When a Flat Point Changes Concavity
A point where a graph has a horizontal tangent may look like a natural place for the function to turn around. Often it is: at a smooth local maximum or minimum, the derivative is zero. But a horizontal tangent does not guarantee a turn. The function may keep increasing or keep decreasing while its graph changes from concave down to concave up, or vice versa.
This situation is called a stationary inflection point. The word “stationary” refers to the zero derivative; “inflection” refers to the change in concavity. To tell what is happening, inspect the signs of both the first and second derivatives on either side of the point. The value of either derivative at the point alone is not enough.
Definitions and Sign Tests
Let \(f\) be defined on an open interval containing \(c\). The point \(x=c\) is a stationary point of \(f\) if \(f\) is differentiable at \(c\) and \(f'(c)=0\). The corresponding point on the graph is \((c,f(c))\).
A point \((c,f(c))\) is an inflection point if \(f\) is continuous at \(c\) and the graph changes concavity as \(x\) passes through \(c\). In common derivative analysis, a change from \(f''<0\) on one side to \(f''>0\) on the other (or the reverse) establishes that change in concavity, provided the function is twice differentiable on those neighboring intervals.
Thus, a stationary inflection point must satisfy both conditions: \(f'(c)=0\), and the concavity changes at \(c\). These conditions answer different questions. The first derivative describes whether the function is increasing or decreasing; the second derivative describes whether its slope is increasing or decreasing.
For a differentiable function, a change in the sign of \(f'\) from positive to negative indicates a local maximum; a change from negative to positive indicates a local minimum. If \(f'\) has the same strict sign on both sides, there is no such turn. A zero of \(f''\) does not by itself establish an inflection point: the sign of \(f''\) must actually change.
Worked Example: A Shifted Cubic
Worked Example: A Shifted Cubic
Analyze \(f(x)=2(x-1)^3+4\) at \(x=1\). Determine whether the point is stationary, whether it is an inflection point, and whether it is a local extremum.
Differentiate:
At \(x=1\), \(f'(1)=6(0)^2=0\), so the point is stationary. Its function value is \(f(1)=2(0)^3+4=4\), giving the graph point \((1,4)\).
For \(x<1\), \(f''(x)=12(x-1)<0\), so the graph is concave down. For \(x>1\), \(f''(x)=12(x-1)>0\), so it is concave up. The concavity changes at \(x=1\), making \((1,4)\) an inflection point.
Meanwhile, \(f'(x)=6(x-1)^2>0\) for every \(x\ne1\). The derivative is positive on both sides, so the function is increasing on each side and does not turn at \(x=1\). Therefore \((1,4)\) is not a local maximum or minimum.
Check the calculations by differentiating \(f'(x)=6(x-1)^2\), which gives \(f''(x)=12(x-1)\), as above. Substitution gives \(f'(1)=0\) and \(f''(1)=0\); the sign changes on the intervals, rather than these values alone, establish the classifications.
Why the First-Derivative Signs Rule Out a Turn
The sign test has a direct justification. Suppose \(f'(c)=0\), and for some positive \(\delta\), \(f'(x)>0\) whenever \(c-\delta<x<c\) and whenever \(c<x<c+\delta\). Then \(f\) is increasing on each side. The Mean Value Theorem also compares points on either side with the value at \(c\).
Choose \(x\) with \(c-\delta<x<c\). The Mean Value Theorem gives a number \(\xi\) between \(x\) and \(c\) such that \(f(c)-f(x)=f'(\xi)(c-x)\). Here \(f'(\xi)>0\) and \(c-x>0\), so \(f(c)>f(x)\).
Choose \(y\) with \(c<y<c+\delta\). The Mean Value Theorem gives a number \(\eta\) between \(c\) and \(y\) such that \(f(y)-f(c)=f'(\eta)(y-c)\). Here \(f'(\eta)>0\) and \(y-c>0\), so \(f(y)>f(c)\).
Values just to the left of \(c\) are less than \(f(c)\), and values just to the right are greater than \(f(c)\). Thus \(c\) is neither a local maximum nor a local minimum. If \(f'\) is negative on both sides instead, the inequalities reverse and the function is decreasing through \(c\), again with no local extremum.
This argument explains why a stationary inflection such as the shifted cubic can be a point where the graph flattens without reversing direction. The derivative vanishes at the point, but its nearby signs still show the direction of change.
Worked Example: A Fifth-Power Function
Worked Example: A Fifth-Power Function
Analyze \(g(x)=x^5\) at \(x=0\).
The derivatives are \(g'(x)=5x^4\) and \(g''(x)=20x^3\). Substitution gives \(g'(0)=0\) and \(g(0)=0\), so the graph has a stationary point at \((0,0)\).
For \(x<0\), \(g''(x)=20x^3<0\), so \(g\) is concave down. For \(x>0\), \(g''(x)=20x^3>0\), so it is concave up. Therefore, the concavity changes at \(0\), and \((0,0)\) is an inflection point.
For every \(x\ne0\), \(g'(x)=5x^4>0\). The function is increasing on both sides of \(0\), so the stationary point is not a local extremum. For example, \(g(-1)=-1<g(0)=0<g(1)=1\), consistent with the sign analysis.
As a check, differentiating \(5x^4\) gives \(20x^3\). Substituting \(x=0\) into the original function and its first derivative gives \(g(0)=0\) and \(g'(0)=0\); the opposite signs of \(g''\) on the two sides verify the inflection.
Worked Example: A Polynomial with a Nearby Critical Point
Worked Example: A Polynomial with a Nearby Critical Point
Consider \(h(x)=x^3+x^4\). Determine the behavior at \(x=0\), taking care to check signs on both sides rather than relying only on the derivative values there.
Differentiate and factor:
At \(x=0\), \(h'(0)=0\) and \(h(0)=0\), so \((0,0)\) is a stationary point. If \(-\frac{1}{2}<x<0\), then \(x<0\) and \(1+2x>0\), so \(h''(x)<0\). If \(x>0\), both factors are positive, so \(h''(x)>0\). The graph changes from concave down to concave up at \(0\), making \((0,0)\) an inflection point.
For \(-\frac{1}{2}<x<0\), the factor \(x^2\) is positive and \(3+4x>1>0\), so \(h'(x)>0\). For \(x>0\), both factors in \(x^2(3+4x)\) are positive, so \(h'(x)>0\) there as well. Thus the function is increasing on both sides of \(0\), and \((0,0)\) is not a local extremum.
Check by differentiating \(h'(x)=3x^2+4x^3\): the result is \(6x+12x^2\), matching \(h''\). Substitution gives \(h(0)=0\), \(h'(0)=0\), and \(h''(0)=0\). The signs around \(0\), not the zero second-derivative value, establish the inflection.
Contrast with a Local Extremum
Compare the examples with \(p(x)=x^2\). Its derivative is \(p'(x)=2x\), which is negative to the left of \(0\) and positive to the right. The function decreases and then increases, so \(x=0\) is a local minimum. Also, \(p''(x)=2>0\) on both sides: the graph remains concave up, with no change in concavity.
This contrast summarizes the two tests. A sign change in \(f'\) identifies a change in direction, while a sign change in \(f''\) identifies a change in concavity. A stationary inflection can have \(f'(c)=0\) and \(f''(c)=0\), just as a local extremum can have \(f'(c)=0\). The surrounding signs decide which behavior occurs.
Common Pitfall: Treating a Zero Derivative as a Minimum
For \(f(x)=x^3\), a mistaken argument might say: “\(f'(0)=0\), so \(x=0\) is a local minimum.” This conclusion does not follow. A zero derivative marks a stationary point, not automatically an extremum.
The correction is to inspect nearby signs. Since \(f'(x)=3x^2>0\) for \(x\ne0\), the function is increasing on both sides of \(0\). Direct values confirm the direction: \(f(-0.1)=-0.001\), \(f(0)=0\), and \(f(0.1)=0.001\). The point is a stationary inflection, not a local minimum.
A related error is to treat \(f''(c)=0\) as proof of an inflection. It is not. For \(p(x)=x^4\), \(p''(x)=12x^2\) is zero at \(0\) but positive on both sides, so concavity does not change. Always test intervals on both sides of the candidate point.
Check Your Understanding
For each question, use signs on intervals around the point rather than relying only on derivative values at the point.
- For \(f(x)=(x+2)^3-1\), find the stationary point and determine whether it is an inflection point or a local extremum.
- For \(g(x)=x^4\), evaluate \(g''(0)\). Does the concavity change at \(0\)? Explain using the signs of \(g''\).
- If \(f'(c)=0\), \(f'\) is negative on both sides of \(c\), and \(f''\) changes sign there, what can you conclude about the behavior at \(c\)?
- Why does \(f'(c)=0\) alone not prove that \(c\) is a local maximum or minimum?
- For \(h(x)=x^3+x^4\), state the signs of \(h'\) and \(h''\) immediately to the left and right of \(0\), and classify \((0,0)\).
