From the Second Derivative to Concavity
A second derivative describes how the first derivative changes. Concavity, however, is expressed through the slopes of chords of the graph. The connection between these ideas is not just a visual pattern: the Mean Value Theorem provides a precise proof.
The proof uses the theorem twice. First, it shows that the sign of \(f''\) determines whether \(f'\) increases or decreases. Then it relates the slopes of two adjacent chords to values of \(f'\). The order of those values gives the required concavity inequality.
Concavity in Terms of Chord Slopes
Let \(I\) be an interval, and let \(f\) be defined on \(I\). For \(a<b\) in \(I\), the slope of the chord joining \((a,f(a))\) and \((b,f(b))\) is
The chord-slope definition compares the slopes on two adjacent subintervals. This lets us express concavity without relying on a graph.
The comparisons concern every triple of ordered points in the interval. If the inequality is strict for every triple, the chord slopes strictly increase or strictly decrease, respectively. The second derivative criterion below establishes these strict inequalities under its hypotheses.
The Mean Value Theorem We Need
The Mean Value Theorem states that if a function \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\), where \(a<b\), then there is some \(c\in(a,b)\) such that
When applying the theorem to \(f'\), continuity and differentiability of \(f'\) are needed. In particular, if \(f\) is twice differentiable throughout an interval, then \(f'\) is continuous on each closed subinterval and differentiable in its interior. These are the conditions that make the first application below valid.
We prove both conclusions. The positive case shows the main steps; changing the sign of \(f''\) reverses the comparisons at each stage.
Proof of the Positive Case
Assume \(f''(t)>0\) throughout the interior of \(I\), and choose any three points \(x<y<z\) in \(I\). We first compare the values of \(f'\) on the two portions of the interval. Next, we use those values to compare the chord slopes.
Take any \(u<v\) in the interior of \(I\). Apply the Mean Value Theorem to \(f'\) on \([u,v]\). Its hypotheses hold because \(f'\) is continuous on \([u,v]\) and differentiable on \((u,v)\). Thus, for some \(w\in(u,v)\), $$ f'(v)-f'(u)=f''(w)(v-u). $$ Since \(f''(w)>0\) and \(v-u>0\), it follows that \(f'(v)-f'(u)>0\), or \(f'(u)<f'(v)\). Therefore \(f'\) is strictly increasing.
Apply the Mean Value Theorem to \(f\) on \([x,y]\). There is a point \(c_1\in(x,y)\) such that $$ \frac{f(y)-f(x)}{y-x}=f'(c_1). $$ Apply it again on \([y,z]\). There is a point \(c_2\in(y,z)\) such that $$ \frac{f(z)-f(y)}{z-y}=f'(c_2). $$ The theorem applies to both subintervals because \(f\) is continuous on each closed subinterval and differentiable in its interior.
Since \(c_1\in(x,y)\) and \(c_2\in(y,z)\), we have \(c_1<y<c_2\), so \(c_1<c_2\). The strict increase of \(f'\) gives \(f'(c_1)<f'(c_2)\). Substituting the chord-slope expressions from Step 2 yields $$ \frac{f(y)-f(x)}{y-x} < \frac{f(z)-f(y)}{z-y}. $$ This holds for every \(x<y<z\) in \(I\), so \(f\) is strictly concave up on \(I\).
For the negative case, suppose \(f''(t)<0\) throughout the interior of \(I\). In Step 1, the Mean Value Theorem gives \(f'(v)-f'(u)=f''(w)(v-u)<0\) whenever \(u<v\). Thus \(f'\) is strictly decreasing. In Step 2, the same two applications to \(f\) give \(c_1<c_2\) and the same chord-slope identities. Now \(f'(c_1)>f'(c_2)\), so
This is precisely strict concavity down according to the chord-slope definition.
Worked Examples
Worked Example: A Quadratic That Is Strictly Concave Up
Show that \(f(x)=x^2\) is strictly concave up on \(\mathbb{R}\), and check the chord-slope inequality for \(x=-1\), \(y=0\), and \(z=2\).
First differentiate: $$ f'(x)=2x,\qquad f''(x)=2. $$ For every real \(x\), \(f''(x)=2>0\). The theorem therefore shows that \(f\) is strictly concave up on \(\mathbb{R}\).
For the stated triple, evaluate the function: $$ f(-1)=1,\qquad f(0)=0,\qquad f(2)=4. $$ The left chord slope is $$ \frac{f(0)-f(-1)}{0-(-1)} =\frac{0-1}{1}=-1. $$ The right chord slope is $$ \frac{f(2)-f(0)}{2-0} =\frac{4-0}{2}=2. $$ Thus \(-1<2\), as required. The derivatives also check directly: \(f'(-1)= -2\) and \(f'(2)=4\), consistent with \(f'\) increasing.
Worked Example: A Cubic on a Restricted Interval
Show that \(f(x)=x^3\) is strictly concave up on \((0,\infty)\), and compare the chord slopes for \(x=1\), \(y=2\), and \(z=4\).
Differentiate: $$ f'(x)=3x^2,\qquad f''(x)=6x. $$ For every \(x\in(0,\infty)\), \(6x>0\). The theorem applies on this interval and proves strict concavity up there.
Compute the function values: $$ f(1)=1,\qquad f(2)=8,\qquad f(4)=64. $$ The first chord slope is $$ \frac{f(2)-f(1)}{2-1} =\frac{8-1}{1}=7. $$ The second chord slope is $$ \frac{f(4)-f(2)}{4-2} =\frac{64-8}{2} =\frac{56}{2}=28. $$ Therefore \(7<28\), which confirms the predicted ordering for this triple. The restriction to positive inputs matters: \(f''(x)=6x\) is not positive on all of \(\mathbb{R}\), so this argument does not establish concavity up on the whole real line.
Worked Example: A Trigonometric Function That Is Strictly Concave Down
Show that \(f(x)=\sin x\) is strictly concave down on \((0,\pi)\), and check three chord slopes inside that interval.
Differentiate: $$ f'(x)=\cos x,\qquad f''(x)=-\sin x. $$ For \(0<x<\pi\), \(\sin x>0\), so \(f''(x)=-\sin x<0\). The theorem proves strict concavity down on \((0,\pi)\).
Choose \(x=\frac{\pi}{6}\), \(y=\frac{\pi}{2}\), and \(z=\frac{5\pi}{6}\). The function values are $$ f\left(\frac{\pi}{6}\right)=\frac12,\qquad f\left(\frac{\pi}{2}\right)=1,\qquad f\left(\frac{5\pi}{6}\right)=\frac12. $$ Since \(y-x=\frac{\pi}{3}\), the first chord slope is $$ \frac{1-\frac12}{\frac{\pi}{3}} =\frac{\frac12}{\frac{\pi}{3}} =\frac{3}{2\pi}. $$ Since \(z-y=\frac{\pi}{3}\), the second chord slope is $$ \frac{\frac12-1}{\frac{\pi}{3}} =\frac{-\frac12}{\frac{\pi}{3}} =-\frac{3}{2\pi}. $$ Hence \(\frac{3}{2\pi}>-\frac{3}{2\pi}\), the required reversed ordering.
What the Criterion Does—and Does Not—Say
The theorem is a sufficient condition: if the second derivative is positive throughout an interval, the chord slopes strictly increase there; if it is negative throughout, they strictly decrease. The condition concerns the sign of \(f''\) at every interior point of the interval under consideration. Checking the sign at only a few sample points is not enough to use the theorem.
It is also important to distinguish concavity from monotonicity. A positive second derivative says that \(f'\) increases. It does not say that \(f'\) itself is positive, so it does not by itself say that \(f\) is increasing.
Common Mistake: Treating \(f''>0\) as \(f'>0\)
A mistaken argument might say: “For \(f(x)=x^2\), \(f''(x)=2>0\), so \(f'(x)>0\) everywhere.” The conclusion is false. In fact, $$ f'(x)=2x,\qquad f'\left(-\frac12\right)=-1<0. $$ The positive second derivative implies that \(f'\) is strictly increasing; it does not imply that \(f'\) is positive. For example, \(f'(-1)=-2\) and \(f'(1)=2\): the derivative increases between these points while still being negative at the first one. The correct conclusion from \(f''>0\) is strict concavity up, proved by comparing adjacent chord slopes.
A related proof error is to apply the Mean Value Theorem to \(f\) and then claim directly that \(f''>0\) orders the resulting derivative values. A single application to \(f\) represents chord slopes as values of \(f'\), but it does not establish how those values compare. The missing step is to apply the theorem to \(f'\) on the interval between the relevant points; that step proves that \(f'\) increases when \(f''>0\).
Why the Two Applications Matter
The proof gives a useful chain of reasoning, not merely a sign rule. The first application of the Mean Value Theorem turns information about \(f''\) into an ordering of values of \(f'\). The second application turns the slopes of chords of \(f\) into values of \(f'\). Because the points that represent the two adjacent chords are ordered, the first ordering transfers to the chord slopes.
Check Your Understanding
Answer these questions using the theorem and its proof.
- What differentiability conditions are needed to apply the Mean Value Theorem first to \(f'\) and then to \(f\)?
- If \(f''<0\) throughout an interval, what inequality must hold between the chord slopes on \([x,y]\) and \([y,z]\) when \(x<y<z\)?
- Why does \(f''>0\) show that \(f'\) increases, rather than that \(f'\) is positive?
- For \(f(x)=x^3\), why does the theorem establish strict concavity up on \((0,\infty)\) but not on all of \(\mathbb{R}\)?
- In the proof, why must the Mean Value Theorem be applied to \(f'\) as well as to \(f\)?
