First Derivative Test

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First Derivative Test: meaning, method and solved example (AP Calculus AB) – Maths PDF


In short: First derivative test decides whether a flat or broken slope marks a peak, a valley or neither, by checking how the sign of f’ behaves just before and just after it.

AP Calculus ABCalculusAP Calculus AB Unit 5

What is First Derivative Test?

The first derivative test reads a graph’s story from the sign of its derivative. Positive means the function is climbing, negative means it is falling, so a switch of sign at a critical point reveals a turning point.

Take a critical number \(c\) where \(f\) is continuous. Positive just before \(c\) and negative just after: \(f(c)\) is a local maximum. Negative before and positive after: \(f(c)\) is a local minimum. Same sign on both sides: nothing special happens at \(c\).

How to spot it

  • The problem wants the stretches where a graph climbs or drops and the spots where it turns around.
  • You are handed a graph or table of f’ and must decide what f is doing.
  • A sign chart or number line with test values is expected in the justification.
  • The second derivative is messy, zero, or not allowed, so signs of f’ are the safer tool.

The method

  1. Compute \(f^{\prime}(x)\) and list the critical numbers by solving \(f^{\prime}(x) = 0\) or finding where it is undefined.
  2. Mark those numbers on a number line; they split it into intervals.
  3. Pick one test value in each interval and record the sign of \(f^{\prime}\).
  4. Apply the first derivative test at each critical number and report the extreme value \(f(c)\).
-13f’ > 0f’ < 0f’ > 0f risesf fallsf riseslocal max, f = 9local min, f = -23test x = -2: f’ = 15test x = 0: f’ = -9test x = 4: f’ = 15
f'(x) = 3(x – 3)(x + 1): the sign changes + to – at x = -1 and – to + at x = 3.

First Derivative Test: solved example

Problem

Classify the critical points of \(f(x) = x^3 – 3x^2 – 9x + 4\).

Solution

\(f^{\prime}(x) = 3x^2 – 6x – 9 = 3(x – 3)(x + 1)\), so the critical numbers are \(-1\) and \(3\).

Test values: \(f^{\prime}(-2) = 15 > 0\); \(f^{\prime}(0) = -9 < 0\); \(f^{\prime}(4) = 15 > 0\).

The sign goes from \(+\) to \(-\) at \(-1\): local maximum \(f(-1) = -1 – 3 + 9 + 4 = 9\). It goes from \(-\) to \(+\) at 3: local minimum \(f(3) = 27 – 27 – 27 + 4 = -23\).

Try it

Question 1

For \(g(x) = x^3\), the derivative \(3x^2\) is zero at \(x = 0\). Is there a local extreme there?

Show the answer

No. \(g^{\prime}\) is positive on both sides of 0, so the sign does not change and 0 is neither a maximum nor a minimum.

Question 2

If \(f^{\prime}(x) = (x – 5)(x + 2)^2\), what happens at \(x = 5\) and \(x = -2\)?

Show the answer

At 5 the sign changes from negative to positive: a local minimum. At \(-2\) the squared factor never changes sign, so there is no extreme.

First Derivative Test vs. Second Derivative Test

The first derivative test looks at signs of \(f^{\prime}\) on both sides of a critical number and works even when the second derivative is zero. The second derivative test looks only at the sign of \(f^{\prime\prime}\) at the point itself and can be inconclusive. Neither replaces the other in every situation.

Second Derivative Test

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