Concavity as Evidence Near a Critical Point
A critical point is a natural place to look for a local maximum or minimum, but the fact that a point is critical does not classify it. Concavity can provide useful evidence: it tells us how the slopes are changing near the point. When the derivative is zero there, this change in slopes can force the function to turn.
The key is to keep two pieces of information distinct. Concavity describes the behavior of slopes; a local maximum or minimum describes the function’s values. Near a critical point, the first can help establish the second, but concavity by itself is not always enough.
A point \(c\) is a critical number of \(f\) if \(c\) is in the domain of \(f\) and either \(f'(c)=0\) or \(f'(c)\) does not exist. The concavity argument developed below applies directly when \(f'(c)=0\) and the relevant derivatives exist nearby. A critical point where \(f'\) does not exist requires separate reasoning.
Why a Stationary Point and Concavity Work Together
Suppose \(f'(c)=0\). If the derivative is increasing on both sides of \(c\), then it must be negative just to the left and positive just to the right: its value at \(c\) is zero, and its values increase as \(x\) passes through \(c\). Negative derivative means the function is decreasing; positive derivative means it is increasing. The function therefore turns from decreasing to increasing, giving a local minimum.
If the derivative is decreasing instead, its values are positive just to the left of \(c\) and negative just to the right. The function turns from increasing to decreasing, giving a local maximum.
The condition concerns concavity in an interval around \(c\), not just the shape suggested by one isolated calculation. In practice, the Second Derivative Test often uses the value \(f''(c)\), under its usual differentiability conditions. This tutorial emphasizes the local shape that explains why the sign is useful.
Why the Conclusion Follows
Here is a justification for the concave-up case. Since \(f''>0\), the derivative \(f'\) is strictly increasing on the interval: for any two points \(a<b\) in it, the Mean Value Theorem applied to \(f'\) gives a point \(d\) between them such that \(f'(b)-f'(a)=f''(d)(b-a)>0\). Thus \(f'(b)>f'(a)\).
Because \(f'(c)=0\) and \(f'\) is strictly increasing, \(f'(x)<0\) when \(x<c\), and \(f'(x)>0\) when \(x>c\). The Mean Value Theorem applied to \(f\) now turns those derivative signs into comparisons with \(f(c)\). For \(x<c\), there is a point \(d\) between \(x\) and \(c\) such that \(f(c)-f(x)=f'(d)(c-x)<0\). Therefore \(f(x)>f(c)\). For \(x>c\), there is a point \(d\) between \(c\) and \(x\) such that \(f(x)-f(c)=f'(d)(x-c)>0\). Therefore \(f(x)>f(c)\) on both sides, proving a strict local minimum.
For the concave-down case, \(f''<0\) makes \(f'\) strictly decreasing. Since \(f'(c)=0\), the derivative is positive on the left and negative on the right. The same Mean Value Theorem comparisons show that \(f(x)<f(c)\) for nearby \(x\ne c\), proving a strict local maximum.
Worked Examples
Worked Example: Concave Up at a Stationary Point
Classify the critical point of \(f(x)=x^2-4x+5\).
First calculate the derivatives:
Set \(f'(x)=0\) to find the critical number:
Since \(f''(x)=2>0\) for every \(x\), the function is concave up around \(x=2\). The stationary-point condition and the concavity condition together imply a strict local minimum. Its value is
So the local minimum is at \((2,1)\). Check by factoring around the point: \(f(x)=(x-2)^2+1\), so \(f(x)>1=f(2)\) for every \(x\ne2\). This confirms the conclusion.
Worked Example: Concave Down at a Stationary Point
Classify the critical point of \(g(x)=-x^2+6x-8\).
Differentiate:
Solve for the critical number:
The second derivative is negative everywhere, so \(g\) is concave down around \(x=3\). Because \(g'(3)=0\), there is a strict local maximum there. Its value is
To check, complete the square: \(g(x)=-(x-3)^2+1\). Since \((x-3)^2>0\) when \(x\ne3\), \(g(x)<1=g(3)\). The point \((3,1)\) is indeed a strict local maximum.
When Concavity Alone Is Not Enough
Concavity does not tell us that the function has a stationary point. A concave-up function may simply be increasing throughout a neighborhood, or decreasing throughout it. Without a zero derivative at \(c\), the slopes need not change from negative to positive there. The point could fail to be an extremum altogether.
Worked Example: Concave Up Without a Local Minimum at the Chosen Point
Consider \(h(x)=x^2+2x\) at \(x=0\). Its derivatives are
The function is concave up, but \(h'(0)=2\), not zero. In fact, for \(x\) close to zero, \(h'(x)=2x+2>0\) whenever \(x>-1\), so the function is increasing across a neighborhood of zero. The value at zero is \(h(0)=0\), while
There are nearby values both greater and less than \(h(0)\), so zero is neither a local maximum nor a local minimum. The concavity statement is correct; the unsupported leap from “concave up” to “local minimum here” is not.
Even at a stationary point, a zero second derivative does not settle the classification. It may mean that the usual sign test is inconclusive, not that there is no extremum. To decide what happens, inspect the actual local behavior rather than treating the value zero as if it were positive or negative.
Worked Example: A Zero Second Derivative with a Local Minimum
Let \(p(x)=x^4\) and examine \(x=0\). The derivatives are
At zero, \(p'(0)=0\) and \(p''(0)=0\), so the usual Second Derivative Test does not classify the point. But \(p''(x)=12x^2>0\) for \(x\ne0\), and directly \(p(x)=x^4>0=p(0)\) for every \(x\ne0\). Thus \((0,0)\) is a strict local minimum. The function is concave up on either side of zero, although \(p''(0)\) itself is zero.
Check the derivative calculation: differentiating \(p'(x)=4x^3\) gives \(p''(x)=12x^2\), and evaluating \(p'\) at zero gives \(0\), as required for a stationary point.
Changing Concavity Can Accompany No Extremum
Sometimes the function changes concavity as it passes through a critical point. That change does not itself establish a maximum or minimum. For an extremum, nearby function values must lie consistently below or consistently above the value at the point. A stationary point with values on both sides of its height is not an extremum.
Worked Example: A Stationary Point Without an Extremum
Consider \(q(x)=x^3\) at zero. Its derivatives are
Since \(q'(0)=0\), zero is a critical number. But \(q''(x)<0\) for \(x<0\) and \(q''(x)>0\) for \(x>0\): the concavity changes across zero. Directly, if \(x<0\), then \(q(x)=x^3<0=q(0)\), while if \(x>0\), then \(q(x)=x^3>0=q(0)\). Values on both sides are respectively below and above the value at zero, so there is neither a local maximum nor a local minimum.
As a check, the derivative \(3x^2\) is positive for every \(x\ne0\). The function is increasing through the stationary point; its slope becomes zero there without changing the direction of its increase.
A Reliable Way to Use Concavity Evidence
Confirm that \(c\) is in the domain and determine whether \(f'(c)=0\). A critical point where \(f'\) is undefined does not meet the stationary-point condition used in the concavity argument above.
Use the sign of \(f''\) on an interval around \(c\), where the required derivatives exist. Positive indicates concave up; negative indicates concave down.
A stationary point with concavity up is evidence for a local minimum; a stationary point with concavity down is evidence for a local maximum.
If the second derivative is zero, concavity changes, or the needed conditions fail, examine derivative signs or compare nearby function values instead.
A common error is to treat a positive second derivative as sufficient on its own. For example, with \(h(x)=x^2+2x\), the incorrect calculation “\(h''(0)=2>0\), so \(x=0\) is a minimum” skips the check that \(h'(0)=2\ne0\). The correction is to verify the stationary-point condition first. Since \(h\) is increasing near zero and has nearby values on both sides of \(h(0)\), zero is not an extremum.
Concavity evidence is useful because it explains how a stationary point can become a turn: the slopes increase through zero for a minimum or decrease through zero for a maximum. But it remains evidence that must be paired with the conditions that make the conclusion valid. If those conditions are absent, a graph can flatten, change concavity, or continue in the same direction without having a local extremum.
Check Your Understanding
For each question, explain which conditions support the conclusion rather than relying on a second-derivative sign alone.
- If \(f'(c)=0\) and \(f''\) is positive throughout an interval around \(c\), what kind of local extremum does \(f\) have at \(c\)?
- Why does concavity up at a point not by itself guarantee a local minimum there?
- For \(r(x)=-x^4\), calculate \(r'(0)\) and \(r''(0)\). Is zero a local maximum, local minimum, or neither?
- What does it tell you if \(f'(c)=0\) but \(f'\) stays positive on both sides of \(c\)?
- Why is a zero second derivative not enough to conclude that a critical point is not an extremum?
