What Is the First Derivative Test?
Critical points, where f′ = 0 or fails to exist, are the only places a function can turn around, but not every critical point is a turn. The first derivative test classifies each one by reading the sign of f′ on either side:
+ to − across c: local maximum; − to + across c: local minimum; no sign change: neither
The logic is nothing but monotonicity: rising into a point and falling out of it makes a peak; falling in and rising out makes a valley; continuing in the same direction makes a flat spot on a hill, like x³ at the origin. Unlike the second derivative test, this one never comes back inconclusive: whenever the signs can be read, the verdict is definitive. This calculator finds the critical points from the symbolic derivative, evaluates f′ just left and right of each, states every verdict with its evidence, and color-codes the points on the graph.
How to Use the First Derivative Test Calculator
Enter a function and a window to search. For each critical point found, the steps quote f′ at a point slightly left and slightly right, then the classification with a one-line reason. The graph overlays f′ as a dashed curve, so you can see each classification as geometry: a local max of f is a downward crossing of the dashed curve through zero, a local min an upward crossing, and a "neither" verdict a touch without crossing. The companion increasing/decreasing intervals calculator presents the same sign data organized by interval rather than by point.
Worked Example
Classify the critical points of f(x) = x⁴ − 4x². The derivative f′(x) = 4x³ − 8x = 4x(x² − 2) vanishes at x = −√2, 0, √2. Sampling signs:
f′(−2) = −16 < 0, f′(−1) = 4 > 0, f′(1) = −4 < 0, f′(2) = 16 > 0
So the pattern is −/+ at −√2 (local minimum, value −4), +/− at 0 (local maximum, value 0), and −/+ at √2 (local minimum, −4 again): the double-well shape, two valleys flanking a hill. The x³ chip shows the third verdict: f′ = 3x² is positive on both sides of 0, so despite the horizontal tangent there is no extremum, the graph pauses and continues upward, the case that traps students who classify by "f′ = 0" alone.
First versus Second Derivative Test
The two tests answer the same question with different information. The second derivative test is faster when it works, one evaluation of f″ per point, but it returns "inconclusive" whenever f″(c) = 0, precisely the flat-but-not-extremal and higher-order cases (x³, x⁴) where classification is most delicate. The first derivative test needs two sign evaluations but is complete: it works at every critical point where f is continuous, including corners and cusps where f″ does not even exist, |x| at 0 is instantly a minimum by its −/+ pattern, a point the second derivative test cannot touch. The practical rule of thumb: use f″ when it is easy and nonzero, fall back to sign charts everywhere else, and remember the two must never disagree, a disagreement means an arithmetic error.
Common Mistakes to Avoid
- Declaring an extremum from f′(c) = 0 alone. The zero only nominates the candidate; the sign change elects it. x³ is the eternal counterexample.
- Testing too far from the critical point. Sample past the next critical point and you read the wrong interval's sign. Stay strictly between neighboring critical points; the tool uses a small adaptive offset.
- Forgetting critical points where f′ is undefined. Corners (|x − 2|) and cusps (x^(2/3)) are classified perfectly by sign changes, but only if they make the candidate list in the first place.
- Confusing the sign of f with the sign of f′. The test reads the derivative's sign. The function's own sign is irrelevant to whether it is turning.
- Assuming local verdicts settle global questions. A local max may be far below the function's supremum elsewhere; global extremes on an interval need the closed-interval comparison.
Real-World Applications
Classification of turning points is decision-making's fine print. In economics, marginal analysis is literally this test: profit peaks where marginal profit crosses from positive to negative, and a zero of marginal profit without a sign change is a shoulder, not an optimum, a distinction with budgetary consequences. Pharmacokinetics classifies the peak of a concentration curve to time doses; engineering classifies stationary points of energy to separate stable equilibria (minima) from unstable ones (maxima), the difference between a ball resting in a bowl and balanced on a dome. Machine learning's loss landscapes are navigated by derivative signs, and saddle-like flat spots, the higher-dimensional cousins of the x³ case, are a genuine obstacle that optimizers must escape. Wherever a system's "best" is sought by setting a derivative to zero, this test is the follow-up question that separates real optima from impostors.
Frequently Asked Questions
What exactly does the first derivative test check?
The sign of f′ immediately left and right of each critical point. Positive-to-negative marks a local maximum, negative-to-positive a local minimum, and same-sign-on-both-sides means no extremum, just a flat or non-smooth point in a monotone stretch.
Why is a sign change necessary for an extremum?
Because a local max requires rising before and falling after (and dually for a min); those are statements about f′'s sign by the monotonicity theorem. A critical point without a sign change sits inside a rising or falling run and cannot beat its neighbors on both sides.
When does this test beat the second derivative test?
Whenever f″(c) = 0 (higher-order flatness, where the second derivative test shrugs) and whenever f″ does not exist (corners, cusps). It also wins when f″ is expensive to compute. Its only cost is evaluating f′ twice more per point.
Can the test fail or be ambiguous?
For continuous f with isolated critical points, no: the signs on the flanking intervals are well-defined and the verdict follows. Pathologies need infinitely many critical points crowding together (like x²sin²(1/x)), where "immediately left and right" stops making sense; elementary functions do not do this.
How close to the critical point should the test values be?
Anywhere strictly between c and its neighboring critical points works, since the sign is constant on that whole stretch. The danger is only overshooting into the next stretch. The calculator offsets by a small fraction of the local critical-point spacing to stay safe.
What happens at a corner like |x| at 0?
The derivative is −1 on the left and +1 on the right: a clean −/+ pattern, hence a local minimum, even though f′(0) itself does not exist. The test only needs one-sided signs, which is exactly why it covers non-smooth candidates the f″ test cannot.
Does "neither" mean the point is unimportant?
Not at all: a same-sign critical point of a smooth function is typically a horizontal inflection, where the curve pauses and the concavity flips, x³'s origin being the model. Such points matter for curve shape and for optimizer behavior; they are simply not extrema.
Can a function have a local max with no critical point?
At interior points of differentiability, no, Fermat's theorem forbids it. Local extrema can, however, occur at domain endpoints or at points of discontinuity, which the test does not cover; those are checked directly from the definition or via one-sided monotonicity.
How do I read the classifications off the dashed derivative curve?
Each critical point is a zero of the dashed curve. If the dashed curve crosses the axis downward there, f has a local max; upward, a local min; if it touches and bounces back, no extremum. The picture makes the test automatic once the correspondence clicks.
Does the test generalize to functions of two variables?
Not literally, since "left and right" has no meaning in the plane; its role is taken by the second-partials (Hessian) test at critical points of f(x, y), handled by the multivariable critical points calculator. The one-variable sign-chart intuition survives as examining f along lines through the point.