Diagnosing Shape from Different Kinds of Evidence
Concavity describes how a function’s slope changes across an interval. In problems, the available evidence may be a formula for \(f\), a sign chart, or a graph of \(f'\) or \(f''\). Sometimes these sources provide different levels of detail: a graph may show an overall trend without revealing the exact value of a derivative at every point, while a formula can identify isolated zeros precisely.
The aim is to use each kind of evidence for what it actually establishes. In particular, a trend in \(f'\) tells us about concavity on an interval, but does not necessarily tell us that \(f''\) is strictly positive or strictly negative at every point there. Likewise, a zero of \(f''\) is a candidate for an inflection point, not a conclusion by itself.
Definitions and the Role of the Second Derivative
Let \(f\) be differentiable on an interval \(I\). We say \(f\) is concave up on \(I\) if \(f'\) is nondecreasing there. We say \(f\) is concave down on \(I\) if \(f'\) is nonincreasing there. In the twice-differentiable setting, these descriptions can be connected to the sign of \(f''\).
The inequalities in this criterion are non-strict. A function can be concave up even if \(f''\) is zero at some points. The strict trend of a graph does not change that fact: strictly increasing does not mean that the derivative is strictly positive at every point.
An inflection point is a point on the graph where the function changes concavity: it is concave one way on one side and concave the other way on the other side. In this tutorial, we require the function to be defined and continuous at the candidate point. A zero of \(f''\), or a point where \(f''\) does not exist, can be a candidate, but the change of concavity on the two sides must still be checked.
Why the Derivative Trend Determines Concavity
The connection between \(f'\) and concavity follows from the Mean Value Theorem. Suppose \(f\) is twice differentiable on an interval and \(f''\ge0\) there. Choose any \(a<b\) in that interval. The Mean Value Theorem applied to \(f'\) gives a point \(c\in(a,b)\) such that
Since \(f''(c)\ge0\) and \(b-a>0\), the right-hand side is nonnegative. Thus \(f'(b)\ge f'(a)\), so \(f'\) is nondecreasing and \(f\) is concave up. If \(f''\le0\), the same argument gives \(f'(b)\le f'(a)\), so \(f\) is concave down.
For the reverse implication, suppose \(f'\) is nondecreasing and \(f''(x)\) exists at an interior point \(x\). For \(h>0\), the quotient \(\bigl(f'(x+h)-f'(x)\bigr)/h\) is nonnegative. For \(h<0\), the numerator is nonpositive and the denominator is negative, so the quotient is again nonnegative. Taking the limit as \(h\) tends to zero gives \(f''(x)\ge0\). The nonincreasing case gives \(f''(x)\le0\) by the same reasoning with the inequalities reversed.
A Reliable Method for Mixed Evidence
When evidence comes in more than one form, first identify what each source tells you. A formula for \(f''\) can establish signs directly. A sign chart records those signs on intervals. A graph of \(f'\) shows whether the slopes of \(f\) are trending up or down. A graph of \(f''\), when its scale and features are clear, shows where the second derivative is positive, negative, or zero.
For a formula, find where \(f''=0\) or is undefined, and note any domain breaks. These locations divide the real line into intervals on which the sign can be tested.
Use a sign chart for a formula, or describe whether \(f'\) increases or decreases on each interval. Keep unknown or ambiguous regions marked as unknown.
Positive or nonnegative \(f''\) corresponds to concave up; negative or nonpositive \(f''\) corresponds to concave down. A trend in \(f'\) gives the same interval-level conclusion.
Compare the concavity immediately to the left and right. A candidate is an inflection point only if concavity changes there and the function is defined and continuous there.
Do not infer an exact pointwise value of \(f''\) from a broad graph trend, or fill in a missing interval of evidence without additional information.
Worked Example: A Formula and Sign Chart Identify Two Inflection Points
Worked Example: A Formula and Sign Chart Identify Two Inflection Points
Let \(f(x)=\frac{x^4}{4}-\frac{x^3}{3}\). Find its concavity intervals and inflection points.
Differentiate once:
Differentiate again and factor:
The candidates where the sign can change are \(x=0\) and \(x=\frac23\). Test the three intervals. For \(x=-1\), \(f''(-1)=(-1)(-5)=5>0\). For \(x=\frac13\), \(f''\left(\frac13\right)=\frac13(1-2)=-\frac13<0\). For \(x=1\), \(f''(1)=1(1)=1>0\).
Concavity changes at both candidates. The function is defined and continuous everywhere, so both are inflection points. Their coordinates require evaluating \(f\):
Thus the points are \((0,0)\) and \(\left(\frac23,-\frac4{81}\right)\). Check the differentiation: the derivative of \(x^4/4\) is \(x^3\), and the derivative of \(-x^3/3\) is \(-x^2\); differentiating again gives \(3x^2-2x\), as used. The sign checks confirm the concavity on all three intervals.
Worked Example: A Trend in \(f'\) Does Not Give Strict Pointwise Signs
Worked Example: A Trend in \(f'\) Does Not Give Strict Pointwise Signs
Consider \(f(x)=\frac{x^5}{20}+\frac{x^4}{12}-\frac{x^3}{6}-\frac{x^2}{2}\). A graph of \(f'\) shows it decreasing up to \(x=1\) and increasing after \(x=1\). Determine the concavity change, and explain what happens at \(x=-1\).
Differentiate to check the graph information:
A second differentiation gives
For \(x<1\), the factor \(x-1\) is negative and \((x+1)^2\) is nonnegative, so \(f''\le0\). For \(x>1\), both factors give a positive product, so \(f''>0\). At \(x=-1\), \(f''(-1)=0\), but the sign is nonpositive on both sides of \(-1\). Thus \(f'\) is decreasing on the whole interval to the left of \(1\), even though its derivative is zero at \(-1\). It is increasing to the right of \(1\).
Concavity changes from down to up at \(x=1\), so \(x=1\) is an inflection point. Its function value is
The inflection point is \(\left(1,-\frac8{15}\right)\). At \(-1\), the function remains concave down on both sides, so there is no inflection point there. As a check, substituting \(-1\) into \(f''\) gives \((-2)(0)^2=0\), confirming a zero without a concavity change.
Worked Example: A Zero of \(f''\) Can Be a False Candidate
Worked Example: A Zero of \(f''\) Can Be a False Candidate
Let \(f(x)=\frac{x^5}{20}-\frac{x^4}{4}+2x^2\). Find the inflection point and explain why the zero of \(f''\) at \(x=2\) is not one.
Differentiating twice gives
The factor \(x-2\) is squared, so it does not change sign across \(2\). The factor \(x+1\) is negative for \(x<-1\) and positive for \(x>-1\). Therefore \(f''<0\) on \((-\infty,-1)\), while \(f''>0\) on \((-1,2)\) and \((2,\infty)\). The function is concave down before \(-1\) and concave up after \(-1\); its concavity does not change at \(2\).
Hence \(x=-1\) is the inflection-point input. Its function value is
The inflection point is \(\left(-1,\frac{17}{10}\right)\). At \(x=2\), \(f''(2)=(3)(0)^2=0\), but the sign remains positive on both sides. This verifies why the zero at \(2\) is not enough to establish an inflection point. The derivatives also check directly: differentiating \(x^5/20\), \(-x^4/4\), and \(2x^2\) gives \(x^4/4\), \(-x^3\), and \(4x\), respectively.
Worked Example: A Strictly Rising Derivative Can Have a Flat Point
Worked Example: A Strictly Rising Derivative Can Have a Flat Point
Let \(f(x)=\frac{x^4}{4}\). Use the formula and the trend of \(f'\) to decide whether \(x=0\) is an inflection point.
The derivatives are
The function \(f'(x)=x^3\) is strictly increasing, so \(f\) is concave up. The formula also shows \(f''(x)\ge0\) everywhere, with \(f''(0)=0\). On either side of \(0\), the concavity remains up; there is no change from up to down or down to up. Thus \(x=0\) is not an inflection point. The corresponding function value is \(f(0)=0\), but \((0,0)\) is not an inflection point. Differentiating \(x^4/4\) confirms both derivatives, and substituting \(x=0\) into \(3x^2\) confirms the zero.
When Evidence Is Incomplete or Conflicting
A graph is evidence, not a license to infer details it does not show. If a graph of \(f'\) clearly trends upward over an interval, conclude that \(f\) is concave up there; do not claim \(f''>0\) at every point from that trend alone. If the graph has limited resolution near a candidate, it may not establish whether the trend changes exactly there. State what the graph does establish and leave the uncertain detail open.
If a formula and a graph appear to conflict, check that they refer to the same function, interval, and derivative. Then verify the formula by differentiating and inspect any scale or omitted interval in the graph. Consistent evidence should support the same interval-level concavity. Do not force a conclusion from sources that genuinely disagree.
Check Your Understanding
For each question, state what the evidence establishes and distinguish interval conclusions from pointwise claims.
- If \(f'\) is increasing on an interval, what can you conclude about the concavity of \(f\)? Does that imply \(f''>0\) at every point where it exists?
- If a sign chart shows \(f''<0\) immediately to the left of \(c\) and \(f''>0\) immediately to the right, what further condition on \(f\) is needed to identify \((c,f(c))\) as an inflection point?
- Why is a point where \(f''=0\) only a candidate for an inflection point?
- Can a function be concave up on an interval while \(f''=0\) at one point in that interval? Explain.
- A graph of \(f'\) appears to decrease on an interval. What is the appropriate concavity conclusion, and what pointwise claim should you avoid making without more evidence?
