What Is The First Derivative Test?
The First Derivative Test is a method for deciding whether a critical point of a function is a local maximum, a local minimum, or neither, by examining the sign of the first derivative $f'$ on either side of that point. A critical point is a value $c$ in the domain where $f'(c) = 0$ or where $f'(c)$ does not exist. These are the only places a smooth function can turn around.
The rule itself is short. Suppose $f$ is continuous at a critical point $c$ and differentiable on an open interval around $c$ (except possibly at $c$ itself). Move from just left of $c$ to just right of $c$ and watch the sign of $f'$:
$$ \begin{aligned} f' : +\ \to\ - &\quad\Rightarrow\quad \text{local maximum at } c\ f' : -\ \to\ + &\quad\Rightarrow\quad \text{local minimum at } c\ f' \text{ no sign change} &\quad\Rightarrow\quad \text{neither (not an extremum)} \end{aligned} $$
Geometrically, $f'$ is the slope of the tangent line. Where $f'$ is positive the graph is rising; where $f'$ is negative it is falling. A local maximum is a point where the graph stops rising and begins to fall, so the slope goes from positive to negative, and a local minimum is the reverse.
If the slope keeps the same sign through $c$, the curve never turns, and $c$ is not a peak or a valley. For the full family of results this belongs to, see applications of derivatives.
How Do You Build A Sign Chart Of $f'$?
A sign chart is a number line marked with every critical point, split into open intervals, with the sign of $f'$ recorded on each interval. It turns the First Derivative Test into a routine you can follow without guessing.
Differentiate. Compute $f'(x)$.
Find every critical point. Solve $f'(x) = 0$, and separately list the points where $f'(x)$ is undefined but $f$ itself is defined. Both kinds are critical points.
Split the number line. Mark the critical points; they divide the domain into open intervals.
Test one point inside each interval. Pick a convenient value strictly between neighbouring critical points, and record whether $f'$ there is positive or negative. Choose a test point close enough that it does not jump past another critical point.
Read the changes. At each critical point apply the sign rule: $+\to-$ gives a maximum, $-\to+$ gives a minimum, no change gives neither.
The same chart also hands you the increasing and decreasing functions intervals for free: $f$ increases wherever $f'>0$ and decreases wherever $f'<0$.
How Do You Use The First Derivative Test? Worked Examples
Each example builds the full sign chart, states the behaviour, and reads off the increasing and decreasing intervals.
Example 1: A cubic with two turning points.
Classify the critical points of $f(x) = x^3 - 3x$.
Differentiate and factor:
$$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$
Setting $f'(x) = 0$ gives critical points $x = -1$ and $x = 1$; $f'$ is defined everywhere, so there are no undefined-derivative points. These split the line into three intervals. Test one point in each:
$$ \begin{aligned} x = -2:\ &f'(-2) = 3(-3)(-1) = 9 > 0 \quad(\text{rising})\ x = 0:\ &f'(0) = 3(-1)(1) = -3 < 0 \quad(\text{falling})\ x = 2:\ &f'(2) = 3(1)(3) = 9 > 0 \quad(\text{rising}) \end{aligned} $$
At $x = -1$ the sign goes $+\to-$, a local maximum, with $f(-1) = -1 + 3 = 2$. At $x = 1$ the sign goes $-\to+$, a local minimum, with $f(1) = 1 - 3 = -2$. The function increases on $(-\infty, -1)$ and $(1, \infty)$ and decreases on $(-1, 1)$.
Final answer: local maximum at $(-1, 2)$, local minimum at $(1, -2)$.
Example 2: A stationary point that is not an extremum.
Classify the critical point of $f(x) = x^3$.
$$f'(x) = 3x^2$$
Setting $f'(x) = 0$ gives the single critical point $x = 0$. Test either side:
$$ \begin{aligned} x = -1:\ &f'(-1) = 3(1) = 3 > 0\ x = 1:\ &f'(1) = 3(1) = 3 > 0 \end{aligned} $$
The slope is positive on both sides, so $f'$ does not change sign at $x = 0$. By the test, $x = 0$ is neither a maximum nor a minimum. The graph flattens for an instant and keeps rising, a horizontal inflection point.
This is exactly the case that catches students who assume $f'(c) = 0$ must mean a peak or a valley. The point is a stationary point, but not every stationary point is an extremum.
Final answer: $x = 0$ is neither; it is a horizontal point of inflection.
Example 3: A critical point where $f'$ is undefined.
Classify the critical point of $f(x) = x^{2/3}$.
$$f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3,x^{1/3}}$$
Here $f'(x)$ is never zero, but it is undefined at $x = 0$, while $f(0) = 0$ is perfectly defined. So $x = 0$ is a critical point of the undefined-derivative kind, the type that vanishes if you only solve $f'(x) = 0$. Test each side:
$$ \begin{aligned} x = -1:\ &f'(-1) = \frac{2}{3(-1)} = -\tfrac{2}{3} < 0 \quad(\text{falling})\ x = 1:\ &f'(1) = \frac{2}{3(1)} = \tfrac{2}{3} > 0 \quad(\text{rising}) \end{aligned} $$
The sign goes $-\to+$, so $x = 0$ is a local minimum, with $f(0) = 0$. The graph comes down to a sharp point and turns back up: a cusp. The tangent line is vertical there, which is why $f'$ fails to exist, yet the First Derivative Test still classifies the point cleanly because it only needs the sign of $f'$ nearby, not its value at $c$.
Final answer: local minimum at $(0, 0)$, a cusp.
What Does The Sign Chart Tell You At A Glance?
Once the chart is built, four sign patterns cover every case. This table is the whole test in one place.
Table: How the sign of $f'$ on each side of a critical point $c$ classifies $c$.
Sign of $f'$ left of $c$ | Sign of $f'$ right of $c$ | Behaviour at $c$ |
|---|---|---|
$+$ | $-$ | Local maximum |
$-$ | $+$ | Local minimum |
$+$ | $+$ | Neither (rising through; horizontal or vertical inflection) |
$-$ | $-$ | Neither (falling through) |
The two "Neither" rows are the ones competitors gloss over. A critical point is a candidate for an extremum, not a guarantee. Only a genuine sign change promotes it to a maximum or minimum. This is why the method pairs naturally with critical points and maxima and minima as a single workflow.
Why Choose The First Derivative Test Over The Second Derivative Test?
Both tests classify critical points, so which one earns the effort. The honest answer is that the First Derivative Test is the more reliable of the two.
It is always conclusive. The second derivative test checks the sign of $f''(c)$: positive means a minimum, negative means a maximum. But when $f''(c) = 0$ the second-derivative test says nothing, and you are sent back to the first derivative anyway. The sign of $f'$ around $c$ always exists to be checked, so the First Derivative Test never returns "inconclusive."
It needs only one derivative. For functions where the second derivative is messy to compute, or where $f'$ is undefined at the critical point (as in the cusp of Example 3), the second-derivative test cannot even start. The first-derivative approach still works, because it never evaluates a derivative at $c$, only the sign near $c$.
It gives the increasing and decreasing intervals as a bonus. The same sign chart that classifies the extrema also describes the shape of the whole graph, which is why it feeds directly into curve sketching.
The trade is a little more work: you test points on both sides rather than plug one number into $f''$. For a quick classification of a friendly polynomial, the second-derivative test can be faster. For anything with a flat second derivative, a cusp, or a corner, the First Derivative Test is the one that finishes.
Who Invented The First Derivative Test?
Finding maxima and minima by watching where a curve turns is older than calculus itself. The idea was worked out before Newton and Leibniz gave the derivative its modern form.
Two figures gave the method its modern language:
Pierre de Fermat (1601–1665, France) introduced the adequality method for locating maxima and minima, the direct ancestor of "set the derivative to zero."
Joseph-Louis Lagrange (1736–1813, born in Turin, worked in France) introduced the prime notation $f'(x)$ still used for the derivative today, which made sign charts and the First Derivative Test easy to write down.
Where Is The First Derivative Test Used In The Real World?
Any question that asks for a "largest" or "smallest" value under changing conditions is a maximum or minimum problem, and the First Derivative Test is how you confirm which one you have found.
Engineering and design: finding the height that maximises the range of a projectile, or the dimensions that minimise the material in a can, both reduce to locating a turning point and confirming it with the sign of $f'$.
Economics: a firm's profit as a function of output rises then falls; the profit-maximising quantity is the critical point where marginal profit changes from positive to negative, the economic reading of a $+\to-$ sign change. This underpins many optimization problems.
Physics and motion: the highest point of a thrown object is where its vertical velocity, the derivative of height, switches from positive to negative.
Biology and medicine: peak drug concentration in the bloodstream is a local maximum of a concentration-versus-time curve, identified the same way.
Data and machine learning: minimising an error function means hunting for a point where its slope changes from negative to positive, the same $-\to+$ pattern the test names.
One rule about the sign of a slope quietly decides the "best" setting across engineering, economics, and science.
What Are The Most Common Mistakes With The First Derivative Test?
These three errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and AP review handouts.
Forgetting the critical points where $f'$ is undefined.
Where it slips in:
A student finds critical points only by solving $f'(x) = 0$, and never checks where $f'$ fails to exist, missing cusps and corners like the one in $f(x) = x^{2/3}$.
Don't do this:
Do not treat $f'(x) = 0$ as the complete list of critical points. A point where $f'$ is undefined but $f$ is defined is still a critical point.
The correct way:
List both kinds. Solve $f'(x) = 0$, then separately find every $x$ in the domain where $f'(x)$ does not exist, and put all of them on the sign chart.
Choosing test points too far from the critical point.
Where it slips in:
With critical points close together, a student picks a convenient round number that actually lies past the next critical point, so the sampled sign belongs to the wrong interval.
Don't do this:
Do not grab a test value without checking it sits strictly between neighbouring critical points. A point on the far side of another critical point reports a false sign.
The correct way:
Pick each test point strictly between two adjacent critical points. If the critical points are $-1$ and $1$, test somewhere in $(-1, 1)$ such as $0$, not $2$.
Reading concavity instead of slope at the critical point.
Where it slips in:
A student mixes the two tests, checking whether the graph is concave up or down (a second-derivative idea) when the First Derivative Test only asks about the sign of the slope.
Don't do this:
Do not classify the point by concavity when you set out to use the First Derivative Test. Concavity is $f''$; the First Derivative Test is $f'$.
The correct way:
Look only at whether $f'$ is positive or negative on each side. A change from positive to negative is a maximum regardless of how the curve bends.
Practice Problems On The First Derivative Test
Work each one, then check against the answer. Answers are verified.
Classify the critical point of $f(x) = x^2 - 4x + 3$.
(Answer: $f'(x) = 2x - 4$, critical point $x = 2$; sign $-\to+$, local minimum at $(2, -1)$.)Classify the critical point of $f(x) = -x^2 + 6x$.
(Answer: $f'(x) = -2x + 6$, critical point $x = 3$; sign $+\to-$, local maximum at $(3, 9)$.)Classify the critical points of $f(x) = x^3 - 12x$.
(Answer: $f'(x) = 3(x-2)(x+2)$; local maximum at $(-2, 16)$, local minimum at $(2, -16)$.)Classify the critical points of $f(x) = x^4 - 4x^3$.
(Answer: $f'(x) = 4x^2(x-3)$, critical points $x = 0, 3$; at $x = 0$ the sign is $-\to-$ so neither, at $x = 3$ the sign is $-\to+$ so a local minimum at $(3, -27)$.)Find the increasing and decreasing intervals of $f(x) = x^3 - 3x^2$.
(Answer: $f'(x) = 3x(x-2)$; increasing on $(-\infty, 0)$ and $(2, \infty)$, decreasing on $(0, 2)$; maximum at $(0, 0)$, minimum at $(2, -4)$.)Classify the critical point of $f(x) = x^{1/3}$.
(Answer: $f'(x) = \tfrac{1}{3}x^{-2/3}$ is undefined at $x = 0$ but positive on both sides, so no sign change; $x = 0$ is neither, a vertical-tangent inflection.)
Where Should You Go Next After The First Derivative Test?
The test is one move in a larger toolkit for reading a graph from its derivatives.
Second derivative test. The faster classifier for friendly functions, and the one you fall back to the first derivative from when it says nothing.
Increasing and decreasing functions. The same sign chart of $f'$, read as intervals rather than single points.
Curve sketching. Combine the sign of $f'$ with the sign of $f''$ to draw a full graph by hand.
If your child is meeting the First Derivative Test for the first time, a live Bhanzu trainer teaches it from the sign-of-the-slope picture up, in the high-school math program.
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