Why Critical Points Are Easy to Misread
A critical point is an important clue when analyzing a curve, but it is not a conclusion about the curve’s behavior. A zero derivative does not automatically mean that the graph turns. A derivative that does not exist does not automatically mean that a point is an extremum. And a number obtained while solving a derivative equation is not a candidate at all unless it belongs to the function’s domain.
These distinctions matter because critical points are used to organize curve analysis. The useful question is not simply, “Where does the derivative vanish?” It is, “Which points in the domain require further inspection, and what does the function actually do near them?”
The requirement \(c\) belongs to the domain is essential: a graph has no point \((c,f(c))\) when \(f(c)\) is undefined. Also, the definition identifies candidates, not necessarily local or global extrema.
A Zero Derivative Is Not Enough
A horizontal tangent means the derivative is zero at that point. It does not tell us whether the function changes from increasing to decreasing, from decreasing to increasing, or continues in the same direction. For example, the graph of \(f(x)=x^3\) has a horizontal tangent at the origin but passes through it while increasing.
Worked Example: A Critical Point That Is Not an Extremum
Consider \(f(x)=x^3\), defined for every real \(x\). Differentiate:
Set the derivative equal to zero:
Since \(0\) is in the domain and \(f'(0)=0\), it is a critical number. Its graph point is \((0,f(0))=(0,0)\). To check whether it is an extremum, compare values on either side. For every \(h>0\),
There are values below and above \(f(0)\) arbitrarily close to \(x=0\). Thus \(f(0)\) is neither a local maximum nor a local minimum. In fact, \(f'(x)=3x^2>0\) for \(x\ne 0\), so the function is increasing on both sides. The derivative test has found a point to inspect, not an extremum.
Check: Substituting \(x=0\) into the derivative gives \(f'(0)=3(0)^2=0\), as required for the critical-number calculation.
What Fermat’s Theorem Actually Says
There is a useful theorem connecting differentiable local extrema with zero derivatives. Its hypotheses matter: the point must be an interior point of the domain, the function must have a local extremum there, and the derivative must exist there.
This is a necessary condition, not a sufficient one. It says that a differentiable local extremum in the interior must have zero derivative. It does not say that every zero derivative gives an extremum. It also does not rule out extrema where the derivative fails to exist, or extrema at the boundary of a restricted domain.
The short proof shows why the interior and differentiability conditions appear. Suppose \(f\) has a local maximum at \(c\). For sufficiently small nonzero \(h\), the points \(c+h\) remain in the domain and \(f(c+h)\leq f(c)\). If \(h>0\), then
If \(h<0\), dividing the same inequality by a negative number reverses its direction, so
Because \(f'(c)\) exists, both one-sided limits of these difference quotients equal the same number. The limit from the right is at most zero, while the limit from the left is at least zero. Therefore their common value must be zero: \(f'(c)=0\). For a local minimum, the inequalities reverse, and the same conclusion follows.
This argument cannot be applied at a boundary point if there are not domain points on both sides. It also cannot be applied when the derivative does not exist. Those are precisely reasons to check such points separately rather than force them into the zero-derivative rule.
Points Where the Derivative Does Not Exist
A point can be in the domain even though the derivative there does not exist. Such a point is a critical-number candidate under the definition above. It may be an extremum, or it may not be. The function’s behavior must still be examined.
Worked Example: An Undefined Derivative at a Minimum
Let \(f(x)=|x|\), with domain \(\mathbb{R}\). For \(x<0\), \(f(x)=-x\), so \(f'(x)=-1\). For \(x>0\), \(f(x)=x\), so \(f'(x)=1\). At \(x=0\), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\); since they differ, \(f'(0)\) does not exist.
The number \(0\) is in the domain, so it is a critical number even though its derivative is undefined. To decide whether it is an extremum, note that for every \(x\),
Thus \(x=0\) is a global minimum, and hence also a local minimum. This example shows why looking only for solutions of \(f'(x)=0\) can miss an extremum.
Check: The function value at the candidate is \(f(0)=|0|=0\), and the inequality above verifies directly that no domain value is smaller.
Worked Example: An Undefined Derivative Without an Extremum
Consider \(g(x)=\sqrt[3]{x}\), defined on \(\mathbb{R}\). For \(x\ne 0\), differentiation gives
This formula is undefined at \(x=0\). The derivative at zero does not exist as a finite number: for \(h\ne 0\),
and this quotient grows without bound as \(h\) approaches zero. Hence \(0\) is a critical number. But \(g(-1)=-1<g(0)=0<g(1)=1\), and more generally the cube-root function is increasing through zero. There are both smaller and larger function values arbitrarily near the candidate, so it is neither a local maximum nor a local minimum.
Check: At the candidate, the difference quotient is \(1/h^{2/3}\), not a finite limit; the derivative is indeed undefined. The values on opposite sides confirm that the point is not an extremum.
Together, these examples show both directions of the warning: an undefined derivative can occur at an extremum, as for \(|x|\), or at a point that is not an extremum, as for \(\sqrt[3]{x}\). Undefinedness alone does not settle the question.
Every Candidate Must Belong to the Domain
Derivative algebra is performed on the part of the real line where the original function is defined and differentiable. Solving an equation obtained from a derivative expression does not restore excluded inputs. In particular, a denominator equal to zero is not a valid derivative value, and cancellation in a formula does not fill a hole in the original function’s domain.
Worked Example: Rejecting a Candidate Outside the Domain
Consider
The original denominator is zero at \(x=1\), so the domain is \(\mathbb{R}\setminus\{1\}\). For \(x\ne 1\), use the quotient rule:
The cancellation is valid only when \(x\ne 1\), the existing domain restriction. The derivative is never zero on the domain and exists at every point of the domain. Therefore there are no critical numbers. The apparent solution \(x=1\), obtained by setting the unsimplified numerator \((x-1)^2\) equal to zero, is not a critical number: \(f(1)\) is undefined.
Indeed, for every \(x\ne 1\), the function simplifies to \(f(x)=x+1\), a line with a missing point at \(x=1\). The missing point is not on the graph and cannot be an extremum of this function.
Check: At any allowed input, such as \(x=0\), the original formula gives \(f(0)=(-1)/(-1)=1\), while \(f'(0)=1\). The simplified expression gives \(0+1=1\) and derivative \(1\) as well. At \(x=1\), the original formula has denominator zero, confirming that this input must be excluded.
A Reliable Check for Common Errors
When finding critical points, keep the function’s domain visible throughout the calculation. A useful sequence is to identify the domain, find where the derivative is zero or undefined, and then retain only inputs that belong to the domain. Finally, inspect what the function does near each retained point before calling it an extremum.
Use the original function to determine which inputs are allowed. Do not let later algebra erase an exclusion.
Locate in-domain points where \(f'(x)=0\) and where the derivative does not exist.
Verify that the function is defined there. For a restricted domain, also check boundary points when extrema are being sought.
Use the function’s nearby values or an appropriate derivative test. Do not infer an extremum merely from a zero or undefined derivative.
A common incorrect calculation is: “\(f'(x)=0\) at \(x=c\), therefore \(f\) has a maximum or minimum at \(c\).” The conclusion does not follow. For \(f(x)=x^3\), \(f'(0)=0\), but values immediately to the left are negative and values immediately to the right are positive. The correction is: “\(c\) is a critical number and must be checked; the derivative equation alone does not classify it.”
A second error is accepting every algebraic root of a derivative numerator. In the rational example, setting \((x-1)^2=0\) produces \(x=1\), but the derivative quotient and the original function are not defined there. The correction is to test each proposed input against the original domain before including it in the critical set.
Why the Distinctions Matter
Critical points help organize curve analysis because they mark locations where the usual derivative information may change or fail. But the definition is deliberately broad: it gathers points that need checking. It does not replace the check. Confusing a candidate with a conclusion can produce false extrema; ignoring points of nondifferentiability can omit real extrema; and ignoring the domain can add points that are not on the graph at all.
Keep three statements separate:
- A zero derivative at a point in the domain makes that point a critical-number candidate.
- An undefined derivative at a point in the domain also makes it a critical-number candidate.
- A critical number is not automatically a local or global extremum.
Fermat’s theorem supplies one necessary condition for an interior differentiable extremum. The other candidates and the boundary of the domain require their own inspection. This careful separation makes curve analysis reliable rather than mechanical.
Check Your Understanding
For each question, distinguish the candidate calculation from the conclusion about extrema.
- For \(f(x)=x^4\), find the critical number. Does the derivative equation alone prove that it is a local minimum?
- For \(f(x)=|x-2|\), identify the point where the derivative does not exist and state whether it is an extremum.
- Why does Fermat’s theorem require an interior point where the derivative exists?
- For \(q(x)=\frac{x^2-9}{x-3}\), why must \(x=3\) be excluded when checking derivative candidates?
- Can a critical number fail to be a local extremum? Give an example and explain the nearby function values.
