Stationary points, concavity and curve sketching

The derivatives of a function describe the shape of its graph. The first derivative gives the gradient, while the second derivative describes how that gradient changes. Together, they let you locate turning points, distinguish maxima from minima, identify points of inflection and produce a reliable curve sketch without plotting many coordinates.

Before you start

You should be able to differentiate standard functions and apply the product, quotient and chain rules. It also helps to be confident solving polynomial equations and inequalities.

Increasing, decreasing and stationary

For a differentiable function (y=f(x)):

[ \begin{aligned} f’(x)&>0 &&\Rightarrow \text{the function is increasing},\ f’(x)&<0 &&\Rightarrow \text{the function is decreasing},\ f’(x)&=0 &&\Rightarrow \text{the point is stationary}. \end{aligned} ]

A stationary point has a horizontal tangent. It may be a local maximum, a local minimum or a stationary point of inflection. The equation (f’(x)=0) locates candidates, but does not classify them.

Worked example: increasing and decreasing intervals

Let

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

Differentiate and factorise:

[ f’(x)=3x^2-6x-9=3(x+1)(x-3). ]

The critical values are (x=-1) and (x=3). Test one value in each interval:

Interval	Test value	Sign of (f’(x))	Behaviour
(x<-1)	(-2)	positive	increasing
(-1<x<3)	(0)	negative	decreasing
(x>3)	(4)	positive	increasing

Therefore (f) is increasing on (( -\infty,-1)) and ((3,\infty)), and decreasing on ((-1,3)).

At (x=-1), the derivative changes from positive to negative, so there is a local maximum. At (x=3), it changes from negative to positive, so there is a local minimum.

Substitute into the original function, not the derivative:

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

The stationary points are a local maximum at ((-1,10)) and a local minimum at ((3,-22)).

The first derivative sign test

The first derivative test is the most dependable classification method because it checks what the curve actually does on either side of a stationary point.

Change in (f’(x))	Classification
positive to negative	local maximum
negative to positive	local minimum
no sign change	stationary point of inflection, provided concavity changes

Worked example: a stationary point that is not a turning point

For (f(x)=x^3),

[ f’(x)=3x^2. ]

The only stationary value is (x=0). Since (3x^2>0) on both sides of zero, the function is increasing before and after the stationary point. It does not turn. The point ((0,0)) is a stationary point of inflection because the graph changes from bending downwards to bending upwards there.

Compare this with (g(x)=x^4). Here (g’(x)=4x^3), which changes from negative to positive at zero. Thus ((0,0)) is a local minimum, even though both examples have a zero second derivative at zero.

The second derivative test

The second derivative is

[ f”(x)=\frac{d}{dx}\left(f’(x)\right). ]

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

[ \begin{aligned} f”(a)&>0 &&\Rightarrow \text{local minimum},\ f”(a)&<0 &&\Rightarrow \text{local maximum},\ f”(a)&=0 &&\Rightarrow \text{test is inconclusive}. \end{aligned} ]

When (f”(a)=0), use the first derivative sign test or inspect concavity on both sides. Never conclude automatically that the point is an inflection.

Worked example: classifying with (f”(x))

Consider

[ f(x)=2x^3-9x^2+12x+1. ]

First find the stationary values:

[ f’(x)=6x^2-18x+12=6(x-1)(x-2). ]

Hence (x=1) or (x=2). Now differentiate again:

[ f”(x)=12x-18. ]

At (x=1), (f”(1)=-6<0), so ((1,6)) is a local maximum. At (x=2), (f”(2)=6>0), so ((2,5)) is a local minimum.

Notice that the (y)-coordinates came from (f(1)) and (f(2)).

Concavity, convexity and points of inflection

The sign of the second derivative describes the direction in which the curve bends:

[ \begin{aligned} f”(x)&>0 &&\Rightarrow \text{the gradient is increasing and the curve bends upwards},\ f”(x)&<0 &&\Rightarrow \text{the gradient is decreasing and the curve bends downwards}. \end{aligned} ]

In this guide, a curve with (f”(x)>0) is called convex or concave upwards, while a curve with (f”(x)<0) is called concave or concave downwards. Textbooks vary in their use of the words concave and convex, so the sign of (f”) is the safest description.

A point of inflection is a point where the concavity changes. A solution of (f”(x)=0) is only a candidate. You must show that (f”(x)) changes sign, or otherwise show that the direction of bending changes.

Worked example: a non-stationary point of inflection

Return to

[ f(x)=2x^3-9x^2+12x+1. ]

Since (f”(x)=12x-18), the candidate is

[ 12x-18=0 \quad\Rightarrow\quad x=\frac32. ]

For (x<\frac32), (f”(x)<0). For (x>\frac32), (f”(x)>0). Concavity changes, so there is a point of inflection. Its coordinates are

[ \left(\frac32,f\left(\frac32\right)\right) =\left(\frac32,\frac{11}{2}\right). ]

Also,

[ f’\left(\frac32\right)=-\frac32\ne0, ]

so this point of inflection is not stationary.

Counterexample: why (f”(a)=0) is insufficient

For (f(x)=x^4),

[ f”(x)=12x^2. ]

Although (f”(0)=0), the second derivative is positive on both sides of zero. Concavity does not change, so ((0,0)) is not a point of inflection. It is a local minimum.

Sketching a curve systematically

A good sketch displays structure, not artistic precision. Work through the following evidence.

1. Domain: identify excluded values, endpoints and restrictions.
2. Intercepts: solve (f(x)=0) for (x)-intercepts and calculate (f(0)) for the (y)-intercept.
3. Symmetry: check whether (f(-x)=f(x)) for even symmetry or (f(-x)=-f(x)) for odd symmetry.
4. End behaviour: inspect the leading term or limits as (x\to\pm\infty).
5. Asymptotes: identify any vertical, horizontal or oblique asymptotes where relevant.
6. Stationary points: solve (f’(x)=0), find coordinates and classify each point.
7. Concavity and inflection: solve (f”(x)=0) and verify sign changes.
8. Join consistently: make the curve obey every sign interval and important feature.

Full worked example: sketching a quartic

Sketch

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

Intercepts and symmetry

[ f(x)=x^2(x-2)(x+2), ]

so the graph meets the (x)-axis at (x=-2,0,2). The root at zero is repeated, so the curve touches rather than crosses the axis there. Since (f(-x)=f(x)), the graph is symmetric about the (y)-axis.

End behaviour

The leading term is (x^4), so (f(x)\to\infty) as (x\to\pm\infty).

Stationary points

[ f’(x)=4x^3-8x=4x(x^2-2). ]

Thus (x=-\sqrt2,0,\sqrt2). The second derivative is

[ f”(x)=12x^2-8. ]

At (x=0), (f”(0)=-8<0), giving a local maximum at ((0,0)). At (x=\pm\sqrt2), (f”(\pm\sqrt2)=16>0), giving local minima at

[ (-\sqrt2,-4)\quad\text{and}\quad(\sqrt2,-4). ]

Points of inflection

[ 12x^2-8=0 \quad\Rightarrow\quad x=\pm\sqrt{\frac23}. ]

The sign of (f”) changes at both values. Since

[ f\left(\pm\sqrt{\frac23}\right) =\frac49-\frac83 =-\frac{20}{9}, ]

the points of inflection are

[ \left(-\sqrt{\frac23},-\frac{20}{9}\right) \quad\text{and}\quad \left(\sqrt{\frac23},-\frac{20}{9}\right). ]

Your sketch should now show a symmetric W-shaped curve: rising at both ends, crossing at (x=-2) and (x=2), touching the axis at the central maximum, and reaching the two minima at height (-4).

A parameter problem

Suppose

[ f(x)=x^3-3kx ]

where (k) is a constant. Determine when the curve has two distinct stationary points.

Differentiate:

[ f’(x)=3x^2-3k=3(x^2-k). ]

Stationary points satisfy (x^2=k). There are two distinct real solutions, (x=\pm\sqrt{k}), precisely when

[ k>0. ]

If (k=0), there is one stationary point at zero. If (k<0), there are no real stationary points. This style of question tests whether you can interpret the number of roots of the derivative, rather than merely differentiate.

Common mistakes

• Solving (f’(x)=0) and reporting only (x)-values. A stationary point needs both coordinates.
• Substituting into (f’(x)) to find the (y)-coordinate. Substitute into the original function.
• Calling every stationary point a turning point. The graph of (x^3) shows why this fails.
• Claiming that (f”(a)=0) proves an inflection. It only supplies a candidate.
• Using the second derivative test without first checking (f’(a)=0).
• Giving increasing and decreasing intervals without testing the sign of (f’).
• Drawing a graph that contradicts its derivative sign table or end behaviour.
• Assuming a repeated root always has multiplicity two. Even multiplicity touches the axis, while odd multiplicity crosses it, possibly with a flat tangent.

Check your understanding

1. Classify two stationary points

For (f(x)=x^3-6x^2+9x+1), find and classify all stationary points.

Answer

[ f’(x)=3(x-1)(x-3),\qquad f”(x)=6x-12. ]

At (x=1), (f”(1)=-6), so ((1,5)) is a local maximum. At (x=3), (f”(3)=6), so ((3,1)) is a local minimum.

2. Test an apparent inflection

Does (f(x)=x^6) have a point of inflection at the origin?

Answer

No. Although (f”(0)=0),

[ f”(x)=30x^4\ge0 ]

on both sides of zero. There is no change in concavity.

3. Find an inflection point

Find the point of inflection of (f(x)=x^3+3x^2-9x), and state whether it is stationary.

Answer

[ f”(x)=6x+6=0 \Rightarrow x=-1. ]

The second derivative changes sign at (-1), and (f(-1)=11), so the point is ((-1,11)). Since

[ f’(-1)=3-6-9=-12\ne0, ]

it is not stationary.

4. Interpret a derivative

If (f’(x)=(x+2)^2(x-1)), on which intervals is (f) increasing?

Answer

The squared factor is non-negative and does not change sign at (x=-2). The sign is therefore controlled by (x-1). The function is increasing for (x>1) and decreasing for (x<1), apart from the isolated stationary value at (x=-2). Thus the maximal open increasing interval is ((1,\infty)).

What to learn next

Stationary points are closely connected to tangents and normals. When a curve is defined implicitly or parametrically, use implicit differentiation or parametric differentiation before applying the same classification methods.