Critical Points & the First Derivative Test
- Find critical points where f′ = 0 or f′ is undefined
- Determine intervals of increase and decrease from the sign of f′
- Classify local maxima and minima with the First Derivative Test
Increasing, decreasing, and critical points
The sign of the first derivative tells you the direction of a function: f is increasing where f′ > 0 and decreasing where f′ < 0. A critical point is an x in the domain where f′(x) = 0 or f′(x) does not exist. These are the only places a function can switch between rising and falling, so they are the candidates for local extrema.
The First Derivative Test
To classify a critical point, examine the sign change of f′ across it. If f′ goes positive → negative (rising then falling), the point is a local maximum. If f′ goes negative → positive, it is a local minimum. If f′ does not change sign, the point is neither — just a flat spot on an otherwise monotonic curve. A sign chart of f′ makes this systematic.
Find and classify the local extrema of f(x) = x³ − 3x².
- 1.Differentiate: f′(x) = 3x² − 6x = 3x(x − 2), zero at x = 0 and x = 2 — the critical points.
- 2.Test the sign of f′ on each interval: for x < 0, f′ > 0 (increasing); for 0 < x < 2, f′ < 0 (decreasing); for x > 2, f′ > 0 (increasing).
- 3.At x = 0, f′ changes + to − → local maximum; f(0) = 0.
- 4.At x = 2, f′ changes − to + → local minimum; f(2) = 8 − 12 = −4.
Build a sign chart for f′: mark the critical points on a number line, test one point in each interval, and label the sign. The pattern of signs reads off intervals of increase/decrease and every extremum at once.
At a critical point x = c, f′ changes from negative to positive. What is x = c?
On which interval is f(x) = x² − 4x increasing?
A critical point is only a candidate for an extremum — you must show a sign change in f′ to confirm it. Where f′ = 0 but does not change sign (as at x = 0 for y = x³), there is no local max or min.
Answer the 2 checkpoints as you read.
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