Stationary points

Understanding where functions change direction

f'(x) = 0 and f''(x) = 0

Tricky Stationary Points

When the second derivative test fails and higher-order analysis is needed

Higher order derivative test

Tricky Stationary Points II

When the second derivative test fails and higher-order analysis is needed

f'(x) sign

Alternative Tests

Other methods when f'(a) = 0 and f''(a) = 0

f''(x) = 0

f''(x) = 0, again

But it's not a stationary point

f ''(x) = 0

Tricky Stationary Points

Meaning of "stationary point" on a curve

If we plot the graph of a simple function, y = f(x), where y is plotted vertically and x is plotted horizontally, then a stationary point is when the derivative dy/dx or f'(c) equals zero. This point is also known as a "critical" point, because something significant happens to the behaviour of the function there.

In the examples immediately below, each has critical point when x = 0 and the tangent to the curve is horizontal there but the curve has either a horizontal inflection point, a local minimum or local maximum here.

Our job is to determine which behaviour the graph has at the critical point.

plus-x3
plus-x4
minus-x4
y = x³
y = x4
y = -x4

When the Second Derivative Test Fails

We are taught early in our study of calculus that when f'(c) = 0 at a critical point c:

  • If f''(c) > 0, then c is a local minimum (concave up)
  • If f''(c) < 0, then c is a local maximum (concave down)
  • But what if f''(c) = 0?
The Test is Inconclusive: When both f'(c) = 0 AND f''(c) = 0, the second derivative test gives us no information. The point could be a maximum, a minimum, or neither (an inflection point)!

Three Classic Examples

EXAMPLE 1: f(x) = x³ — (Horizontal) INFLECTION POINT

First derivative: f'(x) = 3x²

Second derivative: f''(x) = 6x

At x = 0: f'(0) = 0 and f''(0) = 0

Result: At x = 0 the test for the stationary point is inconclusive, see the higher-order derivative test.

EXAMPLE 2: f(x) = x⁴ — LOCAL MINIMUM

First derivative: f'(x) = 4x³

Second derivative: f''(x) = 12x²

Third derivative: f'''(x) = 24x

Fourth derivative: f⁽⁴⁾(x) = 24

At x = 0: f'(0) = 0 and f''(0) = 0

Result: At x = 0 the test for the stationary point is inconclusive, see the higher-order derivative test.

EXAMPLE 3: f(x) = -x⁴ — LOCAL MAXIMUM

First derivative: f'(x) = -4x³

Second derivative: f''(x) = -12x²

At x = 0: f'(0) = 0 and f''(0) = 0

Result: At x = 0 the test for stationary point is inconclusive, see the higher-order derivative test.

Higher order derivative test

Tricky Stationary Points

When the second derivative test fails

Study the app on concavity before this to ensure you understand the language surrounding "concave".

See: Concavity app

We've seen that when f'(c) = 0 at a critical point c:

  • If f''(c) > 0, then c is a local minimum (concave up)
  • If f''(c) < 0, then c is a local maximum (concave down)
  • But what if f''(c) = 0?
The Test is Inconclusive: When both f'(c) = 0 AND f''(c) = 0, the second derivative test gives us no information. The point could be a maximum, a minimum, or neither (an inflection point)!

The higher-order derivative test

When f''(c) = 0, we need to check higher derivatives. Here's the general pattern:

At a critical point c where f'(c) = 0:

And the "nth (order) derivative" is the function we get when differentiating f(x) n times
so f⁽2⁾(c) is the second-order (or just second) derivative
and f⁽4⁾(c) is the fourth(-order) derivative, and so on ..
• The first (higher-order) derivative at the critical point to have a non-zero value, f⁽ⁿ⁾(c) ≠ 0, where n is ODD:
  → c is an inflection point (neither max nor min)

• The first (higher-order) derivative at the critical point to have non-zero value, f⁽ⁿ⁾(c) ≠ 0, where n is EVEN:
and f⁽ⁿ⁾(c) < 0   → f(c) is a local maximum
and f⁽ⁿ⁾(c) > 0   → f(c) is a local minimum
Quick Rule: For f(x) = ±xⁿ at critical point x = c, f'(c) = 0:
• Odd powers (x³, x⁵, x⁷, ...) → inflection points
• Even powers (x⁴, x⁶, x⁸, ...) → extrema (max or min)

Applying the Test

WORKED EXAMPLE: f(x) = x³

f'(0) = 0 ✓ (critical point)

f''(0) = 0 (second derivative test fails)

f'''(0) = 6 ≠ 0 (third derivative is first non-zero)

Conclusion: Since the first non-zero derivative is the third (odd), x = 0 is an inflection point.

WORKED EXAMPLE: f(x) = x⁴

f'(0) = 0 ✓ (critical point)

f''(0) = 0 (second derivative test fails)

f'''(0) = 0 (third derivative also zero)

f⁽⁴⁾(0) = 24 > 0 (fourth derivative is first non-zero)

Conclusion: Since the first non-zero derivative is the fourth (even) and positive, x = 0 is a local minimum.

Why This Matters

Flatter Curves: When f''(c) = 0 at an extremum, the curve is flatter near that point. Compare the sharper minimum of y = x² with the gentler minimum of y = x⁴—both have minima at x = 0, but x⁴ approaches the minimum more gradually.

Inflection Points with Horizontal Tangents: The point (0,0) on y = x³ is special: it's an inflection point where the tangent is horizontal. The curve doesn't change direction (it keeps increasing), but it does change its curvature from concave down to concave up.

Alternatives to the second derivative test

f'(a) = 0 and f''(a) = 0

The Challenge

When you find a critical point at x = a where f'(a) = 0, your first instinct is to use the second derivative test. But what happens when f''(a) = 0 as well? The second derivative test becomes inconclusive.

Don't Panic! There are several reliable alternative methods to determine whether x = a is a maximum, minimum, or neither. You don't always need higher derivatives.

Method 1: The First Derivative Test

This is often the most practical and intuitive method. Instead of looking at f''(a), we examine the sign of f'(x) on both sides of x = a.

First Derivative Test Algorithm:

1. Find f'(x) and verify f'(a) = 0
2. Choose a test point just left of a (call it a - h, where h is small)
3. Choose a test point just right of a (call it a + h)
4. Evaluate f'(a - h) and f'(a + h)

Decision Rules:
• If f' changes from NEGATIVE to POSITIVE → LOCAL MINIMUM
• If f' changes from POSITIVE to NEGATIVE → LOCAL MAXIMUM
• If f' has the SAME SIGN on both sides → INFLECTION POINT

EXAMPLE: f(x) = x⁴ at x = 0

Step 1: f'(x) = 4x³, so f'(0) = 0 ✓

Step 2: f''(x) = 12x², so f''(0) = 0 (inconclusive!)

Step 3: Use the first derivative test:

• Test x = -0.1: f'(-0.1) = 4(-0.1)³ = -0.004 < 0 (negative)

• Test x = +0.1: f'(+0.1) = 4(+0.1)³ = +0.004 > 0 (positive)

Conclusion: f' changes from negative to positive, so x = 0 is a local minimum.

EXAMPLE: f(x) = x³ at x = 0

Step 1: f'(x) = 3x², so f'(0) = 0 ✓

Step 2: f''(x) = 6x, so f''(0) = 0 (inconclusive!)

Step 3: Use the first derivative test:

• Test x = -0.1: f'(-0.1) = 3(-0.1)² = +0.03 > 0 (positive)

• Test x = +0.1: f'(+0.1) = 3(+0.1)² = +0.03 > 0 (positive)

Conclusion: f' stays positive on both sides, so x = 0 is an inflection point (not an extremum).

Method 2: Behaviour of f ''(x) near f ''(x) = 0

If f'(a) = 0 and f''(a) = 0 then

Once you have determined f''(a) = 0, examine values around a.

• If f''(x) > 0 on both sides of x = a → concave up (minimum)
• If f''(x) < 0 on both sides of x = a → concave down (maximum)
• If f''(x) changes sign on either side of x = a → point of inflection

EXAMPLE: f(x) = x⁴ at x = 0

f'(x) = 4x³, so f'(0) = 0 ✓

f''(x) = 12x², so f''(0) = 0

Check f''(x) near x = 0:

• At x = -0.1: f''(-0.1) = 12(-0.1)² = 0.12 > 0

• At x = +0.1: f''(+0.1) = 12(+0.1)² = 0.12 > 0

Conclusion: f''(x) > 0 on both sides, so the curve is concave up near x = 0, making it a local minimum.

EXAMPLE: f(x) = x³ at x = 0

f'(x) = 3x², so f'(0) = 0 ✓

f''(x) = 6x, so f''(0) = 0

Check f''(x) near x = 0:

• At x = -0.1: f''(-0.1) = 6(-0.1) = -0.6 < 0

• At x = +0.1: f''(+0.1) = 6(+0.1) = +0.6 > 0

Conclusion: f''(x) changes sign from negative to positive, so x = 0 is a point of inflection.

Method 3: Function value around f '(a) = 0

For some value of the function at x = a, f(a), and function values close to, and around this stationary point (small h > 0):

If f(a - h) > f(a) and f(a + h) > f(a), then points around x = a are 'above' f(a) so we have a local minimum at x = a.

Similarly, if f(a - h) < f(a) and f(a + h) < f(a), then points around x = a are 'below' f(a) so we have a local maximum at x = a.

Finally, if the sign of f(a - h) - f(a) and f(a + h) - f(a) are different, then we have a (horizontal) point of inflection at x = a.

When in doubt, evaluate the function itself at points near the critical point. This direct approach shows whether nearby values of the function are higher or lower than the value of the function at the critical point.

VALUES TABLE: f(x) = -x⁴

Critical point at x = 0 with f'(0) = 0 and f''(0) = 0

x -0.5 -0.2 -0.1 0 0.1 0.2 0.5
f(x) -0.0625 -0.0016 -0.0001 0 -0.0001 -0.0016 -0.0625

Conclusion: All nearby values are less than f(0) = 0, so x = 0 is a local maximum.

Method 4: using technology to sketch graph

Sometimes the best approach is simply to sketch the function or use graphing technology. Visual inspection can quickly reveal the nature of a critical point.

Visual Clues:

  • Does the graph have a valley at x = a? → Minimum
  • Does the graph have a peak at x = a? → Maximum
  • Does the graph pass through x = a without turning? → Inflection point

Comparison of Methods

WHEN TO USE EACH METHOD

First Derivative Test: Best general-purpose method. Works for all functions and gives clear results. Choose this when derivatives are easy to evaluate.

Sign Chart: Best for systematic analysis and when you want to understand the complete behavior of f' around the critical point. Great for exam solutions.

Table of Values: Best when derivatives are complicated or when you want numerical confirmation. Also useful for approximate/numerical analysis.

Graphical Analysis: Best for intuition and quick checks. Use technology to visualize complex functions.

Complete Worked Example

PROBLEM: Classify the critical point of f(x) = x⁵ - 5x⁴ at x = 0

Verify it's a critical point:

f'(x) = 5x⁴ - 20x³ = 5x³(x - 4)

f'(0) = 0 ✓

Try second derivative test:

f''(x) = 20x³ - 60x²

f''(0) = 0 (inconclusive!)

Solution using First Derivative Test:

f'(x) = 5x³(x - 4)

• At x = -0.1: f'(-0.1) = 5(-0.1)³(-0.1 - 4) = 5(-0.001)(-4.1) ≈ 0.02 > 0

• At x = +0.1: f'(+0.1) = 5(+0.1)³(+0.1 - 4) = 5(+0.001)(-3.9) ≈ -0.02 < 0

Conclusion: f' changes from positive to negative, so x = 0 is a local maximum.

f ''(x) = 0, again

But it's not a stationary point

A different kind of inflection point

We have explored what happens when both f'(a) = 0 and f''(a) = 0 at a point x = a. But what if only f''(a) = 0 while f'(a) ≠ 0?

Key Distinction: When f''(a) = 0 but f'(a) ≠ 0, we still have a point of inflection where concavity changes, but the tangent line is NOT horizontal. The function continues increasing or decreasing through the inflection point.

This is actually the typical case for inflection points! Points where the tangent is horizontal (f'(a) = 0) are special exceptions.

Definition

A point of inflection occurs at x = a if:

1. The function is continuous at x = a
2. The concavity changes at x = a

This typically means f''(a) = 0 AND f''(x) changes sign at x = a

Note: f'(a) can be ANY value (zero or non-zero)

The condition f'(a) = 0 is NOT required for an inflection point. When f'(a) ≠ 0, the inflection point has a sloped tangent line.

Classic Examples

EXAMPLE 1: f(x) = x³ - 3x

First derivative: f'(x) = 3x² - 3

Second derivative: f''(x) = 6x

At x = 0:

• f'(0) = 3(0)² - 3 = -3 ≠ 0 (NOT a stationary point)

• f''(0) = 6(0) = 0 ✓

• For x < 0: f''(x) < 0 (concave down)

• For x > 0: f''(x) > 0 (concave up)

Conclusion: x = 0 is an inflection point where the curve changes from concave down to concave up, while continuing to decrease (since f'(0) = -3 < 0).

EXAMPLE 2: f(x) = sin(x) at x = 0, π, 2π, ...

First derivative: f'(x) = cos(x)

Second derivative: f''(x) = -sin(x)

At x = π:

• f'(π) = cos(π) = -1 ≠ 0 (NOT a stationary point)

• f''(π) = -sin(π) = 0 ✓

• For x slightly less than π: f''(x) > 0 (concave up)

• For x slightly more than π: f''(x) < 0 (concave down)

Conclusion: x = π is an inflection point where concavity changes, and the function is decreasing throughout (f'(π) < 0).

EXAMPLE 3: f(x) = (x - 2)³ + 5

First derivative: f'(x) = 3(x - 2)²

Second derivative: f''(x) = 6(x - 2)

At x = 2:

• f'(2) = 3(2 - 2)² = 0 (This IS a stationary point!)

• f''(2) = 6(2 - 2) = 0 ✓

Note: This is the special case where f'(a) = 0 AND f''(a) = 0 (horizontal inflection point, like x³ shifted right and up).

But consider x = 3:

• f'(3) = 3(3 - 2)² = 3 ≠ 0

• f''(3) = 6(3 - 2) = 6 ≠ 0 (no inflection here)

Visual characteristics

At an inflection point where f'(a) ≠ 0, you'll observe:

  • The curve "straightens out" momentarily as it passes through the point
  • The tangent line crosses the curve (rather than just touching it) i.e. the tangent changes from being above the curve to below it (or vice versa), either side of the inflection point
  • The curve changes from bending one way to bending the other
  • The function continues in the same direction (keeps increasing or keeps decreasing)
  • If the curve represents a road we are looking down on, and we are turning the steering wheel one way to go round a bend, the inflection point is where we stop turning one way and start turning the other - at the inflection point the steering is (for an instant) pointing straight ahead, not turning
Contrast with f'(a) = 0: When f'(a) = 0 at an inflection point, the tangent is horizontal and the curve has a local "flat spot." When f'(a) ≠ 0, there's no flat spot—the curve just smoothly changes its curvature while maintaining its direction.

Finding all inflection points on a curve

To find ALL inflection points of a function (both types):

Step-by-Step Procedure:

1. Find f''(x)
2. Solve f''(x) = 0 to find candidates
3. For each candidate x = a, check if f''(x) changes sign:
   • Test f''(x) at points just left and right of a
   • If signs differ → inflection point ✓
   • If signs same → NOT an inflection point ✗
4. Also check points where f''(x) is undefined (if any)

Note: You do NOT need to check f'(a) unless you specifically want to know if the tangent is horizontal

COMPLETE ANALYSIS: f(x) = x⁴ - 4x³

Step 1: f'(x) = 4x³ - 12x² = 4x²(x - 3)

Step 2: f''(x) = 12x² - 24x = 12x(x - 2)

Step 3: Solve f''(x) = 0: 12x(x - 2) = 0 → x = 0 or x = 2

Test x = 0:

• f''(-0.1) = 12(-0.1)(-0.1 - 2) = 12(-0.1)(-2.1) = 2.52 > 0

• f''(+0.1) = 12(+0.1)(+0.1 - 2) = 12(+0.1)(-1.9) = -2.28 < 0

• Signs change! x = 0 IS an inflection point ✓

• Check: f'(0) = 0 (horizontal tangent at inflection point)

Test x = 2:

• f''(1.9) = 12(1.9)(1.9 - 2) = 12(1.9)(-0.1) = -2.28 < 0

• f''(2.1) = 12(2.1)(2.1 - 2) = 12(2.1)(+0.1) = 2.52 > 0

• Signs change! x = 2 IS an inflection point ✓

• Check: f'(2) = 4(2)²(2 - 3) = 16(-1) = -16 ≠ 0 (sloped tangent!)

Summary: Two inflection points—one with horizontal tangent (x = 0), one with sloped tangent (x = 2).

Common Mistakes to Avoid

MISTAKE 1: Assuming all points where f''(x) = 0 are inflection points

Counterexample: f(x) = x⁴ has f''(0) = 0, but x = 0 is a minimum, not an inflection point. You must verify that f''(x) changes sign.

MISTAKE 2: Thinking inflection points must have f'(x) = 0

Reality: Most inflection points have f'(x) ≠ 0. The horizontal tangent case is the exception, not the rule.

MISTAKE 3: Confusing "change in direction" with "change in concavity"

Clarification: At an inflection point, the concavity changes (the curve bends differently), but the direction may continue unchanged (function keeps increasing or decreasing).