Mathwords logoReference LibraryMathwords

First Derivative Test

First Derivative Test

A method for determining whether a critical point is a minimum, maximum, or neither.

 

Three rules for First Derivative Test: negative-left/positive-right=minimum; positive-left/negative-right=maximum;...

Number line showing f'(x) positive (increasing) left of x=3, negative (decreasing) right, indicating x=3 is a maximum.

 

 

See also

Second derivative test, increasing function, decreasing function, critical number

Key Formula

If f′(x) changes from {+→−at x=c, then f(c) is a local maximum−→+at x=c, then f(c) is a local minimum+→+ or −→−at x=c, then f(c) is neither\text{If } f'(x) \text{ changes from } \begin{cases} + \to - & \text{at } x = c, \text{ then } f(c) \text{ is a local maximum} \\ - \to + & \text{at } x = c, \text{ then } f(c) \text{ is a local minimum} \\ + \to + \text{ or } - \to - & \text{at } x = c, \text{ then } f(c) \text{ is neither} \end{cases}
Where:
  • f′(x)f'(x) = The first derivative of the function f
  • cc = A critical number where f'(c) = 0 or f'(c) is undefined
  • ++ = The derivative is positive (function is increasing)
  • −- = The derivative is negative (function is decreasing)

Worked Example

Problem: Use the First Derivative Test to find and classify the critical points of f(x) = x³ − 3x + 2.
Step 1: Find the first derivative of f(x).
f′(x)=3x2−3f'(x) = 3x^2 - 3
Step 2: Set the derivative equal to zero and solve for the critical numbers.
3x2−3=0  ⟹  x2=1  ⟹  x=−1 or x=13x^2 - 3 = 0 \implies x^2 = 1 \implies x = -1 \text{ or } x = 1
Step 3: Test the sign of f'(x) in each interval created by the critical numbers. Choose test points x = −2, x = 0, and x = 2.
f′(−2)=3(4)−3=9>0,f′(0)=3(0)−3=−3<0,f′(2)=3(4)−3=9>0f'(-2) = 3(4) - 3 = 9 > 0, \quad f'(0) = 3(0) - 3 = -3 < 0, \quad f'(2) = 3(4) - 3 = 9 > 0
Step 4: Apply the First Derivative Test at x = −1. The derivative changes from positive to negative (+ → −), so f(−1) is a local maximum.
f(−1)=(−1)3−3(−1)+2=−1+3+2=4f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4
Step 5: Apply the First Derivative Test at x = 1. The derivative changes from negative to positive (− → +), so f(1) is a local minimum.
f(1)=(1)3−3(1)+2=1−3+2=0f(1) = (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0
Answer: f has a local maximum of 4 at x = −1 and a local minimum of 0 at x = 1.

Another Example

This example shows the 'neither' case, where the derivative has the same sign on both sides of the critical point. Many students expect every critical point to be a max or min, but this demonstrates that is not always the case.

Problem: Use the First Derivative Test to classify the critical point of f(x) = x³ at x = 0.
Step 1: Find the first derivative.
f′(x)=3x2f'(x) = 3x^2
Step 2: Identify the critical number. Setting f'(x) = 0 gives x = 0.
3x2=0  ⟹  x=03x^2 = 0 \implies x = 0
Step 3: Test the sign of f'(x) on each side of x = 0. Choose x = −1 and x = 1.
f′(−1)=3(1)=3>0,f′(1)=3(1)=3>0f'(-1) = 3(1) = 3 > 0, \quad f'(1) = 3(1) = 3 > 0
Step 4: The derivative does not change sign — it is positive on both sides (+ → +). By the First Derivative Test, x = 0 is neither a local maximum nor a local minimum.
Sign pattern: +→+  ⟹  neither max nor min\text{Sign pattern: } + \to + \implies \text{neither max nor min}
Answer: The critical point at x = 0 is neither a local maximum nor a local minimum. The function has an inflection point there instead.

Frequently Asked Questions

What is the difference between the First Derivative Test and the Second Derivative Test?
The First Derivative Test examines the sign change of f'(x) around a critical point, while the Second Derivative Test evaluates f''(c) at the critical point itself. The First Derivative Test always gives a definitive answer (max, min, or neither), but the Second Derivative Test can be inconclusive when f''(c) = 0. However, the Second Derivative Test is often faster since you only need to plug the critical number into f''(x) rather than testing intervals.
When does the First Derivative Test fail or not apply?
The First Derivative Test itself does not fail — it always provides a classification for a critical point as long as f is continuous at that point. However, you must be able to determine the sign of f'(x) in the intervals around the critical point. If the derivative oscillates wildly near the critical point (as in some pathological functions), determining the sign can be difficult in practice.
Do you need to check both sides of the critical point?
Yes. The entire logic of the test depends on comparing the sign of the derivative to the left versus the right of the critical number. Checking only one side tells you whether the function is increasing or decreasing there, but it cannot determine whether the function changes direction at the critical point.

First Derivative Test vs. Second Derivative Test

First Derivative TestSecond Derivative Test
What you computeSign of f'(x) on both sides of the critical pointValue of f''(c) at the critical point
Classifies asLocal max, local min, or neitherLocal max (if f''(c) < 0), local min (if f''(c) > 0), or inconclusive (if f''(c) = 0)
Can be inconclusive?No — always gives a definitive answerYes — when f''(c) = 0
RequiresTesting points in intervals around cComputing the second derivative and evaluating at c
Typical useWhen f''(x) is hard to compute, or when you also need to know where f is increasing/decreasingQuick classification when f''(x) is easy to compute

Why It Matters

The First Derivative Test is a core tool in any Calculus 1 or AP Calculus course. You use it to solve optimization problems — finding maximum profit, minimum cost, shortest distance, and similar applications. It also builds the foundation for sketching accurate graphs of functions by identifying where peaks and valleys occur.

Common Mistakes

Mistake: Assuming every critical point is automatically a local max or min.
Correction: A critical point where the derivative does not change sign (e.g., f(x) = x³ at x = 0) is neither a maximum nor a minimum. Always check the sign on both sides before concluding.
Mistake: Forgetting to include critical points where f'(x) is undefined (not just where f'(x) = 0).
Correction: Critical points occur wherever f'(c) = 0 OR f'(c) does not exist (provided f(c) is defined). For example, f(x) = |x| has a critical point at x = 0 because f'(0) is undefined, and it is a local minimum by the First Derivative Test.

Related Terms