakes to Avoid on Critical Points Problems

Critical points problems break at the exact moment you finish solving f'(x) = 0, find your roots, and move on without checking where f'(x) doesn’t exist.

You did the calculus correctly. You factored cleanly.

You solved for every x-value that makes the derivative zero.

But you forgot the second gate.

And now your answer is incomplete, even though 80% of your work was perfect.

Mistake 1: Forgetting to check where f'(x) doesn’t exist

Most critical points come from solving f'(x) = 0, so students solve that equation, find the roots, and stop. They forget that “f'(x) doesn’t exist” is the second condition for critical points.

On the FE, you’re given a function like f(x) = x^(2/3). You take the derivative f'(x) = (2/3)x^(-1/3).

You solve f'(x) = 0 and find no solutions, so you conclude there are no critical points.

But f'(x) doesn’t exist at x = 0 because you can’t divide by zero in the term x^(-1/3).

That means x = 0 is a critical point, and you missed it.

Your answer reports “no critical points” when there’s actually one. The problem marks you wrong even though your derivative and algebra were correct.

Mistake 2: Not factoring the derivative completely before solving

Students set f'(x) = 0 and try to solve it directly without factoring first. Or they factor out one term but leave a quadratic or higher-degree polynomial unfactored, missing additional critical points.

You take the derivative of f(x) = x⁴ – 4x² and get f'(x) = 4x³ – 8x.

You set it equal to zero and factor out 4x to get 4x(x² – 2) = 0.

You find x = 0 as a critical point, but you don’t factor x² – 2 further, so you miss x = ±√2.

You report one critical point when there are actually three. The answer choices include all of them, and you pick the one that only lists the incomplete set.

Mistake 3: Using f(c) instead of f”(c) in the second derivative test

Students confuse the notation and plug the critical point into the original function f(x) instead of into the second derivative f”(x). Or they plug the value f(c)—the y-coordinate—into the second derivative instead of the x-coordinate c.

You find a critical point at x = 2. The problem asks if it’s a local minimum or maximum.

You compute f(2) = 8 and think “positive, so it’s a minimum.”

But the second derivative test requires f”(2), not f(2). You classified the critical point wrong.

You report a local maximum as a local minimum, or vice versa. Or you say the test is inconclusive when it’s actually definitive.

Mistake 4: Ignoring exponential terms and losing factors

When the derivative includes an exponential multiplied by another factor, like f'(x) = e^x · (x – 3), students see the exponential and assume it complicates things. They either try to set e^x = 0 (which has no solution) or they forget to set the other factor equal to zero.

Your derivative is f'(x) = 5e^(2x)(1 – 2x²). You set the whole thing equal to zero, get confused because e^(2x) is never zero, and you either give up or report no critical points.

But you should’ve set 1 – 2x² = 0 and solved for x to find the critical points.

You miss critical points that exist because you couldn’t factor out the exponential properly.

Rules of Thumb for Critical Points Problems on the FE

You have the structure: take the derivative, check both conditions (f'(x) = 0 and f'(x) DNE), factor completely, classify critical points if needed. These rules protect that structure from the execution errors that cost points—forgetting to check where the derivative doesn’t exist, stopping factoring too early, or misapplying the second derivative test when classifying critical points.

• Always check both gates—f'(x) = 0 and f'(x) = DNE: Critical points come from two conditions, not one. Most students only solve the derivative equals zero and stop. But if your derivative has a denominator, a fractional exponent, or an absolute value, there might be points where it’s undefined and those are critical points too. Check both before you call the problem done.
• Factor the derivative completely before you solve: If you set f'(x) = 0 and try to solve without factoring, you’ll either miss critical points or waste time trying to apply the quadratic formula to something that factors cleanly. Pull out all common factors first. Then factor any remaining polynomials. Don’t stop until every factor is as simple as possible.
• Exponentials don’t create critical points from f'(x) = 0: If your derivative is a product involving e^(something), remember that exponentials are never zero. Ignore the exponential term and set the other factors equal to zero. That’s where your critical points are.
• Use the second derivative test only after you find all critical points: Don’t classify a critical point as a minimum or maximum until you’ve identified all of them. Find every critical point first, then take the second derivative and test each one. If you try to classify as you go, you might miss a point or misapply the test.
• Plug the x-value into f”(x), not into f(x): The second derivative test requires f”(c), where c is the critical point’s x-coordinate. Don’t plug c into the original function. Don’t plug the y-value into the second derivative. Just take f”(c) and check the sign to classify the critical point.

Final Thoughts | Critical Points

Every critical points problem hinges on whether you check both conditions.

Find where f'(x) = 0, and you catch the horizontal tangents—the peaks and valleys where slope pauses.

Check where f'(x) doesn’t exist, and you catch the corners, cusps, and fractional exponents that create critical points without crossing zero.

Miss the second check, and you report an incomplete answer.

The problem marks you wrong even though your derivative was correct, your factoring was clean, and your algebra was perfect.

The workflow removes that fragility.

It forces you to check both gates systematically—not because you remember to, but because the structure requires it.

Step 2 solves f'(x) = 0. Step 3 checks where f'(x) doesn’t exist.

You can’t skip one without breaking the process.

That’s what turns critical points problems from “did I get them all?” into confident execution.

You stop second-guessing whether x = 0 counts as a critical point.

You stop wondering if there’s a fourth root hiding in the factored derivative.

You just work the steps, check both gates, and move on.

Critical points problems are points you should bank every time. The workflow ensures you do.

Want more practice? Check out our complete FE problem library here.

The gap between knowing calculus and passing the FE isn’t more derivatives practice.

It’s knowing which problems actually show up, how they’re worded, and what traps they set.

You can spend hours working random calculus problems from a textbook, hoping they’re relevant.

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