Why polynomial concavity can change more than once
A polynomial may bend in different directions across its graph. Its second derivative provides a systematic way to locate those changes: find where the second derivative is zero, determine its sign between those values, and check whether the sign actually switches at each candidate.
The key caution is that a zero of the second derivative is not automatically an inflection point. Factoring helps distinguish zeros that produce a change in concavity from zeros that do not. This is especially useful for polynomials with several real roots or repeated roots in their second derivatives.
From the second derivative to a sign chart
For a twice-differentiable function, the sign of \(f''(x)\) determines concavity on an interval: \(f''(x)>0\) means the function is concave up there, and \(f''(x)<0\) means it is concave down there. An inflection point is a point on the graph where concavity changes. For a polynomial, \(f''\) exists everywhere, so the candidates to check are the real zeros of \(f''\).
Factoring is helpful because it reveals how the sign behaves at each zero. A factor with an even power does not change sign as it passes through its root; a factor with an odd power does. When several factors are present, consider their combined sign, rather than treating each factor as an independent concavity change.
Why root multiplicity predicts a sign change
Here is a justification of the multiplicity rule. Suppose \(c\) is a real zero of the polynomial \(f''\), and its multiplicity is \(m\). By factoring out all copies of \(x-c\), write
Because \(r\) is continuous and \(r(c)\ne 0\), it keeps the same sign sufficiently close to \(c\). Thus, near \(c\), any change in the sign of \(f''\) must come from \((x-c)^m\). If \(m\) is even, this factor is positive on both sides of \(c\), so \(f''\) does not change sign there. If \(m\) is odd, the factor is negative on one side and positive on the other, so \(f''\) changes sign. Therefore, a zero of \(f''\) has a local concavity change exactly when its multiplicity is odd.
This result concerns a zero of \(f''\), not a zero of \(f\). To give the point on the graph, evaluate the original polynomial at the candidate value. Also, the sign chart still gives the final check: it makes the intervals of concavity explicit and confirms whether the sign changes.
Worked Example: Three inflection points
Worked Example: Three inflection points
Analyze the concavity and find the inflection points of \(f(x)=\frac{x^5}{20}+\frac{x^4}{12}-\frac{x^3}{3}\).
Differentiate twice:
Factor the second derivative:
The real zeros are \(-2\), \(0\), and \(1\). Each has multiplicity one, so each is a candidate for a sign change. Check the signs of the factors on each interval:
For example, on \((0,1)\), the factors \(x\) and \(x+2\) are positive while \(x-1\) is negative, so \(f''(x)<0\). The signs change at all three zeros. The graph is concave down on \((-\infty,-2)\) and \((0,1)\), and concave up on \((-2,0)\) and \((1,\infty)\).
Evaluate \(f\) at each candidate:
The inflection points are \(\left(-2,\frac{12}{5}\right)\), \((0,0)\), and \(\left(1,-\frac{1}{5}\right)\).
Check: Differentiating the stated polynomial gives \(f'(x)=\frac{x^4}{4}+\frac{x^3}{3}-x^2\), and differentiating again gives \(x^3+x^2-2x=x(x+2)(x-1)\), as used in the sign chart. Substitution into the original polynomial gives the three coordinates above.
Worked Example: A repeated zero that is not an inflection point
Worked Example: A repeated zero that is not an inflection point
Analyze the concavity and find the inflection points of \(g(x)=\frac{x^5}{20}-\frac{x^3}{2}+x^2\).
Differentiate twice and factor:
The factorization can be checked by multiplication: \((x-1)^2(x+2)=(x^2-2x+1)(x+2)=x^3-3x+2\). The candidates are \(-2\) and \(1\). At \(-2\), the multiplicity is one; at \(1\), it is two.
For \(x<-2\), the squared factor is positive and \(x+2\) is negative, so \(g''(x)<0\). For \(-2<x<1\), both factors are positive, so \(g''(x)>0\). For \(x>1\), both factors are again positive, so \(g''(x)>0\). Concavity changes at \(-2\), but not at \(1\).
Evaluate the polynomial at the genuine candidate:
Thus the inflection point is \(\left(-2,\frac{32}{5}\right)\). The function is concave down on \((-\infty,-2)\) and concave up on \((-2,1)\) and \((1,\infty)\).
Check: Differentiating \(g\) gives \(g''(x)=x^3-3x+2\), which factors as \((x-1)^2(x+2)\). Substituting \(-2\) into \(g\) gives \(32/5\). The sign remains positive on both sides of \(1\), so \((1,g(1))\) is not an inflection point.
Worked Example: Two changes of concavity
Worked Example: Two changes of concavity
Analyze \(h(x)=x^4-4x^3\).
Differentiate:
The zeros of \(h''\) are \(0\) and \(2\), both simple. If \(x<0\), both \(x\) and \(x-2\) are negative, so \(h''(x)>0\). If \(0<x<2\), \(x\) is positive and \(x-2\) is negative, so \(h''(x)<0\). If \(x>2\), both factors are positive, so \(h''(x)>0\).
Therefore, \(h\) is concave up on \((-\infty,0)\), concave down on \((0,2)\), and concave up on \((2,\infty)\). Both zeros give inflection points. Their coordinates are
The inflection points are \((0,0)\) and \((2,-16)\).
Check: Differentiating \(h(x)=x^4-4x^3\) twice gives \(h''(x)=12x^2-24x=12x(x-2)\). Evaluating \(h\) at the two sign-changing zeros gives \(0\) and \(-16\), respectively.
A common mistake: counting every zero as an inflection point
A tempting shortcut is to solve \(f''(x)=0\) and declare every solution an inflection point. The second example shows why that fails: \(g''(1)=0\), but the factor \((x-1)^2\) has even multiplicity, and \(g''\) is positive on both sides of \(1\).
The implication is invalid because a zero alone does not establish a change in concavity. The correction is to inspect the signs on both sides. Here they are both positive, so there is no inflection point at \(x=1\). A sign chart prevents the mistake by recording the behavior on intervals, not just the candidate values.
How polynomial degree limits the number of inflection points
If \(f\) is a polynomial of degree \(n\ge 2\), then \(f''\) has degree \(n-2\). A nonzero polynomial of degree \(n-2\) has at most \(n-2\) distinct real zeros. Since every polynomial inflection point must occur at a real zero of \(f''\), \(f\) can have at most \(n-2\) distinct inflection points.
This is an upper bound, not a guarantee. Some zeros may have even multiplicity and produce no change in concavity, as in the second example. If \(f''\) is identically zero, then \(f\) is linear or constant: its concavity does not switch at an isolated point, so it has no inflection points under the sign-change definition.
A reliable workflow
Compute \(f''(x)\) carefully and factor it completely over the real numbers.
Put the zeros in increasing order; these divide the real line into intervals.
Use the factors or a test value on each interval to establish whether \(f''\) is positive or negative.
A zero with odd multiplicity changes sign; a zero with even multiplicity does not.
For each sign-changing zero \(c\), calculate \(f(c)\). The inflection point is \((c,f(c))\).
Polynomials have no domain breaks, so every interval in the chart lies within their domain. Factoring the second derivative and checking the signs on each interval gives a complete account of their concavity, even when several candidate points are close together or some roots are repeated.
Check Your Understanding
Use factoring and sign changes in each question; distinguish candidate zeros from genuine inflection points.
- If \(f''(x)=(x+1)^2(x-3)\), which candidate zeros give changes in concavity?
- For \(p(x)=\frac{x^4}{12}-\frac{x^3}{3}\), find and factor \(p''(x)\), then state the intervals of concavity.
- Why does a zero of even multiplicity in \(f''\) fail to produce a sign change?
- If a polynomial has degree \(7\), what is the greatest possible number of distinct inflection points?
- If \(f''(c)=0\), what additional check is needed before calling \((c,f(c))\) an inflection point?
