The First Derivative Test

You've found the critical points; now you classify them. The sign of \(f'\) on either side of a critical point tells you whether the graph rises into it and falls out (a peak), falls in and rises out (a valley), or just flattens and keeps going.

By the end you'll be able to build a sign chart for \(f'\), read off the intervals where \(f\) increases and decreases, and classify each critical point as a local max, local min, or neither.

Predict: is \(x = 0\) a local max or a local min of \(f\)? Decide by reading the sign of \(f'\) just left and just right of it on the number line below — then drag the probe across 0 and watch the tangent flip from uphill to downhill.

Top: \(f(x) = x^3 - 3x^2\). Bottom: a number line shaded by the sign of \(f'(x) = 3x(x-2)\) — green where \(f' > 0\) (f increasing), red where \(f' < 0\) (f decreasing), with the critical points \(x = 0\) and \(x = 2\) in amber. The tangent segment on the curve and the lit-up number-line region always share a color: they are showing the same sign. Drag the dot (or use the slider) to slide the probe; hover or focus the amber critical points for their classification.

x = −1.00 · f′(−1.00) = 9.000 > 0 → f is increasing ↗

f above, sign of f′ below — the probe links the tangent's slope on the curve to the shaded sign region under it
f increasing (f′ > 0) f decreasing (f′ < 0) critical point (f′ = 0)

The First Derivative Test uses the sign of \(f'\) on either side of a critical point to find intervals of increase and decrease and to classify each critical point as a local maximum, local minimum, or neither.

Formal

Where \(f'(x) > 0\), \(f\) is increasing; where \(f'(x) < 0\), \(f\) is decreasing. Critical points — where \(f' = 0\) or \(f'\) is undefined — are the only places \(f\) can switch direction, so they split the domain into intervals of one sign. At a critical point \(c\), read the sign of \(f'\) just left and just right: + then − means \(c\) is a local maximum; − then + means a local minimum; and no sign change (+/+ or −/−) means neither — the curve flattens for an instant and continues the same way. The recipe: (1) compute \(f'\) and find all critical points, (2) mark them on a number line, (3) test one point per interval for the sign of \(f'\), (4) read off the shape.

Visual

Picture the number line you built in the demo: a strip of green (uphill) and red (downhill) segments with amber pins at the critical points. A peak is exactly a pin where green turns to red — you walk up, then down. A valley is red turning green. If the color is the same on both sides, the road just levels off mid-climb — no summit, no valley. The sign chart is a compressed sketch of the whole graph.

Worked example

\(f(x) = x^3 - 3x^2\). Differentiate: \(f'(x) = 3x^2 - 6x = 3x(x - 2)\), so the critical points are \(x = 0\) and \(x = 2\). Test each interval: at \(x = -1\), \(f' = 3(-1)(-3) = 9 > 0\) (↗); at \(x = 1\), \(f' = 3(1)(-1) = -3 < 0\) (↘); at \(x = 3\), \(f' = 3(3)(1) = 9 > 0\) (↗). Sign pattern + → − at \(x = 0\): local max. − → + at \(x = 2\): local min.

Your turn

\(g(x) = x^3 - 6x^2 + 9x\). Differentiate: \(g'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)\), so the critical points are \(x = 1\) and \(x = 3\). Test points: \(g'(0) = 9 > 0\) and \(g'(4) = 9 > 0\), but \(g'(2) = 3(1)(-1) = -3 < 0\). So the sign pattern across the number line is + , − , +. Now finish it: classify \(x = 1\) and \(x = 3\) from the transitions — \(x = 1\) is a local ____ and \(x = 3\) is a local ____.

Reveal the answer

At \(x = 1\) the sign flips + → −, so \(g\) rises then falls: local max. At \(x = 3\) it flips − → +, so \(g\) falls then rises: local min. Same recipe as the demo — only the pin locations moved.

More info — why the test works, and two ways it bites back

The guarantee behind "\(f' > 0\) means increasing" is the MVT from the previous lesson: for any \(a < b\) in the interval, some \(c\) between them has \(f'(c) = \frac{f(b)-f(a)}{b-a}\); if every \(f'\) value is positive, that forces \(f(b) > f(a)\). Two pitfalls to watch: (1) a single arithmetic slip at a test point flips a sign and silently swaps "max" for "min" — sanity-check against a mental sketch; (2) critical points where \(f'\) is undefined (cusps, vertical tangents) still bound intervals and can still be extrema — don't only hunt for \(f' = 0\). For more worked sign charts, see the Paul's Online Notes link under Dive deeper.

Check your understanding

Question 1 of 4

\(f(x) = x^3 - 12x\). Use the First Derivative Test to classify the critical points.

Question 2 of 4

\(g\) has a critical point at \(x = 1\), and near it \(g'(x) = (x-1)^2\). What does the First Derivative Test conclude about \(x = 1\)?

Question 3 of 4

To build a sign chart for \(h(x) = x^4 - 8x^3 + 7\), first differentiate. What is \(h'(x)\), factored, and where are its critical points?

Question 4 of 4

A sign chart shows \(f' < 0\) on \((-\infty, 5)\) and \(f' > 0\) on \((5, \infty)\), with \(f'(5) = 0\). What happens at \(x = 5\)?

Recap

  • \(f' > 0\) on an interval means \(f\) is increasing there; \(f' < 0\) means decreasing. Critical points (\(f' = 0\) or \(f'\) undefined) are the only places the direction can switch.
  • First Derivative Test at a critical point \(c\): sign of \(f'\) goes + → − ⇒ local max; − → + ⇒ local min; no sign change ⇒ neither.
  • Method: find the critical points, mark them on a number line, test one point per interval, then read the shape off the sign chart.
  • Don't skip critical points where \(f'\) is undefined, and remember \(f'(c) = 0\) alone proves nothing — \(x^3\) at 0 has a horizontal tangent and no extremum.

Dive deeper

Sources

  • The First Derivative Test