Why Critical Numbers Are Candidates, Not Conclusions
A derivative describes the local rate of change of a function. When an interior point is a local maximum or minimum and the derivative exists there, the derivative must be zero. This makes zeros of the derivative important when investigating a graph.
But that is only part of the picture. A function can have an extremum where its derivative does not exist, and a point where the derivative is zero need not be an extremum. A point where a formula has a break is different again: if the function is not defined there, that point is not a critical number.
The first step is therefore to check the function’s domain. Only after identifying where the function exists should you use the derivative to locate possible interior features.
If \(f'(c)=0\), the point \((c,f(c))\) is called a stationary point. If \(f'(c)\) does not exist but \(c\) is in the domain, it is still a critical number, but it is not stationary. A critical number is a location to investigate, not a guarantee that the graph turns there.
Why Differentiable Interior Extrema Have Zero Derivative
The reason stationary points are candidates for extrema is Fermat’s theorem. Its conditions 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.
Here is a proof. Suppose first that \(f\) has a local maximum at \(c\). For all sufficiently small \(h\), both \(c+h\) and \(c\) are in the domain, and \(f(c+h)\leq f(c)\). For positive \(h\), the difference quotient is therefore nonpositive. For negative \(h\), it is nonnegative.
Because \(f'(c)\) exists, the limits of these quotients from the right and left must agree. The right-hand limit is nonpositive, and the left-hand limit is nonnegative. Their common value can only be zero, so \(f'(c)=0\). For a local minimum, the inequalities reverse; the two-sided limit must again be zero. This proves the theorem.
The theorem explains why a differentiable interior extremum cannot be overlooked when finding critical numbers. It does not say that every critical number is an extremum. It also does not cover domain endpoints, which are not interior points. When finding extrema on a closed interval, inspect endpoints separately as well as interior critical numbers.
Worked Example: A Stationary Point That Is a Minimum
Worked Example: A Stationary Point That Is a Minimum
Find the critical numbers of \(f(x)=x^2-4x+5\), and determine whether the critical point is a local minimum.
The expression is a polynomial, so its domain is all real numbers. Differentiate:
Set the derivative equal to zero and solve:
The derivative exists everywhere, so \(x=2\) is the only critical number. Its function value is:
Thus the candidate point is \((2,1)\). To verify that it is a minimum, complete the square:
because \((x-2)^2\geq 0\) for every real \(x\), with equality at \(x=2\). Therefore \((2,1)\) is in fact a global minimum. As a check, differentiating the original expression gives \(2x-4\), which is zero at \(x=2\), and substituting \(x=2\) into the original function gives the stated height \(1\).
Critical Numbers Where the Derivative Does Not Exist
A critical number does not have to be a zero of the derivative. If \(c\) is in the domain but \(f'(c)\) does not exist, then \(c\) is also a critical number under the definition. Corners and cusps are familiar situations in which this can happen. Whether such a point is a maximum, a minimum, or neither must be checked separately.
Worked Example: A Nondifferentiable Point That Is a Minimum
Consider \(f(x)=|x-1|\). Find its critical numbers and determine the behavior at the point where its formula changes.
The absolute value is defined for every real \(x\), so the domain is \(\mathbb{R}\). Writing the function piecewise gives:
For \(x<1\), the derivative is \(-1\). For \(x>1\), the derivative is \(1\). At \(x=1\), compare the one-sided difference quotients. Since \(f(1)=0\), for \(h<0\) we have \(f(1+h)=-h\), while for \(h>0\) we have \(f(1+h)=h\). Thus:
The one-sided limits differ, so \(f'(1)\) does not exist. Since \(1\) belongs to the domain, it is a critical number. The function value is \(f(1)=|1-1|=0\). For every real \(x\), \(|x-1|\geq 0\), so this point is a global minimum. As a check, values on either side agree with this conclusion: \(f(0)=1\), \(f(1)=0\), and \(f(2)=1\). The point is a minimum even though it is not stationary.
Stationary Does Not Mean Turning
A derivative of zero tells you that the tangent is horizontal. By itself, it does not tell you whether the function has a local maximum or minimum. A stationary point can lie on a graph that continues through the point without turning.
Worked Example: A Stationary Point Without an Extremum
Consider \(f(x)=x^3\). Find its critical number and check whether the corresponding point is a local extremum.
The domain is \(\mathbb{R}\). Differentiating gives:
This derivative is zero when \(3x^2=0\), which gives \(x=0\). It exists everywhere, so \(0\) is the only critical number and is stationary. The function value is \(f(0)=0^3=0\).
For any \(\varepsilon>0\), however, the function values immediately to either side are:
There are therefore values below \(f(0)\) and values above \(f(0)\) arbitrarily close to \(x=0\). The point is neither a local maximum nor a local minimum. This verifies that a stationary critical number is a candidate for an extremum, not proof of one. As a check, differentiating \(x^3\) gives \(3x^2\), and substituting \(x=0\) gives \(f'(0)=0\), as required for a stationary point.
A Hole Is Not a Critical Number
The domain condition prevents a common mistake: treating a missing point in a formula as a critical number. Even if algebra simplifies a function on either side of a hole, simplification does not put the missing input back into the domain.
Worked Example: A Removable Hole in a Rational Function
For \(f(x)=\dfrac{x^2-1}{x-1}\), identify the domain break and find all critical numbers.
The denominator is zero when \(x-1=0\), so \(x=1\) is excluded. The domain is \(\mathbb{R}\setminus\{1\}\). On that domain, factor and simplify:
For every input in the domain, the derivative is \(f'(x)=1\). It is never zero, and it exists at every point in the domain. Therefore there are no critical numbers. The point \(x=1\) is not a critical number because \(f(1)\) is undefined. The simplified expression would give the limiting height \(2\), but that does not change the original function’s domain. For example, \(f(0)=1\), while \(f(2)=3\); neither substitution uses the excluded input. The derivative check is also direct: differentiating \(x+1\) on the domain gives \(1\), confirming that no domain point has derivative zero or undefined.
A Practical Procedure
Use the following order when locating critical numbers. Keeping the domain check first avoids counting points where the function does not exist.
Record inputs excluded by denominators, radicals, logarithms, or other restrictions in the original function.
Find a derivative formula that is valid for the inputs where the function is defined and differentiable.
Solve \(f'(x)=0\), then retain only solutions in the domain of \(f\).
Retain these as critical numbers only if the function itself is defined at those points.
When looking for extrema on an interval, include its endpoints as candidates. A critical number alone does not determine whether the graph has an extremum or a turning point.
Common Mistakes and What the Candidate List Means
Mistake: cancelling a factor and forgetting the original domain. In the rational example, it would be wrong to write \(f(x)=x+1\) for every real \(x\) and then claim that the original function is defined at \(x=1\). The cancellation is valid only when \(x\ne 1\). The corrected statement is \(f(x)=x+1\) for \(x\ne 1\); the hole remains, and \(1\) is not a critical number.
Mistake: assuming every critical number is an extremum. For \(f(x)=x^3\), \(f'(0)=0\), but values on one side of \(0\) are negative and on the other side are positive. The point is stationary without being a local extremum. A zero derivative is evidence to investigate, not a classification.
Mistake: ignoring a point because the derivative is unavailable. For \(f(x)=|x-1|\), the derivative does not exist at \(1\), but the function does. That makes \(1\) a critical number, and direct comparison shows it is a minimum. The test is not “derivative exists”; it is whether the point is in the function’s domain and either the derivative is zero or it does not exist.
Once the candidates have been found, a later step is needed to determine their behavior. First-derivative sign charts and sign changes provide one way to classify extrema, but the candidate list by itself does not perform that classification. The distinction is useful when organizing a curve analysis: it identifies which interior locations deserve attention, while keeping holes and endpoints in their proper, separate roles.
Check Your Understanding
For each question, pay attention to both the function’s domain and what the derivative information does—and does not—establish.
- State the definition of a critical number used in this tutorial. Why must the point belong to the function’s domain?
- For \(g(x)=x^2-6x\), find the critical number and its function value.
- Why is \(x=1\) not a critical number of \(\dfrac{x^2-1}{x-1}\), even though the simplified expression \(x+1\) has a value there?
- For \(h(x)=x^3\), explain why the critical number \(0\) is not a local maximum or minimum.
- What additional points must be considered when searching for extrema on a closed interval?
