From Curvature to a Theorem
A graph that bends upward has tangent slopes that increase as \(x\) increases. A graph that bends downward has tangent slopes that decrease. The second derivative measures how the first derivative changes, so its sign provides a precise test for these two shapes.
The test is more than a visual rule. The Mean Value Theorem lets us prove the connection: a positive second derivative makes the first derivative increase, and increasing slopes force each intermediate point of the graph to lie below the chord joining two surrounding points. The reverse inequalities give concavity down.
Concavity in Terms of Chords
Let \(f\) be defined on an interval. Choose any three points \(x<z<y\) in that interval. The point on the straight chord from \((x,f(x))\) to \((y,f(y))\) that has horizontal coordinate \(z\) has height
The weights in this expression are positive, and they add to \(1\). The expression is therefore a weighted average of the endpoint heights.
In words, a concave-up graph lies on or below each chord joining two points on the graph. A concave-down graph lies on or above those chords. The strict versions mean the graph lies strictly below or above the chord at every intermediate point.
For a twice-differentiable function, these definitions agree with the description in terms of tangent slopes: concave up corresponds to increasing \(f'\), and concave down to decreasing \(f'\). The theorem below establishes the connection between the second derivative and the chord definition.
The Second Derivative Concavity Theorem
- If \(f''(x)>0\) for every \(x\in I\), then \(f\) is strictly concave up on \(I\).
- If \(f''(x)<0\) for every \(x\in I\), then \(f\) is strictly concave down on \(I\).
- If \(f''(x)\geq 0\) for every \(x\in I\), then \(f\) is concave up on \(I\), with a non-strict chord inequality. If \(f''(x)\leq 0\) throughout \(I\), then \(f\) is concave down.
The strict conclusions require a strict sign throughout the interval. The weak conclusions allow the second derivative to equal zero at some points. In each case, the condition concerns the second derivative over the interval, not just at one selected point.
Why the Theorem Works
We prove the positive case. Let \(f''(x)>0\) throughout \(I\), and select any \(x<z<y\) in \(I\). First, apply the Mean Value Theorem to \(f\) on each of the intervals \([x,z]\) and \([z,y]\). There are points \(c_1\in(x,z)\) and \(c_2\in(z,y)\) such that the two secant slopes are values of the derivative:
Because \(c_1<z<c_2\), we have \(c_1<c_2\). Apply the Mean Value Theorem to \(f'\) on \([c_1,c_2]\). For some \(d\in(c_1,c_2)\),
Both factors on the right are positive: \(f''(d)>0\) by hypothesis, and \(c_2-c_1>0\) because \(c_2>c_1\). Consequently \(f'(c_2)>f'(c_1)\). Substituting the secant-slope equalities gives
Since \(z-x>0\) and \(y-z>0\), multiplication by their product preserves the inequality. Rearranging gives
This is exactly the strict chord inequality for concavity up. Since \(x<z<y\) were arbitrary, it holds for every such triple, so \(f\) is strictly concave up on \(I\).
For the negative case, \(f''(d)<0\) makes \(f'(c_2)-f'(c_1)<0\). The secant-slope inequality and then the chord inequality reverse, proving strict concavity down. If \(f''\geq 0\), the same argument gives non-strict inequalities throughout; if \(f''\leq 0\), it gives the corresponding non-strict inequalities for concavity down.
Worked Examples
Worked Example: A Quadratic That Is Concave Up
Determine the concavity of \(f(x)=3x^2-2x+1\) on \(\mathbb{R}\), and check the chord inequality at \(x=-1\), \(z=0\), and \(y=2\).
Differentiate once:
Differentiate again:
Since \(6>0\) for every real \(x\), the theorem says \(f\) is strictly concave up on \(\mathbb{R}\). For the requested chord check, evaluate the function at all three points:
The chord height at \(z=0\) between the endpoint coordinates \(-1\) and \(2\) is
Thus \(f(0)=1<7\), as required for concavity up. As an additional slope check, \(f'(-1)=6(-1)-2=-8\) and \(f'(2)=6(2)-2=10\); the derivative has a larger value at the larger input. Differentiating these expressions confirms \(f''=6\), consistent with the conclusion.
Worked Example: A Quadratic That Is Concave Down
Determine the concavity of \(g(x)=-2x^2+4x+3\) on \(\mathbb{R}\), and check the chord inequality at \(x=0\), \(z=1\), and \(y=2\).
Its derivatives are
Because \(-4<0\) throughout \(\mathbb{R}\), \(g\) is strictly concave down. Evaluate the three function values:
The chord height at \(z=1\) is
Here \(g(1)=5>3\), so the graph lies above the chord, as concavity down requires. The endpoint derivatives also decrease: \(g'(0)=-4(0)+4=4\), while \(g'(2)=-4(2)+4=-4\). Their order agrees with the negative second derivative. Differentiation verifies the stated value \(g''=-4\).
Worked Example: A Radical Function on Its Domain
Determine the concavity of \(h(x)=\sqrt{x}\) on the interval \((0,\infty)\). The interval matters because the derivatives used in the theorem must exist there.
For \(x>0\), write \(h(x)=x^{1/2}\) and differentiate:
Differentiate again:
For every \(x>0\), \(x^{3/2}>0\), so \(h''(x)<0\). The theorem therefore says \(h\) is strictly concave down on \((0,\infty)\).
For a numerical check, compare the secant slope from \(x=1\) to \(x=4\) with the tangent slopes at the endpoints. The function values are \(h(1)=1\) and \(h(4)=2\), so the secant slope is
The endpoint tangent slopes are \(h'(1)=1/(2\sqrt{1})=1/2\) and \(h'(4)=1/(2\sqrt{4})=1/4\). They decrease from \(1/2\) to \(1/4\), in agreement with \(h''<0\). The Mean Value Theorem also predicts a point \(c\in(1,4)\) where \(h'(c)=1/3\). Solving \(1/(2\sqrt{c})=1/3\) gives \(\sqrt{c}=3/2\), hence \(c=9/4\), which is indeed between \(1\) and \(4\).
Concavity Is Not the Same as Increasing
A common error is to decide concavity from the sign of \(f'\). The first derivative tells whether the function is increasing or decreasing; the second derivative tells whether its slopes are increasing or decreasing. Those are different questions.
The correction is to compute the second derivative:
Since \(p''(x)=2>0\), the theorem shows that \(p\) is concave up on \((0,2)\), even though it is decreasing there. Indeed, \(p'(0)=-4\) and \(p'(2)=0\): the tangent slopes increase while remaining negative until the right endpoint. A function can decrease while bending upward.
Using the Theorem Carefully
To apply the theorem correctly, identify an interval where the function is twice differentiable, calculate \(f''\), and establish its sign throughout that interval. The theorem gives a conclusion on the interval where the condition holds. A second derivative’s sign at a single point is not enough to establish concavity across an interval.
The Mean Value Theorem also explains the role of the hypotheses. Its use on \(f\) and \(f'\) requires continuity on the closed subintervals being considered and differentiability in their interiors. When \(f\) is twice differentiable on an open interval, any closed subinterval chosen strictly inside it meets those requirements.
The result matters because it turns a geometric description of curve shape into an analytic test. Instead of relying only on a sketch, you can use derivatives and a theorem to justify the shape. The crucial distinction is simple: \(f'\) describes the direction of change of the function, while \(f''\) describes how that direction changes.
Check Your Understanding
Use the theorem and the definitions in this tutorial to answer the following questions.
- If \(f''(x)>0\) throughout an interval, what chord inequality must hold for any three points \(x<z<y\) in that interval?
- Why does applying the Mean Value Theorem to \(f'\) connect the sign of \(f''\) to the change in \(f'\)?
- A function has \(f'(x)<0\) and \(f''(x)>0\) throughout an interval. Is it increasing or decreasing there, and is it concave up or down?
- For \(q(x)=-x^2+6x\), calculate \(q''(x)\) and state its concavity on \(\mathbb{R}\).
- What condition on \(f''\) gives a non-strict conclusion of concavity up rather than strict concavity up?
