If you want to find where a function reaches its maximum or minimum, you need critical points. They are the only places an extremum can occur — but not every critical point is an extremum. Finding them is the first step; classifying them comes next.

What Is a Critical Point?

A critical point of f(x) is any x-value where either f'(x) = 0 or f'(x) does not exist (but f(x) does exist). Critical points are the only candidates for local maxima and minima — though not every critical point is an extremum.

Step 1 — Finding Critical Points

📋 Method
Step 1Compute f'(x)
Step 2Set f'(x) = 0 and solve. These are your stationary points.
Step 3Find where f'(x) is undefined (but f(x) is still defined). These are also critical points.

Step 2 — Classifying Critical Points

Once you have the critical points, you classify them using either the first or second derivative test.

First Derivative Test: f' changes + to − → local maximum f' changes − to + → local minimum f' does not change sign → neither (saddle point) Second Derivative Test: f''(c) > 0 → local minimum (concave up) f''(c) < 0 → local maximum (concave down) f''(c) = 0 → inconclusive (use first derivative test)

Example 1 — f(x) = x³ − 3x + 2

📋 Find and classify all critical points
f'(x)= 3x² − 3 = 3(x−1)(x+1)
Critical ptsf'(x) = 0 → x = 1 and x = −1
f''(x)= 6x. f''(1) = 6 > 0 → local min. f''(−1) = −6 < 0 → local max
Valuesf(1) = 0 (local min). f(−1) = 4 (local max)

Example 2 — f(x) = x⁴ (Second Derivative Test Fails)

📋 Find and classify x = 0
f'(x)= 4x³ = 0 → x = 0
f''(0)= 0 → Second derivative test inconclusive
First deriv testf' < 0 for x < 0 and f' > 0 for x > 0 → sign changes − to + → local minimum

Example 3 — f(x) = x^(2/3) (Undefined Derivative at 0)

📋 Find all critical points
f'(x)= (2/3)x^(−1/3) = 2/(3x^(1/3))
f'(x) = 0?Never — the numerator is always 2
f' undefined?At x = 0. And f(0) = 0 exists → x = 0 is a critical point
Classifyf' < 0 for x < 0, f' > 0 for x > 0 → local minimum (cusp)
⚠ Common Mistake

Never forget critical points where f'(x) is undefined — like |x| at x=0 or x^(2/3) at x=0. These cusps and corners are often local extrema and students miss them by only setting f'(x) = 0.

Frequently Asked Questions

Is every critical point a local extremum?▾
No. A critical point where f'(c) = 0 but f' does not change sign is neither a max nor a min. Example: f(x) = x³ at x = 0 has f'(0) = 0, but x = 0 is an inflection point — the function continues increasing through it.
When should I use first vs second derivative test?▾
Use the second derivative test when f''(c) is easy to compute and non-zero — it's faster. Use the first derivative test when f''(c) = 0 (inconclusive), when f' is hard to differentiate again, or when you need to understand the whole sign pattern of f'.

Global vs Local Extrema

A local maximum is the highest point in some neighbourhood — other parts of the function may be higher. A global maximum is the highest point over the entire domain. On a closed interval, the global extrema occur either at critical points or at the endpoints. You must evaluate f at all critical points AND both endpoints, then compare.

📋 Find global extrema of f(x) = x³ − 3x on [−2, 2]
f'(x) = 03x² − 3 = 0 → x = ±1 (both in [−2,2])
Evaluate allf(−2)=−2, f(−1)=2, f(1)=−2, f(2)=2
Global max= 2, achieved at x = −1 and x = 2
Global min= −2, achieved at x = 1 and x = −2

Critical Points of Trig Functions

📋 Find critical points of f(x) = 2sin(x) + sin(2x) on [0, 2π]
f'(x)= 2cos(x) + 2cos(2x) = 2cos(x) + 2(2cos²x−1) = 4cos²x + 2cos x − 2
Factor= 2(2cos²x + cos x − 1) = 2(2cos x − 1)(cos x + 1)
Solvecos x = 1/2 → x = π/3, 5π/3. cos x = −1 → x = π
Critical ptsx = π/3, π, 5π/3

The Bridge to Optimisation

Finding critical points is not an end in itself — it is the first step in any optimisation problem. The answer to every unconstrained optimisation problem — in profit, geometry, physics, engineering — always lies at a critical point (for unconstrained problems) or at a boundary (for constrained problems on closed domains).

The procedure is always: find f', set it to zero, solve for x, classify with second derivative test or sign chart, evaluate at remaining candidates, report the answer with units.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026