A Reliable Workflow for Polynomial Concavity
For a polynomial, the second derivative is another polynomial, so it is defined and continuous for every real input. Its zeros are useful candidates for changes in concavity: they divide the real line into intervals on which the sign of the second derivative can be checked.
The key is to treat these zeros as candidates, not as automatic inflection points. A zero of the second derivative may be where concavity changes, but it may also be a point where the second derivative merely touches zero and keeps the same sign. The sign on the intervals to either side decides.
For polynomial problems, the second derivative exists everywhere. Thus the practical test is to find the real zeros of \(f''\), determine its sign between consecutive zeros, and check whether the sign changes across each candidate. When it does, the corresponding point on the graph is an inflection point.
Why the Sign of the Second Derivative Determines Concavity
Here is the reason behind the sign test. Suppose \(f''(x)>0\) on an interval. The Mean Value Theorem applied to \(f'\) shows that \(f'\) is strictly increasing there: for any \(u<v\) in the interval, some \(c\in(u,v)\) satisfies
Now choose any \(x<z<y\) in the interval. Applying the Mean Value Theorem to \(f\) on \([x,z]\) and \([z,y]\) gives points \(u\in(x,z)\) and \(v\in(z,y)\) such that
Since \(u<v\) and \(f'\) is strictly increasing, the first slope is smaller than the second. Multiplying by the positive quantities \(z-x\) and \(y-z\), then rearranging, gives
The right side is the height of the chord from \((x,f(x))\) to \((y,f(y))\) at input \(z\). So the graph lies below that chord: it is strictly concave up. If \(f''<0\), the inequalities reverse and the graph is strictly concave down. This justifies using the sign of \(f''\) to read concavity.
A Step-by-Step Method
Calculate \(f'\), then \(f''\), keeping the algebra explicit.
Find all real solutions of \(f''(x)=0\). For a polynomial, these are the only possible places where its sign can change.
Order the candidate inputs, choose a test value in each resulting interval, and evaluate the sign of \(f''\).
Label the intervals concave up or down. A candidate is an inflection input only if the concavity changes across it.
For each confirmed inflection input \(c\), calculate \(f(c)\). The inflection point is \((c,f(c))\).
Worked Examples
Worked Example: A Quartic with Two Inflection Points
Find the concavity intervals and inflection points of \(f(x)=x^4-4x^2\).
Step 1: Differentiate twice.
Step 2: Factor and find the candidates. Set \(f''(x)=0\):
The two candidates are \(x=-\sqrt{2/3}\) and \(x=\sqrt{2/3}\). They divide the real line into three intervals.
Step 3: Test the sign. Since \(f''(x)=4(3x^2-2)\), test \(x=-1,0,1\), one in each interval:
Therefore \(f\) is concave up on \((-\infty,-\sqrt{2/3})\) and \((\sqrt{2/3},\infty)\), and concave down on \((-\sqrt{2/3},\sqrt{2/3})\). The sign changes at both candidates, so both give inflection points.
Step 4: Find their coordinates. At either candidate, \(x^2=2/3\), so \(x^4=4/9\). Hence
The inflection points are \((-\sqrt{2/3},-20/9)\) and \((\sqrt{2/3},-20/9)\). The graph changes from concave up to concave down at the first, then from concave down to concave up at the second.
Check. Differentiating \(f(x)=x^4-4x^2\) gives \(f'(x)=4x^3-8x\), and differentiating again gives \(f''(x)=12x^2-8\), the expression factored above.
Worked Example: A Zero of the Second Derivative That Is Not an Inflection
Determine the concavity and test the candidate for \(f(x)=x^4\).
Step 1: Differentiate twice and solve for zeros.
The only solution of \(f''(x)=0\) is \(x=0\). It is a candidate, but that alone does not classify it.
Step 2: Check the sign on either side. For every \(x<0\) and every \(x>0\), \(x^2>0\), so
The function is concave up on both \((-\infty,0)\) and \((0,\infty)\). The sign does not change at zero, so there is no inflection point there.
Step 3: Interpret the point. Since \(f(0)=0\), the candidate lies at \((0,0)\). It is on the graph, but the graph is concave up on both sides of it. It is therefore not an inflection point.
Check. Differentiating \(x^4\) gives \(4x^3\), and differentiating \(4x^3\) gives \(12x^2\), confirming the second derivative and its nonnegative sign.
Worked Example: Three Inflection Points and Four Concavity Intervals
Analyze the concavity and inflection points of \(f(x)=\frac{x^5}{5}-x^3\).
Step 1: Differentiate twice.
Step 2: Find and order the candidates. Solving \(2x(2x^2-3)=0\) gives \(x=0\) or \(x^2=3/2\). Thus the ordered candidates are \(-a,0,a\), where \(a=\sqrt{3/2}\). They divide the real line into four intervals.
Step 3: Determine the signs. The factor \(2\) is positive. The factor \(x\) is negative to the left of zero and positive to its right; \(2x^2-3\) is positive when \(|x|>a\) and negative when \(|x|<a\). Combining these signs gives
The sign changes at each of \(-a,0,a\), so all three candidates are inflection inputs.
Step 4: Calculate the points. At \(x=0\), \(f(0)=0\). For \(x=a\), use \(a^2=3/2\) and \(a^5=a^3a^2\):
Since \(a^3=(3/2)\sqrt{3/2}\), this is \(-\frac{21}{20}\sqrt{3/2}\). The function is odd, so \(f(-a)=\frac{21}{20}\sqrt{3/2}\). The points are therefore
Reading from left to right, the graph changes from concave down to up, then up to down, then down to up.
Check. Differentiating \(f(x)=x^5/5-x^3\) gives \(x^4-3x^2\); differentiating once more gives \(4x^3-6x=2x(2x^2-3)\), as used in the sign analysis.
A Common Mistake: Calling Every Candidate an Inflection
A frequent error is to solve \(f''(x)=0\) and immediately label every solution an inflection input. The calculation identifies candidates, not the final classification. For \(f(x)=x^4\), for example, one might write
The last implication is not justified. The correction is to check the signs on either side: \(12x^2>0\) for \(x<0\) and for \(x>0\). Concavity is up on both sides, so \((0,0)\) is not an inflection point. A candidate becomes an inflection point only when the concavity changes there.
Reading the Result as Graph Shape
Concavity describes how the slope changes as the input increases. Where \(f''>0\), the derivative \(f'\) increases; where \(f''<0\), it decreases. This provides a way to interpret a sign chart in terms of the graph: concave up means the slopes are increasing, and concave down means they are decreasing.
For a polynomial, a well-supported conclusion records both the intervals and the transitions between them. At a confirmed inflection input, include the function value to locate the point on the graph. Do not infer the exact appearance of the whole graph from concavity alone: this analysis identifies its bending behavior, not every other graph feature.
Check Your Understanding
For each question, use the sign of the second derivative and distinguish candidates from confirmed inflection points.
- For \(f(x)=x^3-3x\), calculate \(f''\), identify the concavity intervals, and find any inflection point.
- Why is solving \(f''(x)=0\) not enough by itself to establish an inflection point?
- For \(f(x)=x^6\), what is the sign of \(f''\) on either side of zero, and is zero an inflection input?
- If \(f''\) changes from negative to positive at \(x=c\), how does the graph's concavity change as \(x\) passes through \(c\)?
- What additional calculation is needed to give the coordinates of an inflection point after its input has been confirmed?
