When a Parameter Changes Concavity
A polynomial family depends on one or more parameters as well as on the variable \(x\). Changing a parameter can move the zeros of \(f''\), bring zeros together, or remove real zeros altogether. Those changes can alter whether, and where, the graph has inflection points.
The key is not merely to solve \(f''(x)=0\). A zero of the second derivative is a candidate: it is an inflection point only if the concavity changes there. A parameter value can create a zero without creating an inflection point, especially when two simple zeros merge into a repeated zero.
For a polynomial family, the practical task is to find the parameter values for which \(f''\) has real roots, then determine whether its sign changes at each root. The following result makes that last step precise.
Root Multiplicity and Sign Change
A real root \(c\) of a polynomial \(g\) has multiplicity \(m\) if \(g(x)\) can be written as \((x-c)^m q(x)\), where \(q\) is a polynomial and \(q(c)\ne 0\). The factor \(q\) stays nonzero and keeps the same sign sufficiently close to \(c\). The factor \((x-c)^m\) determines whether the sign switches as \(x\) crosses \(c\).
This statement applies when \(f''\) is not the zero polynomial. If \(f''\) is identically zero, it has no isolated sign-changing zero: it is zero everywhere.
Justification. By the definition of multiplicity, near \(c\) we can write \(f''(x)=(x-c)^m q(x)\), with \(q(c)\ne 0\). Since \(q\) is continuous, it has the same sign on both sides of \(c\) sufficiently close to \(c\). If \(m\) is even, \((x-c)^m\) is positive on both sides, so \(f''\) has the same sign on both sides. If \(m\) is odd, \((x-c)^m\) is negative to the left and positive to the right, so \(f''\) has opposite signs on the two sides. Therefore the sign changes exactly when \(m\) is odd, as claimed.
This gives an efficient way to analyze polynomial families. Factor \(f''\) when possible, find its real roots for the parameter values under consideration, and inspect the multiplicity of each root. At a boundary parameter value, roots may merge and their multiplicity may become even; that case must be checked rather than inferred from nearby parameter values.
A Reliable Parameter-Analysis Method
Compute \(f''(x)\), keeping the parameter visible.
Solve \(f''(x)=0\) for real \(x\), and determine which parameter values allow those roots.
Use root multiplicity or test the sign of \(f''\) on either side of each candidate.
For each sign-changing root \(c\), give \((c,f(c))\). Analyze parameter values where roots merge or disappear separately.
Worked Example: Two Roots Merge at One Parameter Value
Worked Example: Two Roots Merge at One Parameter Value
Consider \(f_a(x)=x^4+a x^3\), where \(a\) is any real parameter. Find the inflection points for each value of \(a\).
Differentiate twice:
If \(a\ne 0\), the roots are \(x=0\) and \(x=-a/2\), and they are distinct. Each factor occurs once at each root, so each root has multiplicity one. Both are odd-multiplicity roots and therefore sign-changing zeros of \(f_a''\).
The corresponding graph coordinates are found by substitution. At \(x=0\), \(f_a(0)=0\), giving \((0,0)\). At \(x=-a/2\),
Thus, when \(a\ne 0\), the inflection points are \((0,0)\) and \(\left(-a/2,-a^4/16\right)\). The second derivative vanishes at both stated \(x\)-coordinates: \(6(0)(a)=0\), and \(6(-a/2)(-a+a)=0\). Their distinct linear factors confirm the sign changes.
At the boundary value \(a=0\), the two roots merge: \(f_0''(x)=12x^2\). The root \(x=0\) has multiplicity two, so there is no sign change. Indeed, \(12x^2\) is positive on both sides of zero. There is no inflection point when \(a=0\). This verifies that a zero of \(f''\) at the merged location does not by itself make an inflection point.
Worked Example: A Threshold for the Existence of Inflection Points
Worked Example: A Threshold for the Existence of Inflection Points
Consider \(g_a(x)=x^4+a x^2\). Determine exactly which parameter values give inflection points.
Its derivatives are
If \(a<0\), solving \(g_a''(x)=0\) gives \(x^2=-a/6\), so the two real roots are \(x=\pm\sqrt{-a/6}\). They are distinct simple roots: the derivative of \(g_a''\) is \(24x\), which is nonzero at either root. Hence both roots have odd multiplicity and are inflection points. Between them \(g_a''<0\), while outside them \(g_a''>0\), confirming the sign changes.
For either root \(r=\pm\sqrt{-a/6}\), we have \(r^2=-a/6\) and \(r^4=a^2/36\). Therefore
For \(a<0\), the two inflection points are consequently \(\left(-\sqrt{-a/6},-5a^2/36\right)\) and \(\left(\sqrt{-a/6},-5a^2/36\right)\). Substitution gives \(g_a''(r)=12(-a/6)+2a=-2a+2a=0\) at either candidate.
If \(a=0\), then \(g_a''(x)=12x^2\), which has a double root at zero but does not change sign. If \(a>0\), then \(12x^2+2a>0\) for every real \(x\), so there are no roots and no inflection points. Thus inflection points occur exactly when \(a<0\).
Worked Example: A Discriminant Boundary
Worked Example: A Discriminant Boundary
Consider \(h_a(x)=x^4/4+a x^3/3+x^2/2\). Find the parameter ranges for which this family has inflection points, and classify the boundary case.
Differentiating gives
The quadratic \(h_a''\) has discriminant \((2a)^2-4(3)(1)=4(a^2-3)\). If \(a^2<3\), it has no real roots. Since its leading coefficient is positive, it is positive for every real \(x\), so there are no inflection points.
If \(a^2>3\), the two roots are
They are distinct, and hence simple roots of the quadratic. The sign of \(h_a''\) is positive outside the two roots and negative between them, so both roots are sign-changing zeros. The inflection points are \((r_-,h_a(r_-))\) and \((r_+,h_a(r_+))\), where, for either root \(r\),
These coordinates use the roots just found and the original function. Substituting either root into \(h_a''\) gives zero by the quadratic formula; their simplicity and the upward-opening quadratic verify the sign changes.
At \(a^2=3\), the quadratic factors as \(h_a''(x)=3(x+a/3)^2\). There is one real root, \(r=-a/3\), with multiplicity two. Thus \(h_a''\) is nonnegative on both sides and does not change sign. Although \(h_a''(r)=0\), the point is not an inflection point. The family has two inflection points when \(|a|>\sqrt{3}\), and none when \(|a|\le\sqrt{3}\).
Why Boundary Values Need Their Own Check
Parameter ranges often meet at a boundary where the equation \(f''(x)=0\) has a repeated root. On one side of the boundary there may be two simple roots; on the other side there may be no real roots. At the boundary, those roots can coincide in an even-multiplicity root. The examples show why the boundary cannot be classified by simply extending the conclusion from either neighboring range.
A common mistake is to treat every solution of \(f''(x)=0\) as an inflection point. For instance, at \(a=0\) in the family \(g_a(x)=x^4+a x^2\), the incorrect calculation stops at \(g_0''(x)=12x^2\), \(g_0''(0)=0\), and declares an inflection point. The correction is to check the sign on both sides: \(12x^2>0\) for \(x<0\) and for \(x>0\). Concavity does not change, so zero is not an inflection point.
For polynomial families, multiplicity is a useful shortcut, but a sign check remains a good verification. Keep the roles of the equations distinct: \(f''(x)=0\) locates candidates, while a change in the sign of \(f''\) decides whether a candidate is an inflection point.
Check Your Understanding
For each question, distinguish a zero of the second derivative from a sign-changing zero.
- For \(f_a(x)=x^4+a x^3\), what happens to the two candidate roots when \(a=0\), and why is the resulting zero not an inflection point?
- In \(g_a(x)=x^4+a x^2\), which parameter values give two inflection points?
- Why does a root of multiplicity three in \(f''\) imply a sign change?
- For \(h_a(x)=x^4/4+a x^3/3+x^2/2\), what happens at \(a^2=3\)?
- What additional check is needed after solving \(f''(x)=0\) in a polynomial family?
