Does the First Derivative Test always work at every critical point?

No—the test requires f' to change sign at the critical point. If f' is positive on both sides (or negative on both sides) of a critical point, the test is inconclusive for that point, and there is no local extremum there. The test works, but it tells you that no extremum exists in this case.

Can I apply the First Derivative Test if f' is undefined at the critical point?

Yes, as long as f' exists in some open interval around c (except possibly at c itself). The test checks sign changes in that interval. For example, if f'(x) = \sqrt[3]{x}, then f' is undefined at x = 0, but you can still test the sign of f' just to the left and right of zero.

How is the First Derivative Test different from the Second Derivative Test?

The First Derivative Test checks whether f' changes sign by evaluating f' at two nearby points, while the Second Derivative Test evaluates f''(c) directly at the critical point—if f''(c) > 0 then c is a local minimum, if f''(c) < 0 then c is a local maximum. The FDT is more robust because it works even when f''(c) = 0, whereas the Second Derivative Test is inconclusive in that case.

What if the critical point is at the boundary of the domain?

You can still apply the First Derivative Test if the function is defined on an open or half-open interval; test the sign of f' on whichever side of the critical point lies within the domain. For example, if the domain is [0, 10] and a critical point is at x = 0, test f' for x > 0 to determine if there is a local extremum at the boundary.

First Derivative Test

Q: What if the critical point is at the boundary of the domain?

You can still apply the First Derivative Test if the function is defined on an open or half-open interval; test the sign of f' on whichever side of the critical point lies within the domain. For example, if the domain is [0, 10] and a critical point is at x = 0, test f' for x > 0 to determine if there is a local extremum at the boundary.

Classify critical points as local maxima or minima by checking whether the derivative changes sign before and after the point.

f' \text{ changes } + \to - \implies \text{local max at } c

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What each symbol means

What First Derivative Test takes

Symbol	Meaning
$f$	The function you are analyzing; you examine the sign of its derivative $f'$ on either side of a critical point to classify whether the function reaches a peak, a valley, or neither at that point.
$c$	The critical point where $f'(c) = 0$ or $f'$ is undefined; the test determines whether $f$ has a local maximum or minimum there by observing whether $f'$ changes sign at $c$ .

When to use it

Use the First Derivative Test when you have found critical points and need to determine whether each is a local maximum, local minimum, or neither.

Level

Usually taught in: Calculus I

Worked examples

1. Classify extrema of a cubic polynomial

Problem

Find the local extrema of

f(x) = x^3 - 3x

using the First Derivative Test.

$f'(x) = 3x^2 - 3$
Differentiate the function using the power rule on each term: the derivative of $x^3$ is $3x^2$ and the derivative of $-3x$ is $-3$ .
$3x^2 - 3 = 0$
Set $f'(x) = 0$ to find critical points where the derivative is zero.
$3x^2 = 3$
Add 3 to both sides.
$x^2 = 1$
Divide both sides by 3.
$x = \pm 1$
Take the square root of both sides, giving $x = -1$ or $x = 1$ .
$f'(-2) = 3(-2)^2 - 3 = 12 - 3 = 9 > 0$
Test the sign of $f'$ at $x = -2$ (to the left of $x = -1$ ); this shows $f$ is increasing there.
$f'(0) = 3(0)^2 - 3 = -3 < 0$
Test the sign of $f'$ at $x = 0$ (between $-1$ and $1$ ); this shows $f$ is decreasing there.
$f'(2) = 3(2)^2 - 3 = 12 - 3 = 9 > 0$
Test the sign of $f'$ at $x = 2$ (to the right of $x = 1$ ); this shows $f$ is increasing there.

Answer: $\text{Local maximum at } x = -1 \text{; local minimum at } x = 1$

At $x = -1$ , the derivative changes from $+$ to $-$ , indicating the function switches from increasing to decreasing—this is a local maximum. At $x = 1$ , the derivative changes from $-$ to $+$ , indicating the function switches from decreasing to increasing—this is a local minimum. This illustrates the core of the First Derivative Test.

2. Classify extrema with negative leading coefficient and fraction test points

Problem

Determine the local extrema of

h(x) = -2x^3 + 9x^2 - 12x + 2

using the First Derivative Test.

$h'(x) = -6x^2 + 18x - 12$
Differentiate using the power rule: the derivative of $-2x^3$ is $-6x^2$ , of $9x^2$ is $18x$ , and of $-12x$ is $-12$ .
$-6x^2 + 18x - 12 = 0$
Set $h'(x) = 0$ to find critical points.
$x^2 - 3x + 2 = 0$
Divide both sides by $-6$ to simplify the equation.
$(x - 1)(x - 2) = 0$
Factor the quadratic trinomial into two linear factors.
$x = 1 \text{ or } x = 2$
Apply the zero product property to identify the critical points.
$h'(0) = -6(0)^2 + 18(0) - 12 = -12 < 0$
Test the sign of $h'$ at $x = 0$ (to the left of $x = 1$ ); the derivative is negative.
$h'(1.5) = -6(2.25) + 18(1.5) - 12 = -13.5 + 27 - 12 = 1.5 > 0$
Test the sign of $h'$ at $x = 1.5$ (between $1$ and $2$ ); note that $(1.5)^2 = 2.25$ and the derivative is positive.
$h'(3) = -6(9) + 18(3) - 12 = -54 + 54 - 12 = -12 < 0$
Test the sign of $h'$ at $x = 3$ (to the right of $x = 2$ ); the derivative is negative.

Answer: $\text{Local minimum at } x = 1 \text{; local maximum at } x = 2$

Even though the leading coefficient is negative and we needed to handle fractional test points, the First Derivative Test applies exactly as before. At $x = 1$ , the derivative changes from $-$ to $+$ (local minimum). At $x = 2$ , the derivative changes from $+$ to $-$ (local maximum). This shows the test works universally regardless of polynomial form.

3. Optimize speed on a highway leg of a road trip

Problem

During a road trip with a highway leg and a city leg, a driver's speed (in mph) on the highway is modeled by

v(t) = -\frac{1}{2}t^2 + 5t

for

0 \leq t \leq 10

hours. Use the First Derivative Test to find when the speed reaches a maximum on this highway leg.

$v'(t) = -t + 5$
Differentiate the function using the power rule: the derivative of $-\frac{1}{2}t^2$ is $-t$ and the derivative of $5t$ is $5$ .
$-t + 5 = 0$
Set $v'(t) = 0$ to find when the rate of change of speed is zero, indicating a critical point.
$t = 5$
Solve for $t$ by subtracting 5 from both sides and multiplying by $-1$ .
$v'(2) = -2 + 5 = 3 > 0$
Test the sign of $v'$ at $t = 2$ (before $t = 5$ ); the positive value means speed is increasing.
$v'(7) = -7 + 5 = -2 < 0$
Test the sign of $v'$ at $t = 7$ (after $t = 5$ ); the negative value means speed is decreasing.

Answer: $t = 5 \text{ hours}$

At $t = 5$ , the derivative changes from $+$ to $-$ , so the speed reaches a local maximum at exactly 5 hours into the highway leg. Before this point, the driver accelerates; after this point, the driver decelerates. This application of the First Derivative Test helps the driver understand their driving pattern and when peak speed is achieved.

Common mistakes

Where First Derivative Test usually goes wrong

The mistake	Why it is wrong	The fix
Finding $f'(x) = 0$ and concluding ' $x = 2$ is a local maximum' without testing the sign of $f'$ on either side.	Finding where $f'(x) = 0$ locates critical points, but critical points can be local maxima, local minima, or neither—the First Derivative Test requires you to check the sign change of $f'$ to determine which.	After finding critical points, always test $f'$ at two nearby points: one slightly left and one slightly right of the critical point, then compare their signs to apply the test.
Testing $f'$ at only one point near the critical point, such as writing ' $f'(1.1) > 0$ , so $x = 1$ is a local minimum.'	The First Derivative Test requires you to know the sign of $f'$ on both sides of the critical point; testing only one side gives incomplete information and may lead to a wrong conclusion.	Always evaluate $f'$ at test points on both the left and right of the critical point, then compare the signs to see if there is a sign change.
Concluding 'there are no local extrema' when a critical point is found and $f'$ has the same sign on both sides.	This is actually correct reasoning—if the derivative does not change sign at a critical point, then the First Derivative Test shows there is no local extremum there. However, students sometimes think they have failed to apply the test rather than successfully determining that no extremum exists at that point.	Recognize that when $f'$ does not change sign at a critical point (for example, positive on both sides), the test is successful and correctly identifies that point as neither a local maximum nor a local minimum.

Tips and when to use something else

Always find all critical points first by solving $f'(x) = 0$ and identifying where $f'(x)$ is undefined; you can only apply the First Derivative Test where $f'$ is defined in an interval around the critical point.
Use the Second Derivative Test instead if $f''(c) \neq 0$ ; it is often faster because you only evaluate one derivative at the critical point, rather than testing an interval.
Remember that the First Derivative Test tells you about local extrema only, not global maxima or minima on a given interval; you must also check endpoints and behavior at infinity.
If $f'$ does not change sign at a critical point, that point is not a local extremum (it may be a saddle point or inflection point); the test has successfully classified it as neither a peak nor a valley.

Related formulas

Critical PointsCritical points mark where a function's derivative is zero or undefined, revealing local extrema and where the graph changes direction.Second Derivative TestThe Second Derivative Test determines whether a critical point is a local minimum or maximum using the concavity at that point.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.Concavity and Inflection PointsPinpoint where a function's curve changes from bending upward to downward, revealing the transition point between concave up and concave down.Higher Order DerivativesHigher order derivatives describe how the rate of change is changing; use them to find concavity and acceleration in functions.

Frequently asked questions

Does the First Derivative Test always work at every critical point?: No—the test requires $f'$ to change sign at the critical point. If $f'$ is positive on both sides (or negative on both sides) of a critical point, the test is inconclusive for that point, and there is no local extremum there. The test works, but it tells you that no extremum exists in this case.
Can I apply the First Derivative Test if $f'$ is undefined at the critical point?: Yes, as long as $f'$ exists in some open interval around $c$ (except possibly at $c$ itself). The test checks sign changes in that interval. For example, if $f'(x) = \sqrt[3]{x}$ , then $f'$ is undefined at $x = 0$ , but you can still test the sign of $f'$ just to the left and right of zero.
How is the First Derivative Test different from the Second Derivative Test?: The First Derivative Test checks whether $f'$ changes sign by evaluating $f'$ at two nearby points, while the Second Derivative Test evaluates $f''(c)$ directly at the critical point—if $f''(c) > 0$ then $c$ is a local minimum, if $f''(c) < 0$ then $c$ is a local maximum. The FDT is more robust because it works even when $f''(c) = 0$ , whereas the Second Derivative Test is inconclusive in that case.
What if the critical point is at the boundary of the domain?: You can still apply the First Derivative Test if the function is defined on an open or half-open interval; test the sign of $f'$ on whichever side of the critical point lies within the domain. For example, if the domain is $[0, 10]$ and a critical point is at $x = 0$ , test $f'$ for $x > 0$ to determine if there is a local extremum at the boundary.

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Reviewed 2026-09-18