Application of Derivatives · Lesson 4 of 7

Testing for Local Extrema

“Classify critical points using derivative signs and choose a valid test when a shortcut is inconclusive.”

Learning Objectives

• Apply the first derivative test at continuous critical points. • Apply the second derivative test at suitable stationary points. • Handle nondifferentiable extrema and zero second derivatives. • Distinguish a stationary point, a turning point, and an inflection point. • Report local extreme values as well as their input locations.

Turn a Candidate into a Classification

Solving $f^{'} (x) = 0$ finds places worth investigating, but does not tell you whether the graph has a peak, a valley, or neither. To classify a critical point, ask how the function behaves just before and just after it. The sign of the derivative turns this graphical question into a procedure.

The first derivative test concerns a function continuous at an interior critical point, with derivative signs controlled on the neighbouring intervals. It can work even when the derivative does not exist at the point itself. What matters is the behaviour on the two sides and the continuity connecting them.

Sign just left of $c$	Sign just right of $c$	Conclusion at $c$
Positive	Negative	Local maximum
Negative	Positive	Local minimum
Positive	Positive	No local extremum
Negative	Negative	No local extremum

In the first row, values rise towards $c$ and fall after it, so $f (c)$ is a local peak. In the second row, values fall towards $c$ and rise after it, giving a valley. In the last two rows the direction does not reverse, so the graph passes through without turning.

Example — A cubic with a peak and a valley

Problem
Find and classify all local extrema of $f (x) = x^{3} - 3 x + 3$ .

1.Differentiate: $f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1)$ . The stationary candidates are $- 1$ and 1.
2.For $x < - 1$ the derivative is positive, for $- 1 < x < 1$ it is negative, and for $x > 1$ it is positive.
3.At $- 1$ , the sign changes from positive to negative: a local maximum. Its value is $f (- 1) = - 1 + 3 + 3 = 5$ .
4.At 1, the sign changes from negative to positive: a local minimum. Its value is $f (1) = 1 - 3 + 3 = 1$ .
5.The inputs $- 1, 1$ are locations; the outputs 5 and 1 are the corresponding local extreme values.

Classifying two stationary points— The sign applies throughout an interval, not just at one test point.

The sign changes produce a peak and a valley— A positive-to-negative change gives the peak; a negative-to-positive change gives the valley.

Example — A repeated derivative root without a turn

Problem
Classify the critical point of $f (x) = 2 x^{3} - 6 x^{2} + 6 x + 5$ .

1.The derivative is $f^{'} (x) = 6 (x - 1)^{2}$ , zero only at $x = 1$ .
2.For inputs on either side of 1, the squared factor is positive. Thus the derivative stays positive rather than changing sign.
3.The input 1 is neither a local maximum nor a local minimum. Its output is $f (1) = 7$ .
4.For this particular cubic, $f^{''} (x) = 12 (x - 1)$ changes sign at 1, so concavity changes there: it is a stationary inflection point.

No extremum does not automatically mean inflection

An inflection point is where the curve changes concavity. A derivative that does not change sign proves “no local extremum” under the first derivative test; it does not alone prove a concavity change. For example, $x^{3}$ has one at zero, but $x$ has no change in concavity there.

The Second Derivative as a Shortcut

When a function is sufficiently smooth, its second derivative can classify a stationary point without a full sign chart. A positive second derivative corresponds to a valley shape; a negative one corresponds to a peak shape. The stationary-point condition is essential.

Second derivative testLaTeX

f^{'} (c) = 0, f^{''} (c) > 0 f^{'} (c) = 0, f^{''} (c) < 0 \Rightarrow local minimum, \Rightarrow local maximum .

Use the test when the function is twice differentiable in a neighbourhood of the stationary point. If the second derivative is zero, the test gives no conclusion.

The reasoning can be seen through the first derivative. A positive second derivative at a stationary point under these conditions makes the first derivative change from negative to positive there. A negative one gives the opposite change. This is another way to detect turning behaviour, not an unrelated rule.

Example — Three critical points in one quartic

Problem
Find local extrema of $f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 12$ .

1.Differentiate and factor: $f^{'} (x) = 12 x (x - 1) (x + 2)$ , so the candidates are $x = - 2, 0, 1$ .
2.Differentiate again: $f^{''} (x) = 36 x^{2} + 24 x - 24$ . At $- 2$ it is 72, at 0 it is $- 24$ , and at 1 it is 36.
3.The positive values give local minima at $- 2$ and 1. The negative value gives a local maximum at 0.
4.Evaluate the original function: $f (- 2) = 48 - 32 - 48 + 12 = - 20$ , $f (1) = 3 + 4 - 12 + 12 = 7$ , and $f (0) = 12$ .
5.There are two local minimum values at different inputs. Local valleys need not have equal heights.

When a Test Does Not Apply

A failed test is not a failed function. If the second derivative is unavailable or zero, return to the first derivative signs or compare function values directly. Treating “inconclusive” as “no extremum” would lose genuine maxima and minima.

Example — A corner handled by the first derivative test

Problem
Find the local minimum of $f (x) = 3 + ∣ x ∣$ .

1.The function is continuous at 0, but its derivative does not exist there, so 0 is a critical point.
2.For $x < 0$ , $f (x) = 3 - x$ and $f^{'} (x) = - 1$ . For $x > 0$ , $f (x) = 3 + x$ and $f^{'} (x) = 1$ .
3.The derivative changes from negative to positive across 0. There is a local minimum at 0, with value $f (0) = 3$ .
4.The second derivative test cannot be used at the corner. The first derivative test succeeds because its side conditions hold.

A corner minimum survives without a derivative— The left branch decreases and the right branch increases. The central value is the local minimum.

Example — A zero second derivative requires another test

Problem
Apply the second derivative test to $f (x) = 2 x^{3} - 6 x^{2} + 6 x + 5$ at its stationary point.

1.As before, $f^{'} (x) = 6 (x - 1)^{2}$ , so the stationary point is at 1.
2.Since $f^{''} (x) = 12 (x - 1)$ , $f^{''} (1) = 0$ . The second derivative test is inconclusive.
3.The first derivative is positive on both sides, proving no local extremum. The separate concavity change establishes that this example has a stationary inflection point.
4.The conclusion came from further analysis, not from $f^{''} (1) = 0$ alone.

Function at input 0	First derivative there	Second derivative there	Actual behaviour
$x^{4}$	0	0	Local minimum
$- x^{4}$	0	0	Local maximum
$x^{3}$	0	0	Neither; stationary inflection

These three functions show why zero second derivative carries no classification by itself. To find all local extrema, list every critical point in the domain, test each one, and calculate the associated output values. Keep domain endpoints for the separate absolute-extrema comparison.

Choosing a test

Use the first derivative test when a factored derivative makes signs easy, or when a corner is involved. Use the second test when its values at stationary points are quick to evaluate and nonzero. Check the assumptions whichever test you choose.

Check Your Understanding

Use these questions to check both your calculations and your interpretation of the ideas. Read the conditions carefully before choosing an answer.

Quick check

Across $c$ , the derivative changes from positive to negative. What occurs?

Quick check

If $f^{'} (c) = 0$ and $f^{''} (c) > 0$ under the second-test conditions, what follows?

Quick check

What follows from $f^{'} (c) = f^{''} (c) = 0$ alone?

Quick check

Why is 0 a local minimum of $3 + ∣ x ∣$ ?

Quick check

For $f (x) = x^{3} - 3 x + 3$ , what is the local maximum value?

Quick check

What additional property defines an inflection point?

Practice Problems

Find all local extrema of $x^{3} - 6 x^{2} + 9 x + 15$ , giving values and input locations.
Find the local extrema of $sin x + cos x$ on $(0, π /2)$ and of $sin x - cos x$ on $(0, 2 π)$ .
Classify the critical point of $x /2 + 2/ x$ on $x > 0$ .
Find and classify the critical points of $1/ (x^{2} + 2)$ .
Find the local extrema of $x 1 - x$ on $(0, 1)$ , checking the domain while differentiating.
Use first derivative signs to classify critical points of $(x - 2)^{4} (x + 1)^{3}$ . Separately check which stationary point is an inflection point.
Explain why $x^{4}$ , $- x^{4}$ , and $x^{3}$ have different classifications at 0 even though their first two derivatives there are zero.

Key Takeaways

• A positive-to-negative derivative change gives a local maximum; the reverse gives a local minimum. • A nonzero second derivative classifies a stationary point under the test’s smoothness conditions. • A zero second derivative is inconclusive, so use another method. • An undefined derivative can occur at a genuine extremum. • Inflection requires a concavity change and cannot be inferred from absence of an extremum alone.

Previous · Lesson 3

Understanding Maximum and Minimum Values

Next · Lesson 5

Building and Solving Optimization Models

Turn a Candidate into a Classification

The Second Derivative as a Shortcut

When a Test Does Not Apply

Check Your Understanding

Quiz

Practice Problems

Key Takeaways