Introduction
The first derivative describes how a function changes. Its sign tells us whether the function is increasing or decreasing, but there is another question: is its rate of change itself increasing or decreasing?
That question leads to concavity. When \(f'\) increases as \(x\) increases, the slopes of \(f\) become larger, and the graph bends upward. When \(f'\) decreases, those slopes become smaller, and the graph bends downward. This connection is especially useful when finding or interpreting \(f''\) is difficult, or when \(f''\) does not exist.
Concavity and Chord Slopes
Let \(f\) be defined on an interval \(I\). For two distinct points \(a,b\in I\), the secant slope is
It measures the average change in \(f\) between the two points. For \(a<b<c\), the chord-slope description of concavity compares the slopes over the two adjacent intervals.
The inequalities are not required to be strict. A function can be concave up while being linear on part of its domain, for example. This distinction matters when we connect concavity to monotonicity: the relevant derivative condition is nondecreasing or nonincreasing, not necessarily strictly increasing or strictly decreasing.
The First-Derivative Criterion
Here is the key relationship. Assume \(I\) is an open interval and \(f\) is differentiable throughout \(I\). If \(f'\) is nondecreasing on \(I\), then \(f\) is concave up there. If \(f'\) is nonincreasing, then \(f\) is concave down. In fact, under these assumptions the implications work in reverse too.
A function \(g\) is nondecreasing on \(I\) if \(g(x)\le g(y)\) whenever \(x<y\) in \(I\); nonincreasing means \(g(x)\ge g(y)\) whenever \(x<y\). The theorem compares values of \(f'\), not the sign of \(f'\). Its justification uses the Mean Value Theorem: for a function continuous on a closed interval and differentiable in its interior, some interior derivative equals the secant slope across the interval.
Why the Criterion Works
First suppose that \(f'\) is nondecreasing. Choose any three points \(a<b<c\) in \(I\). Apply the Mean Value Theorem separately on \([a,b]\) and \([b,c]\). There are points \(u\in(a,b)\) and \(v\in(b,c)\) such that
Because \(u<v\) and \(f'\) is nondecreasing, \(f'(u)\le f'(v)\). The left secant slope is therefore no larger than the right secant slope, which is precisely the chord-slope condition for concavity up.
For the reverse implication, suppose \(f\) is concave up. Fix \(x<y\) in \(I\). For any \(h>0\) small enough that \(x+h<y\), the chord-slope condition gives
Let \(h\) decrease to zero. Since \(f\) is differentiable at \(x\), the left side tends to \(f'(x)\), so \(f'(x)\le [f(y)-f(x)]/(y-x)\). Applying the same chord-slope condition to \(x<y-h<y\), and then letting \(h\) decrease to zero, gives \([f(y)-f(x)]/(y-x)\le f'(y)\). Combining these inequalities yields
In particular, \(f'(x)\le f'(y)\). Since this holds for every \(x<y\) in \(I\), \(f'\) is nondecreasing. Reversing the inequalities proves the concave-down case as well. This establishes both directions of the criterion.
Worked Examples
Worked Example: A Cubic with Different Concavity on Either Side of Zero
Determine the concavity of \(f(x)=x^3\) using the monotonicity of \(f'\).
Differentiate:
Take \(x<y\le 0\). Then \(x^2>y^2\), except when \(x=y=0\), which is excluded by \(x<y\). Thus \(3x^2>3y^2\), so \(f'\) is decreasing on \((-\infty,0]\). The criterion shows that \(f\) is concave down there.
Now take \(0\le x<y\). Then \(x^2<y^2\), so \(3x^2<3y^2\). Thus \(f'\) is increasing on \([0,\infty)\), and \(f\) is concave up there.
As a check, the adjacent chord slopes obey the corresponding inequalities on each interval by the theorem. The result is concave down to the left of zero and concave up to the right; the derivative’s direction of change, rather than its sign, determines this classification.
Worked Example: Concavity When the Second Derivative Fails to Exist
Let
Find its concavity, including at the join \(x=0\). For \(x<0\), \(f'(x)=0\). For \(x>0\), \(f'(x)=-x\). At zero, the left difference quotient is \(0\), and the right difference quotient is \(-h/2\) for \(h>0\), which tends to \(0\). Hence \(f'(0)=0\), and overall
This derivative is nonincreasing: it is constant for \(x\le0\), decreases as \(x\) increases on \(x>0\), and its values on the right are at most its value \(0\) at the join. Therefore \(f\) is concave down on the whole real line.
The second derivative cannot be used at zero. On the left, the derivative \(f'\) has slope \(0\); on the right, it has slope \(-1\). Thus \(f''(0)\) does not exist. The first-derivative criterion still applies because \(f\) itself is differentiable and \(f'\) is nonincreasing. As a check, on \(x\le0\) the function is constant, and on \(x>0\) its derivative becomes more negative, consistent with slopes that do not increase.
Worked Example: A Positive Derivative Does Not Mean Concave Up
Determine the concavity of \(f(x)=\arctan x\) on \((0,\infty)\), and explain why the sign of \(f'\) alone is not enough.
The derivative is
For \(0<x<y\), we have \(x^2<y^2\), so \(1+x^2<1+y^2\). Both denominators are positive; taking reciprocals reverses the inequality:
Therefore \(f'\) is decreasing on \((0,\infty)\), and \(f\) is concave down there. At the same time, \(f'(x)>0\), so \(f\) is increasing. These are compatible conclusions: the function rises while its slopes get smaller. Checking the derivative formula confirms the distinction: positivity describes the direction of change of \(f\), whereas the decrease of \(f'\) determines concavity.
A Common Mistake: Confusing Derivative Sign with Concavity
A common error is to conclude that a function is concave up whenever \(f'(x)>0\). That condition says the function is increasing; it says nothing by itself about whether the slopes increase or decrease.
A separate mistake is to demand strict increase of \(f'\) before concluding concavity up. The criterion requires only that \(f'\) be nondecreasing. A constant derivative on an interval is allowed; it corresponds to a linear part of the function, which satisfies the non-strict chord inequality.
How to Apply the Criterion
When \(f''\) is awkward to calculate or does not exist, work directly with \(f'\). The following sequence keeps the reasoning focused on the relevant condition.
State where \(f\) is differentiable and where you want to determine its concavity.
Use the derivative formula, including the value at any join if the function is piecewise defined.
Show that \(f'\) is nondecreasing or nonincreasing on the interval. An algebraic comparison, a graph, or a known monotonicity argument may be useful.
Nondecreasing \(f'\) gives concavity up; nonincreasing \(f'\) gives concavity down. Include the interval and do not require strict monotonicity.
Why This Viewpoint Matters
The first-derivative criterion describes concavity through the behavior of slopes, which is exactly what the chord definition measures. It also works when the derivative has a corner or another failure of differentiability: the second derivative may be unavailable at such a point even though the first derivative remains monotone.
The key distinction is simple but important: \(f'\) being positive or negative describes whether \(f\) rises or falls. The first derivative being increasing or decreasing describes whether the slopes rise or fall, and therefore determines concavity. Keep these questions separate, and use non-strict inequalities unless strict behavior has actually been established.
Check Your Understanding
For each question, focus on whether the first derivative is increasing or decreasing, not merely on its sign.
- If \(f'\) is nondecreasing on an interval, what concavity does the first-derivative criterion guarantee?
- Why does \(f'(x)>0\) alone not determine whether \(f\) is concave up or concave down?
- For \(f(x)=x^3\), compare \(f'\) at two negative inputs to determine concavity on the negative half-line.
- Can a function be concave down on an interval where its first derivative is constant? Explain using the non-strict criterion.
- What feature of the piecewise example shows that the second derivative is unavailable at zero, and why can the first-derivative criterion still be applied?
