Reading Concavity from a Second-Derivative Graph
A graph of \(f''\) is a graph of the rate at which the slope of \(f\) changes. Where \(f''\) is positive, the slopes of \(f\) increase as \(x\) increases, and the curve bends upward. Where \(f''\) is negative, those slopes decrease, and the curve bends downward. This makes the sign of \(f''\) a direct guide to concavity.
The graph of \(f''\) does not itself trace the graph of \(f\). Instead, it supplies information about how \(f\) bends. To use it, first mark the locations where the sign could change, then determine the sign on each resulting interval.
If \(f''\geq 0\) throughout an interval, \(f\) is concave up there; if \(f''\leq 0\), it is concave down there. Strict positivity or negativity gives strict concavity. The distinction matters when the graph of \(f''\) touches or crosses the horizontal axis.
Partition the Number Line Before Reading Signs
For a given graph of \(f''\), mark every \(x\)-value where \(f''(x)=0\), as well as every point where \(f''\) is undefined or the displayed graph has a break. These values divide the domain into intervals. On each interval, determine whether the graph lies above or below the horizontal axis.
The sign is about the vertical position of the graph of \(f''\), not its direction of travel. A graph can be descending while remaining above the axis; on that whole interval \(f''>0\), so \(f\) is concave up. Whether \(f''\) itself is increasing or decreasing is not the test for the concavity of \(f\).
Why the Sign of the Second Derivative Controls Bending
The connection follows from the Mean Value Theorem. Suppose \(f''(x)>0\) throughout an interval. For any two points \(u<v\) in that interval, apply the Mean Value Theorem to \(f'\) on \([u,v]\). There is some \(w\in(u,v)\) such that
Both factors on the right are positive, so \(f'(v)>f'(u)\). Thus \(f'\) is increasing: the slopes of \(f\) increase from left to right. To see how that places the curve relative to a chord, take \(a<x<b\) in the interval. Apply the Mean Value Theorem to \(f\) on \([a,x]\) and \([x,b]\). For some \(u\in(a,x)\) and \(v\in(x,b)\),
Since \(u<v\) and \(f'\) is increasing, the first slope is smaller than the second. Multiplying by the positive quantities \(x-a\) and \(b-x\), and rearranging, gives
The right side is the height of the chord joining \((a,f(a))\) to \((b,f(b))\) at \(x\). So the graph lies below its chords: it is concave up. If \(f''<0\), the same argument reverses the inequalities, giving concavity down.
Worked Example: Two Sign Changes
Worked Example: Two Sign Changes
Suppose \(f\) is defined and twice differentiable on all of \(\mathbb{R}\), and the graph of its second derivative is \(f''(x)=(x+2)(x-1)\). Determine the concavity intervals and the inflection-point candidates.
First find the zeros:
These values split the real line into \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\). Test one value in each interval:
Therefore \(f\) is concave up on \((-\infty,-2)\), concave down on \((-2,1)\), and concave up on \((1,\infty)\). The sign changes at both \(x=-2\) and \(x=1\). Since \(f\) is defined and continuous at both values, both are inflection-point locations. The graph of \(f''\) identifies their \(x\)-coordinates, but this information alone does not give the corresponding \(y\)-coordinates.
Check: The test values are positive, negative, positive in order, matching the stated concavity pattern. Also, \(f''(-2)=0\) and \(f''(1)=0\), as required for these two crossings.
A Zero Is a Candidate, Not a Conclusion
A zero of \(f''\) is worth checking because the sign might change there. But the value \(f''(c)=0\), by itself, does not prove that the concavity changes. The graph could cross the axis, or it could touch the axis and turn back. Compare the signs on both sides of the zero.
More generally, a possible change can occur where \(f''\) is undefined, even if \(f''\) has no zero there. Such a location also requires checking the signs on either side. For an inflection point at \(x=c\), use the convention that \(f\) must be defined and continuous at \(c\), and its concavity must change across \(c\). A sign change across a point outside the domain does not give a point on the graph of \(f\).
Worked Example: A Zero Without a Concavity Change
Worked Example: A Zero Without a Concavity Change
Suppose \(f\) is defined and twice differentiable on \(\mathbb{R}\), and \(f''(x)=(x-1)^2(x+2)\). Find the concavity intervals and decide which zero, if any, gives an inflection point.
The zeros are \(x=-2\) and \(x=1\). They divide the line into \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\). Evaluate \(f''\) at one point in each interval:
The function is concave down on \((-\infty,-2)\) and concave up on both \((-2,1)\) and \((1,\infty)\). The signs change at \(-2\), so it is an inflection-point location. At \(1\), the sign is positive on both sides: the graph of \(f''\) reaches zero there but does not cross from one sign to the other. Thus \(x=1\) is not an inflection-point location. The two intervals of concavity up can be combined as \((-2,\infty)\), since the sign remains nonnegative through \(x=1\).
Check: Substitution gives \(f''(-2)=0\) and \(f''(1)=0\). The sign tests show a change at \(-2\) and no change at \(1\), which is exactly why only the first zero qualifies.
Worked Example: An Undefined Point and the Domain
Worked Example: An Undefined Point and the Domain
Consider \(f(x)=x\ln|x|-x-\ln|x|\), defined for \(x\neq 0\). Determine its concavity and inflection points. This example includes a point where \(f''\) is undefined, so the domain must be checked as well as the signs.
For \(x\neq 0\), differentiate each term:
Differentiate again:
The second derivative is zero at \(x=-1\) and undefined at \(x=0\). Since \(x^2>0\) for \(x\neq 0\), the sign of \(f''\) is the sign of \(x+1\). Test one point in each domain interval:
Thus \(f\) is concave down on \((-\infty,-1)\), and concave up on \((-1,0)\) and \((0,\infty)\). At \(x=-1\) the sign changes, and \(f\) is defined and continuous there. Its graph contains the point
so \((-1,1)\) is an inflection point. Although \(f''\) is undefined at \(0\), \(f\) is also undefined there. Consequently, there is no point of the graph at \(x=0\), and \(0\) is not an inflection point.
Check: Differentiating the stated \(f'\) gives \(1/x+1/x^2=(x+1)/x^2\). The sign tests confirm the concavity change at \(-1\); the domain check rules out an inflection point at \(0\).
What a Second-Derivative Graph Gives a Sketch
A second-derivative graph gives the concavity pattern, and its sign changes can locate inflection points along the horizontal axis. It does not by itself provide every feature needed to place and draw the original curve. For a fuller sketch, bring together the relevant information about:
- the domain and any breaks in it;
- intercepts, symmetry, and end behavior;
- the sign and zeros of \(f'\), which indicate where the function rises or falls;
- the values of \(f\) needed to place points at particular heights; and
- any limits or asymptotic behavior that constrain the graph near excluded values or at the ends of the domain.
Use these data together rather than treating the curve of \(f''\) as a sketch of \(f\). In a curve-analysis table, record concavity on each interval and mark confirmed inflection points separately from unconfirmed candidates. A zero of \(f''\) with no sign change should not be marked as an inflection point.
Check Your Understanding
For each question, use sign information on intervals rather than relying only on zeros.
- If \(f''\) is negative throughout an interval, what does that tell you about the concavity of \(f\) there?
- Suppose \(f''(c)=0\) and \(f''\) is positive on both sides of \(c\). Is \(c\) an inflection-point location? Explain.
- If \(f''\) changes sign at \(c\) but \(f\) is not defined at \(c\), can \((c,f(c))\) be an inflection point?
- For \(f''(x)=(x+3)(x-2)\), find the intervals of concavity by testing one point in each interval.
- Name two kinds of information, other than concavity, that help complete a curve sketch.
