How can a derivative help locate a local maximum or minimum?
Quick answer
Find critical points where the derivative is zero or undefined, then check the derivative’s sign on each side. A change from positive to negative means the function rises then falls, so there is a local maximum; negative to positive means it falls then rises, so there is a local minimum. A critical point with no sign change is not a local extremum. [S1]
The question, the details, the next step.
How do I use a derivative to locate a local maximum or minimum?
Start by finding critical points: interior points in the domain where the derivative is zero or does not exist. For a continuous function differentiable near an interior point, a local maximum or minimum can occur only at a critical point. But that makes each critical point a candidate, not a conclusion. [S1]
Then use the first derivative test. Check the sign of the derivative on either side of each candidate. If f′ changes from positive to negative, the function rises then falls and has a local maximum. If it changes from negative to positive, the function falls then rises and has a local minimum. If there is no sign change, the test does not identify a local extremum there. [S1]
Can you work through an example and show how the sign check supports the answer?
Take f(x) = x². Its derivative is f′(x) = 2x, so the critical point is x = 0.
Check the derivative on either side: for x < 0, 2x is negative; for x > 0, 2x is positive. Thus f′ changes from negative to positive at 0. The function falls as it approaches 0 from the left and rises after 0, so f has a local minimum there. Since f(0) = 0, the minimum point is (0, 0). This sign change justifies the conclusion; f′(0) = 0 by itself would only identify a candidate. [S1]
What if the derivative is zero but does not change sign?
Then the first derivative test does not identify a local maximum or minimum at that point. For example, f(x) = x³ has f′(x) = 3x², which is zero at 0 but positive on both sides. The function keeps increasing through 0, so that critical point is neither a local maximum nor a local minimum. [S1]
Also, a derivative may be undefined at a point in the function’s domain; that point can be a critical-point candidate too. Check the function’s behavior on both sides rather than assuming the derivative must equal zero. [S1]
What’s a simple routine I can reuse on a homework problem?
Use this sequence: identify the domain and the interval you are studying; find interior points where f′ is zero or undefined; divide the interval at those candidates; test the derivative’s sign in each resulting interval; then state the sign change and the conclusion. If you need the extremum point, substitute its x-value into the original function.
For the example, a concise written conclusion is: “f′ is negative to the left of 0 and positive to its right, so f has a local minimum at (0, 0).” This describes behavior near the point, not necessarily the lowest value over a larger domain. [S1]
A practical checklist
Identify the function’s domain and the interval being considered.
Find interior points where f′ is zero or undefined.
Check the sign of f′ on each side of every candidate.
Classify positive-to-negative as a local maximum and negative-to-positive as a local minimum.
If there is no sign change, do not label the candidate a local extremum.
Substitute the x-coordinate into f to give the extremum point when needed.
A few more useful answers
Does every local maximum or minimum have to occur where f′(x) = 0?
Not necessarily: a critical point can also occur where the derivative is undefined. Under the stated continuity and differentiability conditions, an interior local extremum must be a critical point. [S1]
Does a local minimum mean the function’s lowest value everywhere?
No. A local minimum describes behavior near that point. It does not establish the smallest value over the entire domain.
Sources
References used for this guide.
- [S1] OpenStax Calculus Volume 1, §4.5: Derivatives and the Shape of a Graph openstax.org · Checked