IB Mathematics: Analysis and Approaches SL/HL — Topic 5 Calculus

Stationary Points and Their Nature

Locate stationary points analytically, then classify maxima, minima and horizontal points of inflexion using first- and second-derivative tests.

AA SL/HL · SL 5.8

Learning goal

Find stationary points and justify their nature using first-derivative sign analysis or the second-derivative test.

Syllabus link

IB Mathematics AA SL/HL: SL 5.8. Stationary points, local extrema, horizontal points of inflexion and derivative tests.

Big idea

Finding \(f'(x)=0\) locates a stationary point; classification is a separate step.

Key relationship

At a stationary point \(x=a\), \(f'(a)=0\).

What is a stationary point?

A stationary point is a point on the graph where the tangent is horizontal. Therefore its derivative is zero.

Definition

\[\boxed{f'(a)=0}.\]

The full stationary point is \((a,f(a))\), not just the \(x\)-value.

Four stationary point types: local maximum, local minimum, increasing horizontal point of inflexion and decreasing horizontal point of inflexion — A stationary point can be a turning point or a horizontal point of inflexion.

Stationary point versus turning point

A local maximum or minimum is a turning point. A horizontal point of inflexion is stationary but is not a turning point, because the function keeps moving in the same overall direction.

We will build the method in stages: locate the coordinates, use first-derivative signs to classify them, then introduce the second derivative as a quicker test for maxima and minima. Optimisation is treated in the next lesson.

How to find stationary points

Three-step locating method

Differentiate to find \(f'(x)\).
Solve \(f'(x)=0\) to find every stationary \(x\)-value.
Substitute each value into \(f(x)\) to find the full coordinates.

Only then classify each point.

Worked example 1: finding a stationary point

Find the stationary point of \(f(x)=2x^2-8x+7\).

Solution

Step 1: differentiate.

\[f'(x)=4x-8.\]

Step 2: solve \(f'(x)=0\).

\[4x-8=0\quad\Rightarrow\quad x=2.\]

Step 3: find the corresponding y-coordinate.

\[\begin{aligned}f(2)&=2(2)^2-8(2)+7\\&=8-16+7\\&=-1.\end{aligned}\]

Therefore the stationary point is \(\boxed{(2,-1)}\).

We have located the point, but we have not yet determined whether it is a maximum, a minimum or a horizontal point of inflexion. To do that, study how the function behaves on either side of the stationary value.

TI-Nspire: generate f′ and solve f′(x)=0

For calculator-allowed work, the TI-Nspire CX can generate \(y=f'(x)\) directly from \(y=f(x)\) and solve \(f'(x)=0\). Analytical differentiation remains essential; this is a useful check and a method for technology-based questions.

Watch this tutorial on YouTube

Stationary points from the graph of f′

Read stationary values from the graph of the derivative, whether the question supplies \(f'\) graphically or you generate its graph with technology. Substitute into \(f\) for the corresponding y-coordinates.

Watch this tutorial on YouTube

The first derivative sign test

The sign of \(f'(x)\) tells us the direction of motion of the function:

\[\begin{gathered}\begin{gathered}f'(x)>0\\\Rightarrow f\text{ increasing}\end{gathered}\\\begin{gathered}f'(x)<0\\\Rightarrow f\text{ decreasing}.\end{gathered}\end{gathered}\]

Local maximum

\(+\to0\to-\): increasing then decreasing.

Local minimum

\(-\to0\to+\): decreasing then increasing.

No sign change

\(+\to0\to+\) or \(-\to0\to-\): horizontal tangent without turning.

Local maximum and minimum, and increasing and decreasing horizontal inflexions, with the derivative signs on each side. — The first derivative test studies the sign immediately before and after the stationary value.

Standard stationary-point sign tables

First derivative sign table
\(x\)	\(-\infty\)	\(a\)	\(+\infty\)
\(f'(x)\)	+	0	−
\(f(x)\)		Max

First derivative sign table
\(x\)	\(-\infty\)	\(a\)	\(+\infty\)
\(f'(x)\)	−	0	+
\(f(x)\)		Min

First derivative sign table
\(x\)	\(-\infty\)	\(a\)	\(+\infty\)
\(f'(x)\)	+	0	+
\(f(x)\)

No sign change: decreasing on both sides
\(x\)	\(-\infty\)	\(a\)	\(+\infty\)
\(f'(x)\)	−	0	−
\(f(x)\)

Constructing a full sign table

You may already have met sign tables when solving inequalities in algebra. Here the same process is applied to \(f'(x)\).

Factor \(f'(x)\) completely, including the denominator when necessary.
Put each zero and excluded value in the top \(x\)-row.
Add one sign row for each factor, then combine the signs to obtain the \(f'(x)\)-row.
Use the tall \(f(x)\)-row to show where the function increases or decreases and hence classify each stationary point.

Use a double vertical guide at a vertical asymptote and split the variation arrows into separate branches. A zero is aligned directly below its \(x\)-value; signs lie between the guides and the table stays open at the bottom.

First derivative sign tables

A full lesson on constructing and interpreting sign tables.

Watch this tutorial on YouTube

Applying the first derivative test

Worked example 2: classify using the first derivative sign test

Find and classify all stationary points of

\[f(x)=x^3-3x^2-9x+5.\]

Solution

1. Find the stationary \(x\)-values.

\[\begin{aligned}f'(x)&=3x^2-6x-9\\&=3(x+1)(x-3).\end{aligned}\]

\[\begin{gathered}f'(x)=0\\\Rightarrow x=-1\text{ or }x=3.\end{gathered}\]

2. Find the coordinates.

\[f(-1)=10,\qquad f(3)=-22.\]

So the stationary points are \((-1,10)\) and \((3,-22)\).

3. Determine the sign of \(f'(x)\). Use the fully factored form and show how the signs of the factors combine.

First derivative sign table: -1 → 3
\(x\)	\(-\infty\)	\(-1\)		\(3\)	\(+\infty\)
\(3\)	+		+		+
\(x+1\)	−	0	+		+
\(x-3\)	−		−	0	+
\(f'(x)\)	+	0	−	0	+
\(f(x)\)		Max		Min

At \(x=-1\), \(f'(x)\) changes from positive to negative, so \((-1,10)\) is a local maximum.

At \(x=3\), \(f'(x)\) changes from negative to positive, so \((3,-22)\) is a local minimum.

Graph of x cubed minus 3 x squared minus 9 x plus 5 with a local maximum at negative 1 comma 10 and a local minimum at 3 comma negative 22 — Graphical check for Worked Example 2: the cubic has a local maximum at \((-1,10)\) and a local minimum at \((3,-22)\).

Classifying two stationary points

This tutorial uses \(f(x)=x^3+6x^2-15x+3\): differentiate, factor \(f'\), find the coordinates, construct separate factor rows, and classify the local maximum and minimum.

Watch this tutorial on YouTube

Extending the method: a vertical asymptote

Worked example 3: factor rows and two separate branches

Find and classify the stationary points of \(f(x)=3x+\frac{12}{x}\), where \(x\ne0\).

Solution

\[\begin{aligned}f'(x)&=3-\frac{12}{x^2}\\&=\frac{3x^2-12}{x^2}\\&=\frac{3(x+2)(x-2)}{x^2}.\end{aligned}\]

The stationary values are \(-2\) and \(2\), while \(x=0\) is a vertical asymptote and must be included as a domain break, not a stationary point.

First derivative sign table: -2 → 0 → 2
\(x\)	\(-\infty\)	\(-2\)		\(0\)		\(2\)	\(+\infty\)
\(3\)	+		+		+		+
\(x+2\)	−	0	+		+		+
\(x-2\)	−		−		−	0	+
\(x^2\)	+		+	0	+		+
\(f'(x)\)	+	0	−	undefined	−	0	+
\(f(x)\)		Max				Min

Since \(f(-2)=-12\) and \(f(2)=12\), \((-2,-12)\) is a local maximum and \((2,12)\) is a local minimum.

Sign table with a vertical asymptote

Work through \(f(x)=3x+12/x\), factor \(f'\), mark the vertical asymptote with a double guide, and classify both stationary points.

Watch this tutorial on YouTube

When the sign does not change: horizontal points of inflexion

What must be shown?

At an isolated stationary point \(x=a\), first establish \(f'(a)=0\). If \(f'\) has the same sign immediately before and after \(a\), the function continues in the same direction and the point is not a turning point.

For the standard examples below, \(+\to0\to+\) gives an increasing horizontal point of inflexion and \(-\to0\to-\) gives a decreasing one. The graphs and the sign changes in \(f''\) confirm the change of concavity. Zero gradient alone does not prove an inflexion.

Worked example 4: an increasing horizontal point of inflexion

Let

\[f(x)=(x-1)^3+2.\]

Solution

\[f'(x)=3(x-1)^2.\]

Thus \(f'(x)=0\) only when \(x=1\), and \(f(1)=2\). The stationary point is \((1,2)\). Since \(3(x-1)^2>0\) on both sides of \(x=1\), the function is increasing before and after the point.

First derivative sign table: 1
\(x\)	\(-\infty\)	\(1\)	\(+\infty\)
\(3\)	+		+
\((x-1)^2\)	+	0	+
\(f'(x)\)	+	0	+
\(f(x)\)

The sign pattern is \(+\to0\to+\), so

Answer \((1,2)\) is an increasing horizontal point of inflexion.

Optional confirmation using \(f''\). Since \(f''(x)=6(x-1)\), the second derivative changes from negative to positive at \(x=1\), confirming a change of concavity.

Graph of f of x equals open parenthesis x minus 1 close parenthesis cubed plus 2 with an increasing horizontal point of inflexion at 1 comma 2 — The graph remains increasing through the horizontal tangent at \((1,2)\).

Worked example 5: a decreasing horizontal point of inflexion

Let

\[g(x)=-(x+2)^3+1.\]

Solution

\[g'(x)=-3(x+2)^2.\]

Thus \(g'(x)=0\) only when \(x=-2\), and \(g(-2)=1\). The stationary point is \((-2,1)\). Since \(-3(x+2)^2<0\) on both sides of \(x=-2\), the function is decreasing before and after the point.

First derivative sign table: -2
\(x\)	\(-\infty\)	\(-2\)	\(+\infty\)
\(-3\)	−		−
\((x+2)^2\)	+	0	+
\(g'(x)\)	−	0	−
\(g(x)\)

The sign pattern is \(-\to0\to-\), so

Answer \((-2,1)\) is a decreasing horizontal point of inflexion.

Optional confirmation using \(g''\). Since \(g''(x)=-6(x+2)\), \(g''(x)>0\) for \(x<-2\) and \(g''(x)<0\) for \(x>-2\). The concavity therefore changes from concave up to concave down.

Graph of g of x equals negative open parenthesis x plus 2 close parenthesis cubed plus 1 with a decreasing horizontal point of inflexion at negative 2 comma 1 — The graph remains decreasing through the horizontal tangent at \((-2,1)\).

A quicker alternative: the second derivative test

The second derivative describes concavity: \(f''(x)>0\) means concave up, and \(f''(x)<0\) means concave down. At a stationary point this lets us classify a local maximum or minimum from a single value of \(f''\).

Second derivative test

At a stationary point \(x=a\), where \(f'(a)=0\):

if \(f''(a)>0\), the point is a local minimum;
if \(f''(a)<0\), the point is a local maximum;
if \(f''(a)=0\), the test is inconclusive — use first-derivative signs (and concavity if needed).

Second Derivative Test

Apply the second derivative test to \(f(x)=2x^3-21x^2+60x+8\). Solve \(f'(x)=0\), find \((2,60)\) and \((5,33)\), then classify the first as a local maximum and the second as a local minimum.

Watch this tutorial on YouTube

Worked example 6: three stationary points using the second derivative test

Find and classify all stationary points of

\[f(x)=x^4-4x^2.\]

Solution

First,

\[f'(x)=4x^3-8x=4x(x^2-2).\]

Hence

\[\begin{gathered}f'(x)=0\\\Rightarrow x=-\sqrt2,\ 0,\ \sqrt2.\end{gathered}\]

The corresponding coordinates are

\[\begin{gathered}f(-\sqrt2)=-4\\f(0)=0\\f(\sqrt2)=-4.\end{gathered}\]

Now differentiate again:

\[f''(x)=12x^2-8.\]

At \(x=-\sqrt2\), \(f''(-\sqrt2)=16>0\), so \((-\sqrt2,-4)\) is a local minimum.

At \(x=0\), \(f''(0)=-8<0\), so \((0,0)\) is a local maximum.

At \(x=\sqrt2\), \(f''(\sqrt2)=16>0\), so \((\sqrt2,-4)\) is a local minimum.

Graph of x to the fourth minus 4 x squared with local minima at negative square root 2 comma negative 4 and square root 2 comma negative 4, and a local maximum at 0 comma 0 — The second derivative is positive at \(x=\pm\sqrt2\) and negative at \(x=0\), distinguishing the two minima from the central maximum.

Second derivative test: x ln x

The method also applies to non-polynomial functions. For \(f(x)=x\ln x\), \(x>0\), apply the product rule, solve \(f'(x)=0\), find exact coordinates and use \(f''\) to establish a local minimum.

Watch this tutorial on YouTube

Why \(f''(a)=0\) is inconclusive

All three functions satisfy \(f'(0)=0\) and \(f''(0)=0\), but their stationary points have different natures:

\(f(x)=x^4\): local minimum at \((0,0)\);
\(f(x)=x^3\): increasing horizontal point of inflexion at \((0,0)\);
\(f(x)=-x^4\): local maximum at \((0,0)\).

Therefore \(f''(a)=0\) does not mean “point of inflexion”. Return to the first derivative sign test; use the graph or concavity to confirm an inflexion when needed.

When f″(a)=0

Compare \(f(x)=x^3-6x^2+12x-5\) and \(f(x)=x^4+1\). In each case the second derivative test is inconclusive; a first-derivative sign table distinguishes an increasing horizontal inflexion from a local minimum.

Watch this tutorial on YouTube

Bringing the methods together

Both classification methods begin with the same locating step: solve \(f'(x)=0\) and find the coordinates.

Decision map for classifying a stationary point using the second derivative then first derivative signs if necessary — Use \(f''(a)\) for a quick decision when it is non-zero; if it is zero, return to the first-derivative sign test.

Decision process

Solve \(f'(x)=0\) and find every stationary coordinate.
Evaluate \(f''(a)\) if convenient.
If \(f''(a)>0\): minimum. If \(f''(a)<0\): maximum.
If \(f''(a)=0\): inconclusive. Study signs of \(f'\) on either side.
If the sign of \(f'\) does not change, the stationary point is not a turning point; classify the horizontal point of inflexion using the established behaviour/concavity.

Common mistakes

Giving only the stationary x-value instead of the full coordinate.
Saying “maximum” or “minimum” without a valid derivative test.
Assuming every solution of \(f''(x)=0\) is a point of inflexion.
Assuming every stationary point is a turning point.
Failing to distinguish increasing and decreasing horizontal points of inflexion.
Skipping factor rows, so the sign of a product or quotient is not justified.
Treating a vertical asymptote as a stationary point instead of marking a double guide.

Practice questions

Instructions

For each question, show how the stationary points are found and justify their nature. Use the requested method where specified. Whenever you construct a first derivative sign table, factor the derivative and include a separate row for each factor. Give exact coordinates unless told otherwise.

Find and classify the stationary point of \(f(x)=3x^2-12x+5\). Use the second derivative test.
Find and classify all stationary points of \(f(x)=x^3-6x^2+9x+1\). Use a full first derivative sign table, including factor rows.
Find and classify all stationary points of \(f(x)=x^4-2x^2\). Use the second derivative test.
Show that \(f(x)=(x-2)^3+8\) has an increasing horizontal point of inflexion. State its coordinates.
Show that \(g(x)=-2(x+1)^3+4\) has a decreasing horizontal point of inflexion. State its coordinates.
For \(f(x)=x^4\), the second derivative test is inconclusive at \(x=0\). Use the first derivative sign test to classify the stationary point.
For \(f(x)=-x^4\), the second derivative test is inconclusive at \(x=0\). Use the first derivative sign test to classify the stationary point.
Find and classify all stationary points of \(f(x)=x^4-8x^2+3\).
Find and classify all stationary points of \(f(x)=\frac14x^4-x^3\). Distinguish between a turning point and a horizontal point of inflexion.
Find and classify all stationary points of \(f(x)=x^5-5x^3\). Use a combination of first- and second-derivative reasoning.

Answer key

1. Quadratic

\[f'(x)=6x-12=0\Rightarrow x=2.\]

\[f(2)=3(2)^2-12(2)+5=-7.\]

\[f''(x)=6>0.\]

Therefore \((2,-7)\) is a local minimum.

2. Cubic: first derivative test

\[\begin{aligned}f'(x)&=3x^2-12x+9\\&=3(x-1)(x-3).\end{aligned}\]

The stationary x-values are 1 and 3. Substitution gives \(f(1)=1-6+9+1=5\) and \(f(3)=27-54+27+1=1\).

Factor rows and classification for question 2
\(x\)	\(-\infty\)	\(1\)		\(3\)	\(+\infty\)
\(3\)	+		+		+
\(x-1\)	−	0	+		+
\(x-3\)	−		−	0	+
\(f'(x)\)	+	0	−	0	+
\(f(x)\)		Max		Min

At \(x=1\), \(f'\) changes from positive to negative: \((1,5)\) is a local maximum. At \(x=3\), it changes from negative to positive: \((3,1)\) is a local minimum.

3. Quartic: second derivative test

\[\begin{aligned}f'(x)&=4x^3-4x\\&=4x(x-1)(x+1).\end{aligned}\]

So \(x=-1,0,1\), giving points \((-1,-1),(0,0),(1,-1)\).

\[f''(x)=12x^2-4.\]

Since \(f''(\pm1)=8>0\), the points \((-1,-1)\) and \((1,-1)\) are local minima. Since \(f''(0)=-4<0\), \((0,0)\) is a local maximum.

4. Increasing horizontal point of inflexion

\[f'(x)=3(x-2)^2.\]

The only stationary x-value is 2 and \(f(2)=8\). The derivative is positive on both sides of 2, so the function remains increasing.

\[f''(x)=6(x-2).\]

This changes from negative to positive at 2, confirming concavity changes from down to up. Hence \((2,8)\) is an increasing horizontal point of inflexion.

5. Decreasing horizontal point of inflexion

\[g'(x)=-6(x+1)^2.\]

The stationary value is \(-1\), giving \(g(-1)=4\). The derivative is negative on both sides, so the function remains decreasing.

\[g''(x)=-12(x+1).\]

This changes from positive to negative at \(-1\). Hence \((-1,4)\) is a decreasing horizontal point of inflexion.

6. Positive quartic

\[f'(x)=4x^3=0\Rightarrow x=0.\]

Here \(f(0)=0\) and \(f''(0)=0\), so the second derivative test is inconclusive.

The sign of the first derivative changes from negative to positive at \(x=0\). Therefore \((0,0)\) is a local minimum.

7. Negative quartic

\[f'(x)=-4x^3=0\Rightarrow x=0.\]

Here \(f(0)=0\) and \(f''(0)=0\), so the second derivative test is inconclusive.

The sign of the first derivative changes from positive to negative at \(x=0\). Therefore \((0,0)\) is a local maximum.

8. Two minima and one maximum

\[\begin{aligned}f'(x)&=4x^3-16x\\&=4x(x-2)(x+2).\end{aligned}\]

Thus \(x=-2,0,2\). The stationary points are \((-2,-13),(0,3),(2,-13)\).

\[f''(x)=12x^2-16.\]

Since \(f''(\pm2)=32>0\), the two points \((-2,-13)\) and \((2,-13)\) are local minima. Since \(f''(0)=-16<0\), \((0,3)\) is a local maximum.

9. One horizontal point of inflexion and one minimum

\[f'(x)=x^3-3x^2=x^2(x-3).\]

So \(x=0,3\), giving \((0,0)\) and \((3,-27/4)\). At 0 the derivative is negative on both sides, so the function continues decreasing.

\[f''(x)=3x^2-6x=3x(x-2).\]

The second derivative changes from positive to negative at 0, confirming that \((0,0)\) is a decreasing horizontal point of inflexion. At 3 the first derivative changes from negative to positive, so \((3,-27/4)\) is a local minimum.

10. Mixed stationary points

\[f'(x)=5x^4-15x^2=5x^2(x^2-3).\]

Thus \(x=-\sqrt3,0,\sqrt3\). The coordinates are \((-\sqrt3,6\sqrt3)\), \((0,0)\), \((\sqrt3,-6\sqrt3)\).

\[\begin{aligned}f''(x)&=20x^3-30x\\&=10x(2x^2-3).\end{aligned}\]

At \(-\sqrt3\), \(f''=-30\sqrt3<0\), so \((-\sqrt3,6\sqrt3)\) is a local maximum. At \(\sqrt3\), \(f''=30\sqrt3>0\), so \((\sqrt3,-6\sqrt3)\) is a local minimum.

At 0 the second derivative test is inconclusive. Since \(f'<0\) on both sides of 0, the pattern is \(-\to0\to-\). Also \(f''\) changes from positive to negative near 0. Hence \((0,0)\) is a decreasing horizontal point of inflexion.

Summary

A stationary point satisfies \(f'(a)=0\). Calculate \(f(a)\) to find its full coordinate.
In a full first derivative sign table, factor \(f'\), show each factor row and combine the signs.
A double vertical guide marks a vertical asymptote and separates the branches.
At a stationary value, \(+\to-\) gives a local maximum and \(-\to+\) gives a local minimum.
At a stationary point, \(f''<0\) means concave down and a local maximum; \(f''>0\) means concave up and a local minimum.
If \(f''(a)=0\), return to the first derivative sign test.
For the standard examples in this lesson, \(+\to0\to+\) and \(-\to0\to-\) describe increasing and decreasing horizontal inflexions respectively; the graphs or concavity confirm their nature.