Why concavity constrains a curve
A curve can rise while bending downward, or fall while bending upward. Its direction of travel is described by the sign of its first derivative; its concavity is described by how that slope changes. The second derivative makes this change precise.
This distinction is useful when checking a sketch. If the second derivative is positive throughout an interval, the tangent slopes cannot decrease as you move from left to right. A drawing that shows them decreasing cannot represent a function with that second-derivative condition, regardless of where the curve is positioned vertically.
Concavity and slope changes
Let \(f\) be a function defined on an interval \(I\). A function is concave upward (also called convex) on \(I\) if, between any two points of its graph, the graph lies on or below the straight chord joining them. It is concave downward if the graph lies on or above each such chord.
For example, if \(a<b\) are in \(I\), the chord between \((a,f(a))\) and \((b,f(b))\) has height at \(x\in[a,b]\) equal to the linear interpolation of the endpoint values. Concavity upward means that the function’s value at \(x\) is no greater than this chord height.
When \(f''(x)>0\) everywhere on an interval, the slopes are strictly increasing there; when \(f''(x)<0\) everywhere, they are strictly decreasing. The weak inequalities matter: \(f''\geq 0\) allows the second derivative to equal zero at some points, and still guarantees nondecreasing slopes.
Why the second derivative controls the slope
The first derivative \(f'(x)\) gives the slope of the tangent to the graph at \(x\). The second derivative is the derivative of that slope function. The connection between its sign and the change in slope follows directly from the Mean Value Theorem.
To prove the first statement, choose any \(x<y\) in \(I\). Apply the Mean Value Theorem to \(f'\) on \([x,y]\). There is a point \(c\in(x,y)\) such that the difference in slopes is the second derivative at \(c\), multiplied by the distance between the points:
Since \(f''(c)\geq 0\) and \(y-x>0\), their product is nonnegative. Thus \(f'(y)-f'(x)\geq 0\), or \(f'(x)\leq f'(y)\). This holds for every \(x<y\), so \(f'\) is nondecreasing. If \(f''>0\) throughout, the product is positive and \(f'(x)<f'(y)\). The case \(f''\leq 0\) follows from the same equation with the inequality reversed.
The slope condition also explains the chord description of concavity. Suppose \(f''\geq 0\), and choose \(a<x<b\) in \(I\). The Mean Value Theorem applied on \([a,x]\) and \([x,b]\) gives points \(u\in(a,x)\) and \(v\in(x,b)\) such that
Here \(u<v\). Because \(f'\) is nondecreasing, \(f'(u)\leq f'(v)\), so the slope of the first secant is no greater than the slope of the second. Multiplying by the positive quantity \((x-a)(b-x)\) and rearranging gives
The right-hand side is exactly the chord height at \(x\). Therefore the graph lies on or below the chord, as required for concavity upward. Reversing the inequalities proves the corresponding chord condition for concavity downward.
Worked Example: Positive second derivative and a chord check
Worked Example: Positive second derivative and a chord check
Consider \(f(x)=x^3\) on the interval \((0,2)\). Find the first and second derivatives, determine how the tangent slopes change, and compare the graph with the chord joining its values at \(0\) and \(2\).
Differentiate once:
Differentiate again:
For every \(x\in(0,2)\), \(6x>0\). Thus \(f'\) is strictly increasing and the graph is concave upward on this interval. To check the chord description, the endpoint values are \(f(0)=0\) and \(f(2)=8\). The chord between them has height \(4\) at \(x=1\), while the curve has value \(f(1)=1\). Therefore \(1\leq 4\), as concavity upward requires. The derivative check also gives \(f'(1)=3\), between \(f'(0)=0\) and \(f'(2)=12\), consistent with increasing slopes.
Check: Differentiating the stated function gives \(f'(x)=3x^2\) and \(f''(x)=6x\); on the specified open interval the second derivative is positive, and the computed curve value lies below the chord.
Concavity does not determine whether a curve rises or falls
A positive second derivative means slopes increase, not that the slopes are positive. A curve with increasing slopes might still have negative slopes over part or all of an interval. Similarly, a negative second derivative means slopes decrease, not that the function itself is decreasing.
Worked Example: A curve that rises while bending downward
Let \(f(x)=-x^2+2x+3\) on \((0,2)\). Determine its concavity and check whether it rises or falls on the interval \(0<x<1\).
Its first derivative is
Its second derivative is
Since \(f''(x)=-2<0\), the tangent slopes decrease throughout \((0,2)\), and the curve is concave downward. But for \(0<x<1\), \(f'(x)=2-2x>0\), so the curve is rising on that part of the interval. For instance, \(f'(0.25)=1.5\) and \(f'(0.75)=0.5\): both slopes are positive, but the later slope is smaller. The function rises with a diminishing slope.
Check: The derivative values satisfy \(1.5>0.5\), as required by the negative second derivative, while both are positive, confirming that concavity downward does not imply that the function is falling.
Using slope constraints to test a sketch
Suppose a proposed sketch includes tangent slopes at several positions. Their values must respect the second-derivative sign across the interval between them. Under \(f''\geq 0\), a later slope cannot be smaller than an earlier one. Under \(f''\leq 0\), a later slope cannot be larger. This is a local-to-global check: the sign condition applies throughout the interval, so it restricts every pair of points in it.
Worked Example: Testing tangent slopes against concavity upward
A proposed graph is claimed to satisfy \(f''(x)>0\) throughout \((0,2)\). Its labelled tangent slopes are \(f'(0.5)=1\), \(f'(1)=3\), and \(f'(1.5)=2\). Can the sketch be consistent with the claim?
Because \(f''>0\) throughout the interval, \(f'\) must be strictly increasing. Since \(0.5<1<1.5\), the three labelled slopes must satisfy
The first two labels give \(1<3\), which is consistent. But the final two give \(3<2\), which is false. Thus no twice-differentiable function can have both the stated positive second derivative throughout the interval and all three labelled slopes. At least one part of the sketch or its claim must be wrong.
For comparison, \(g(x)=x^2+x\) has \(g'(x)=2x+1\) and \(g''(x)=2>0\). Its slopes at \(0.5,1,1.5\) are \(2,3,4\), in increasing order.
Check: Differentiating \(g\) gives the stated derivatives, and substituting the three inputs gives \(2(0.5)+1=2\), \(2(1)+1=3\), and \(2(1.5)+1=4\). These values satisfy the required strict increase.
A practical concavity check
When examining a curve on an interval, keep the second-derivative condition separate from information about the curve’s height or direction. The following sequence focuses on the constraints relevant to concavity:
Check the claimed condition throughout the interval in question; a sign that holds only at one point does not impose the same interval-wide conclusion.
A nonnegative second derivative requires nondecreasing slopes; a nonpositive second derivative requires nonincreasing slopes.
For positive second derivative, check that no later tangent slope is smaller. For negative second derivative, check that no later tangent slope is larger.
For concavity upward, the curve lies on or below each chord; for concavity downward, it lies on or above each chord.
Common mistake: confusing slope with its rate of change
A frequent error is to infer the sign of \(f'\) directly from the sign of \(f''\). For example, the claim “\(f''<0\), so \(f'<0\)” is not a valid deduction. A negative second derivative says that \(f'\) decreases as \(x\) increases. It does not specify whether \(f'\) starts out positive, negative, or zero.
The quadratic in the previous example shows the error concretely. There, \(f''=-2<0\), but at \(x=0.25\), \(f'(0.25)=1.5>0\). The incorrect conclusion would predict a negative slope; the direct derivative calculation gives a positive slope. The correction is to use \(f''\) to describe how slopes change, and use \(f'\) itself to determine whether the function rises or falls at a point.
Check Your Understanding
Answer each question using the interval condition stated.
- If \(f''(x)\geq 0\) throughout an interval, what ordering must hold between \(f'(a)\) and \(f'(b)\) whenever \(a<b\) in that interval?
- If \(f''(x)<0\) throughout an interval, can a tangent slope be positive at some point? Explain what the sign condition actually requires.
- A sketch claims \(f''>0\) on \((0,3)\) but labels \(f'(1)=5\) and \(f'(2)=4\). Is the information consistent? Why?
- For \(f(x)=x^3\) on \((0,2)\), compute \(f''(x)\) and state whether the tangent slopes increase or decrease.
- Describe the position of a concave-down graph relative to the chord joining any two of its points.
