When Is a Curve Sketch Impossible?
A curve sketch is not determined by a handful of labels. It must satisfy all the stated information at once: values, derivative signs, critical points, concavity, and limiting behavior. Sometimes two claims are enough to make that impossible. In other cases, a supposed conflict disappears once an unstated assumption—such as continuity at a point—is examined.
The purpose of a contradiction check is therefore not merely to notice that a sketch looks unusual. It is to identify the precise claims that cannot all be true, and to state the conditions that make the conflict valid.
Extrema, Critical Points, and Their Hypotheses
Let \(f\) be defined on a domain containing an interval around \(c\). The point \(c\) is a local maximum point if there is some \(\delta>0\) such that \(f(x)\leq f(c)\) whenever \(x\) is in the domain and \(|x-c|<\delta\). A local minimum point is defined by reversing the inequality: \(f(x)\geq f(c)\) for all such \(x\). The inequalities are strict for \(x\ne c\) if the extremum is strict.
An interior point \(c\) is called a critical point if \(f'(c)=0\) or \(f'(c)\) does not exist. A critical point is a possible location of a local extremum, not a guarantee of one. Conversely, if \(f\) is differentiable at an interior local extremum, then its derivative must be zero. This is Fermat's theorem.
For a local minimum, when \(h>0\) is sufficiently small, \(f(c+h)-f(c)\geq 0\), so the right-hand difference quotient is nonnegative. When \(h<0\) is sufficiently small, the numerator is still nonnegative but the denominator is negative, so the left-hand difference quotient is nonpositive. If the derivative exists, both one-sided limits must be the same number. That number is both nonnegative and nonpositive, so \(f'(c)=0\). At a local maximum, the inequalities reverse and lead to the same conclusion.
This theorem lets us reject a proposed differentiable local extremum with a nonzero derivative. But derivative signs on a punctured neighborhood require more care: they constrain nearby function values relative to \(f(c)\) only if continuity at \(c\) is available.
Why Continuity Matters for Derivative-Sign Claims
The Mean Value Theorem provides a direct way to test the claim “\(f\) has a local maximum at \(c\), while \(f'(x)>0\) immediately to its right.” The theorem says that if \(f\) is continuous on \([c,y]\) and differentiable on \((c,y)\), then for some \(\xi\in(c,y)\),
If \(y>c\) and \(f'(\xi)>0\), this difference is positive. But a local maximum requires \(f(y)-f(c)\leq 0\). The claims conflict. Continuity on the closed interval is essential to apply the theorem; in particular, continuity at \(c\) cannot simply be presumed from differentiability at nearby points.
Worked Example: Testing a Claimed Maximum Against a Positive Derivative
A proposed sketch claims that \(f\) is continuous at \(0\), has a local maximum at \(0\), and satisfies \(f'(x)>0\) for every \(x\) with \(0<x<1\). Check whether these claims can all hold.
Choose any \(y\) with \(0<y<1\), for example \(y=\tfrac12\). Continuity at \(0\) and differentiability on \((0,\tfrac12)\) allow the Mean Value Theorem on \([0,\tfrac12]\). Thus there is a \(\xi\in(0,\tfrac12)\) such that
Therefore \(f(\tfrac12)>f(0)\), contradicting the local-maximum condition \(f(\tfrac12)\leq f(0)\). The conflict uses continuity at \(0\), the positive derivative to its right, and the claimed local maximum. In fact, any one point \(y\) in the stated right-hand interval is enough to produce the contradiction.
Check: The equation gives \(f(\tfrac12)-f(0)>0\), whereas the claimed maximum requires \(f(\tfrac12)-f(0)\leq 0\). These inequalities cannot both hold.
The same reasoning applies to a local minimum with \(f'(x)<0\) immediately to the right: continuity at the minimum and the Mean Value Theorem imply that nearby right-hand values are smaller than \(f(c)\), contradicting the minimum. A derivative sign on the left can be tested in the same way, using an interval whose right endpoint is \(c\).
A Concrete Pitfall: Ignoring a Discontinuity
It is incorrect to conclude from \(f'(x)>0\) for \(x\ne 0\) near \(0\) alone that \(f(x)>f(0)\) for \(x>0\). Without continuity at \(0\), the function value there can break the apparent pattern.
Worked Example: A Local Maximum Despite Positive Derivatives Nearby
Define \(f(x)=x\) for \(x\ne 0\), and set \(f(0)=1\). For every \(x\ne 0\), the function is locally the identity function, so \(f'(x)=1>0\). Yet for \(0<|x|<\tfrac12\), \(f(x)=x<\tfrac12<1=f(0)\). Thus \(0\) is a strict local maximum.
Incorrect calculation: “Since \(f'(x)>0\) on both sides of \(0\), the function must increase through \(0\), so \(f(h)>f(0)\) for small \(h>0\).” This conclusion treats the derivative behavior away from \(0\) as if it controlled the possibly discontinuous value \(f(0)\). Here, for example, \(f(\tfrac14)=\tfrac14<1=f(0)\), directly disproving it.
Correction: To use the Mean Value Theorem between \(0\) and \(h\), one needs continuity on \([0,h]\), as well as differentiability on \((0,h)\). This function is not continuous at \(0\), so that application is unavailable. If continuity at \(0\) is added to the proposed constraints, the positive derivative to the right does contradict a local maximum, as the previous example showed.
Check: The derivative calculation is valid for \(x\ne 0\), and the displayed inequality verifies the local maximum directly. There is no contradiction once continuity at \(0\) is removed.
Nonzero Derivatives at Differentiable Extrema
Fermat's theorem gives another efficient contradiction test. If a sketch asserts both that \(f\) is differentiable at an interior point \(c\) and that \(c\) is a local extremum, then it must also be consistent with \(f'(c)=0\). A claimed nonzero derivative at that same point is enough to reject the combined claims.
Worked Example: A Claimed Minimum with a Nonzero Derivative
Suppose a proposed graph has a local minimum at \(x=2\), is differentiable there, and is labelled \(f'(2)=-3\). These constraints are incompatible.
Because \(x=2\) is an interior local minimum and \(f\) is differentiable there, Fermat's theorem requires
The derivative label instead gives \(f'(2)=-3\). Since \(-3\ne 0\), both statements cannot be true. The minimal conflicting pair of claims is “differentiable local minimum at \(2\)” and “\(f'(2)=-3\).” If differentiability is removed, a local minimum with a corner is possible; if the extremum claim is removed, a nonzero derivative is possible.
Check: The required value \(0\) and the stated value \(-3\) disagree, so the proposed graph cannot satisfy both claims.
Concavity Restricts the Order of Derivative Values
Second-derivative information can make derivative values at different points incompatible. If \(f\) is twice differentiable on an open interval containing \([a,b]\) and \(f''(x)>0\) throughout that interval, then \(f'\) is strictly increasing there. In particular, for \(a<b\), it must be true that \(f'(a)<f'(b)\).
To justify this, apply the Mean Value Theorem to \(f'\) on \([a,b]\). The hypotheses hold because \(f'\) is continuous on \([a,b]\) and differentiable on \((a,b)\). For some \(\xi\in(a,b)\),
Both factors on the right are positive: \(f''(\xi)>0\) by assumption and \(b-a>0\). Therefore \(f'(b)-f'(a)>0\), which proves \(f'(a)<f'(b)\). The interval hypothesis makes the needed continuity of \(f'\) at the endpoints explicit.
Worked Example: Increasing Concavity with a Decreasing Derivative
A proposed sketch claims that \(f''(x)>0\) on an open interval containing \([1,2]\), but also states \(f'(1)=3\) and \(f'(2)=1\). Test the constraints.
The Mean Value Theorem applied to \(f'\) gives some \(\xi\in(1,2)\) such that
But the endpoint values give
The same difference cannot be both positive and negative. Thus the positive second derivative and the two stated derivative values cannot all hold. The two derivative values alone are possible, and positive second derivative alone is possible; it is their combination that is inconsistent.
Check: The theorem requires \(f'\) to be continuous on \([1,2]\) and differentiable on \((1,2)\); these follow from the stated twice-differentiability on an open interval containing the closed interval. The computed difference \(-2\) contradicts the required positive difference.
Limits Must Be Consistent at the Same Endpoint
A final basic check concerns limiting behavior. A function cannot have two different limits as \(x\) approaches the same endpoint from the same direction. For example, a sketch cannot claim both \(\lim_{x\to\infty}f(x)=2\) and \(\lim_{x\to\infty}f(x)=5\). The direction matters: a left-hand limit and a right-hand limit at a finite point need not be equal.
Worked Example: Two Different Claimed Limits at Infinity
Suppose a proposed description states that \(\lim_{x\to\infty}f(x)=2\) and also that \(\lim_{x\to\infty}f(x)=5\). These two claims are incompatible.
By the first limit, there is a threshold beyond which \(|f(x)-2|<1\), so \(f(x)\) lies in the interval \((1,3)\). By the second, there is a threshold beyond which \(|f(x)-5|<1\), so \(f(x)\) lies in \((4,6)\). Choose \(x\) larger than both thresholds. Then \(f(x)\) would have to lie in both intervals, which is impossible because they do not overlap.
More generally, if two proposed finite limits are \(L_1\ne L_2\), use \(\varepsilon=|L_1-L_2|/3\). For sufficiently large \(x\), both limit claims would imply \(|f(x)-L_1|<\varepsilon\) and \(|f(x)-L_2|<\varepsilon\). The triangle inequality would then give
which is impossible since \(|L_1-L_2|>0\). Thus the two limits must agree if both exist.
Check: In the specific case, \((1,3)\cap(4,6)=\varnothing\). No value of \(f(x)\) can satisfy both eventual bounds.
How to Isolate the Smallest Conflict
When a sketch contains many claims, avoid rejecting it on the basis of a vague impression. Separate the information into statements and look for the shortest argument that forces incompatible conclusions. The smallest conflicting set is a collection of claims that cannot all hold, but becomes consistent as soon as any one of the claims in that collection is removed. “Smallest” here means minimal by removal of claims, not necessarily the set with the fewest words.
Record the point or interval, the derivative or limit information, and any continuity or differentiability assumptions that are stated.
Use Fermat's theorem for differentiable interior extrema, the Mean Value Theorem for derivative signs or ordered derivative values, and uniqueness of limits for competing endpoint claims.
Do not apply the Mean Value Theorem across a point of possible discontinuity. For its application to \(f'\), ensure the required continuity and differentiability of \(f'\).
Show the inequalities or equations that cannot both hold, then identify which claims produced them.
Ask whether removing one of the identified claims eliminates the contradiction. This distinguishes a direct conflict from unrelated information elsewhere in the sketch.
Why These Checks Matter
Derivative and limit labels are constraints, not decorative annotations. A single missing hypothesis can change the verdict: positive derivatives on both sides of a point do not settle the function value at that point if continuity has not been established. On the other hand, once the hypotheses are in place, a short theorem-based argument can rule out a sketch without needing to reconstruct the entire curve.
Use these checks locally and precisely. A contradiction between two claims does not imply that every other feature of the proposed curve is wrong. It identifies exactly what must be revised: perhaps the extremum label, the derivative value, the concavity statement, or the asserted limit. That precision is especially useful when the sketch includes many features that are individually plausible.
Check Your Understanding
For each question, identify the relevant assumptions as well as the conclusion.
- A function is continuous at \(c\), has a local minimum at \(c\), and satisfies \(f'(x)<0\) for every \(x\) just to the right of \(c\). Which theorem gives a contradiction?
- Can \(f'(x)>0\) on both sides of \(0\) coexist with a local maximum at \(0\) if continuity at \(0\) is not stated? Explain.
- A differentiable function has a local maximum at an interior point \(c\). What must \(f'(c)\) equal?
- If \(f''>0\) on an open interval containing \([a,b]\), where \(a<b\), which inequality must hold between \(f'(a)\) and \(f'(b)\)?
- Why can a function not have two distinct finite limits as \(x\to\infty\)?
