Why a Piecewise Join Needs a Two-Sided Check
A piecewise-defined function can follow one formula to the left of a boundary and another formula to the right. At the boundary, the function may be continuous, have a corner, or even fail to connect. Its second derivative may also be undefined there. None of those facts, by itself, determines whether the graph changes concavity.
To analyze a join at \(x=c\), examine the graph on the two sides separately. First check whether the function is continuous at \(c\). Then determine the concavity on an interval just to the left and an interval just to the right. The join is an inflection point only if the function is continuous there and its concavity changes across the boundary.
Definitions and the Boundary Test
A function is concave up on an interval when its slopes increase as \(x\) increases; when the relevant derivatives exist, this is indicated by \(f''(x)>0\) throughout the interval. It is concave down when its slopes decrease; \(f''(x)<0\) throughout the interval indicates this behavior. These tests apply on intervals where the required derivatives exist, not merely at a single point.
In this tutorial, an inflection point is a point on a function’s graph where the function is continuous and its concavity changes from up to down or from down to up. The function does not have to be differentiable at that point. In particular, a corner can be an inflection point if the function is continuous there and the concavity really changes.
When \(f''\) exists on each side, its sign is usually the most direct way to classify concavity. If \(f''\) does not exist on part of a side, use the behavior of the slopes or another valid description of the graph’s shape. A value of \(f''\) at the boundary is not needed for this test.
Why Comparing the Two Sides Works
The second derivative describes how slopes change as \(x\) moves through an interval. If \(f''>0\) throughout an interval, then \(f'\) is increasing there, so the graph is concave up. If \(f''<0\) throughout an interval, then \(f'\) is decreasing, so the graph is concave down. These are interval statements: they describe behavior near the boundary on each side, not a property that must be evaluated at the boundary itself.
Suppose \(f\) is continuous at \(c\), is concave up on \((c-\delta,c)\), and is concave down on \((c,c+\delta)\) for some \(\delta>0\). Then the graph approaches and leaves the same point \((c,f(c))\), while its concavity differs on the two sides. By the definition above, \((c,f(c))\) is an inflection point. Reversing “up” and “down” gives the same conclusion. If the concavity matches on both sides, the defining change has not occurred, so the boundary is not an inflection point.
This reasoning also explains why checking only \(f''(c)\) cannot settle the question. The second derivative might not exist at a piecewise join, and even when it does exist, its value at one point does not tell us whether concavity changes on a surrounding interval.
Worked Examples at Piecewise Boundaries
Worked Example: A Continuous Join Where Concavity Changes
Consider
The boundary is \(x=0\). From the left, \(\lim_{x\to0^-}x^2=0\); from the right, \(\lim_{x\to0^+}(-x^2)=0\); and \(f(0)=0^2=0\). Thus \(f\) is continuous at \(0\).
For \(x<0\), the left-hand formula gives \(f'(x)=2x\) and \(f''(x)=2>0\), so the function is concave up. For \(x>0\), the right-hand formula gives \(f'(x)=-2x\) and \(f''(x)=-2<0\), so the function is concave down.
The one-sided derivatives at zero are also both zero: the left derivative is \(\lim_{h\to0^-}h^2/h=\lim_{h\to0^-}h=0\), and the right derivative is \(\lim_{h\to0^+}(-h^2)/h=\lim_{h\to0^+}(-h)=0\). The first derivative exists there, but the second derivative does not, because \(f'\) has different formulas on the two sides, \(2x\) and \(-2x\). The concavity changes from up to down, and the graph is continuous. Therefore \((0,0)\) is an inflection point. The check is precisely the boundary test: continuity holds and the one-sided concavities are opposite.
Worked Example: A Corner Without an Inflection Point
Consider
At \(x=0\), the left-hand limit is \(0\), the right-hand limit is \(0\), and \(g(0)=0\), so the function is continuous. For \(x<0\), \(g'(x)=2x\) and \(g''(x)=2>0\). For \(x>0\), \(g'(x)=2x+1\) and \(g''(x)=2>0\). Thus the function is concave up on both sides.
There is a corner: the left derivative at zero is \(\lim_{h\to0^-}h^2/h=0\), while the right derivative is \(\lim_{h\to0^+}(h^2+h)/h=\lim_{h\to0^+}(h+1)=1\). Since these are unequal, \(g'(0)\) does not exist. But the concavity does not change; both side formulas have positive second derivative. Therefore \((0,0)\) is not an inflection point. The check is that continuity holds but the one-sided concavities match.
Worked Example: Opposite Concavities but a Discontinuous Join
Consider
For \(x<0\), \(h''(x)=2>0\), so the function is concave up. For \(x>0\), \(h''(x)=-2<0\), so it is concave down. The side concavities are opposite, but at the boundary the left-hand limit is \(\lim_{x\to0^-}x^2=0\), while the right-hand limit is \(\lim_{x\to0^+}(1-x^2)=1\). Since these limits differ, \(h\) is not continuous at \(0\); in particular, neither side limit equals \(h(0)=7\).
There is no continuous joining point on the graph at \(x=0\), so the boundary is not an inflection point under the stated definition. The check shows why opposite concavity alone is insufficient: continuity at the boundary is also required.
A Reliable Procedure for Piecewise Joins
Use the following sequence whenever a piecewise formula changes at an interior point \(x=c\).
Record the boundary value \(c\), and note which formula applies on each side, including whether either formula includes \(c\).
Compute the left-hand and right-hand limits and compare both with \(f(c)\). If the limits do not agree with the function value, the boundary is not an inflection point.
Find \(f''\) from the applicable formula on each open side and determine its sign. If that test is unavailable, use another justified description of slope behavior.
If the function is continuous and the concavity changes from up to down or down to up, the join is an inflection point. If concavity matches, it is not.
It is not necessary for \(f'(c)\) or \(f''(c)\) to exist for this procedure. Differentiability at the join and concavity on the adjacent intervals are different questions. A missing derivative may signal a point to investigate, but it does not prove that concavity changes.
Common Mistake: Treating Every Join as an Inflection
A frequent error is to reason, “The formula changes here, so the second derivative is undefined; therefore, this must be an inflection point.” That conclusion skips both essential checks. For example, in the corner example above, \(g'(0)\) does not exist and the formula changes, yet \(g''=2\) on both sides. The graph is concave up on both sides, so there is no inflection point.
The opposite shortcut is also unreliable: rejecting a point because the derivative does not exist. The first worked example has an inflection point even though \(f''(0)\) does not exist. The relevant evidence is the function’s continuity and its concavity on adjacent intervals, not whether a derivative exists at the join.
Check Your Understanding
For each question, explain how continuity and the concavity on the two sides affect the conclusion.
- If a piecewise function is continuous at \(x=2\), concave down just to the left, and concave up just to the right, is \(x=2\) an inflection-point candidate that qualifies as an inflection point?
- A function has \(f''(x)>0\) on both sides of a continuous join, but \(f'(c)\) does not exist. Is the join an inflection point? Why?
- Why does an undefined second derivative at a boundary not by itself establish a change in concavity?
- On the two sides of a join, the function is concave up and concave down, respectively, but its one-sided limits at the join are unequal. Does the boundary qualify as an inflection point under the definition in this tutorial?
- What three checks should you report when justifying an inflection point at a piecewise boundary?
