Why Critical Numbers Matter
When searching for absolute extrema on a closed interval, you need a manageable set of points to examine. The earlier result on why absolute extrema occur at candidate points identifies interior points where the derivative is zero or undefined as candidates. This tutorial focuses on finding those points accurately.
The essential check is that a candidate must belong to the function’s domain. An expression that makes a derivative undefined does not automatically give a critical number: the original function must be defined there. Also, when working on a closed interval, interior critical numbers are distinct from the interval’s endpoints.
A critical number is an input, not the corresponding function value. If \(c\) is a critical number, \(f(c)\) is the function value at that input. In an absolute-extrema problem, you will later compare function values at the relevant candidates; here, the task is to identify the interior critical numbers.
Why These Points Are Candidates
The key result is Fermat’s theorem: if a function has a local maximum or minimum at an interior point \(c\), and \(f\) is differentiable at \(c\), then \(f'(c)=0\). This explains why solving \(f'(x)=0\) finds possible locations of interior extrema when the derivative exists. If the derivative does not exist, the point can still be a candidate, provided the function itself is defined there.
The differentiable case follows from the signs of the difference quotients. Suppose \(c\) is a local maximum and \(f\) is differentiable at \(c\). For sufficiently small positive \(h\), \(f(c+h)\leq f(c)\), so
For sufficiently small negative \(h\), the numerator is still nonpositive, but \(h\) is negative, so
Since the derivative exists, both one-sided limits equal \(f'(c)\). The first inequality makes the right-hand limit nonpositive, and the second makes the left-hand limit nonnegative. Their common value must therefore be zero. The argument for a local minimum reverses the inequalities and reaches the same conclusion. This result gives a necessary condition, not a guarantee: a point where \(f'(c)=0\) need not be an extremum.
A Reliable Search Procedure
Use the following procedure to find interior critical numbers on \([a,b]\). First determine where \(f\) is defined. Then differentiate on the parts of that domain where the derivative formula applies. Find inputs where the derivative equals zero, and identify domain inputs where the derivative does not exist. Finally, retain only points that lie in both the function’s domain and the interval’s interior.
Determine which inputs make the original function \(f(x)\) defined. Do this before treating any derivative-expression denominator or radical as a source of critical numbers.
Differentiate and solve the equation, keeping track of restrictions that apply to the derivative formula.
Look for corners, cusps, vertical tangents, or other failures of differentiability. Keep a point only if \(f\) itself is defined there.
For an absolute-extrema problem on \([a,b]\), retain only candidates in \((a,b)\). The endpoints \(a\) and \(b\) are checked separately.
Worked Examples
Worked Example: Finding Critical Numbers of a Polynomial
Find the interior critical numbers of \(f(x)=x^3-3x^2+2\) on \([0,3]\).
The polynomial is defined for every real \(x\). Differentiate using the power rule:
Factor the derivative and solve:
The interval’s interior is \((0,3)\). The value \(0\) is an endpoint, so it is not an interior critical number for this search. The value \(2\) lies in \((0,3)\), and the polynomial is defined there. Thus the interior critical number is \(2\).
Check: \(f'(2)=3(2)^2-6(2)=12-12=0\), as required. The input \(x=0\) also makes the derivative zero, but it remains an endpoint and is treated separately.
Worked Example: Filtering a Derivative Equation by the Domain
Find the interior critical numbers of \(g(x)=x+\frac{4}{x}\) on \([-3,3]\).
The original function is undefined at \(x=0\), so its domain is all real numbers except \(0\). For \(x\neq0\), differentiate:
Set the derivative equal to zero and solve, respecting \(x\neq0\):
Both values lie in the interval’s interior and in the domain. They are the interior critical numbers. The derivative expression is undefined at \(x=0\), but \(g(0)\) is also undefined, so \(0\) is not a critical number.
Check: \(g'(-2)=1-\frac{4}{(-2)^2}=1-1=0\), and \(g'(2)=1-\frac{4}{2^2}=1-1=0\). Both proposed critical numbers satisfy the derivative equation.
Worked Example: A Defined Point Where the Derivative Fails to Exist
Find the interior critical numbers of \(h(x)=|x-1|\) on \([0,2]\).
For \(x<1\), \(h(x)=1-x\), so \(h'(x)=-1\). For \(x>1\), \(h(x)=x-1\), so \(h'(x)=1\). Neither derivative is zero on its interval. At \(x=1\), the function is defined: \(h(1)=0\). But the left-hand derivative is \(-1\) and the right-hand derivative is \(1\), so the derivative does not exist at \(1\). Since \(1\in(0,2)\), it is an interior critical number.
Check: the two one-sided derivative values disagree, confirming that \(h'(1)\) does not exist; the function value \(h(1)=|1-1|=0\) confirms that the candidate belongs to the function’s domain.
Worked Example: Rejecting a Hole in the Function’s Domain
Find the interior critical numbers of \(q(x)=\frac{x^2-1}{x-1}\) on \([0,2]\).
The original expression is undefined at \(x=1\), so \(1\) is excluded from the domain. For \(x\neq1\), the quotient rule gives
This derivative is never zero where it is defined. At \(x=1\), the derivative formula is undefined, but the original function is undefined there too. Therefore \(1\) is not a critical number, and there are no interior critical numbers for this function on \([0,2]\).
Check: for every domain input, \(q'(x)=1\neq0\); the only input excluded by the derivative expression is \(1\), which is also excluded from the function’s domain. No interior input remains that meets either critical-number condition.
Common Pitfall: Solving Before Checking the Domain
A common error is to label every input that makes a derivative formula undefined as a critical number. For \(q(x)=\frac{x^2-1}{x-1}\), someone might write, “The derivative is undefined at \(x=1\), so \(x=1\) is critical.” That conclusion is wrong because \(q(1)\) does not exist. A critical number must be an input in the function’s domain.
The correction is to separate two questions: where does the derivative fail to exist, and is the original function defined at that input? Only points that pass the domain check can be critical numbers. Likewise, a derivative equation may produce values outside the interval being studied; those values are not interior critical numbers for that interval.
What to Keep Separate
When finding absolute extrema on \([a,b]\), do not combine the interior critical-number search with the endpoint check. The interior search considers points in \((a,b)\) where the function is defined and either the derivative is zero or it does not exist. The endpoints \(a\) and \(b\) are included in the later comparison because they belong to the closed interval, whether or not they satisfy the derivative conditions.
Also distinguish an undefined derivative from an undefined function. If the function is defined but has a corner at an input, that input can be an interior critical number. If both the function and derivative are undefined there, it cannot. This domain check is especially important for rational expressions, radicals, and formulas that have been simplified algebraically: cancellation does not restore an input that was excluded by the original formula.
Check Your Understanding
For each question, identify the interior critical numbers requested and explain any domain restriction.
- Find the interior critical numbers of \(f(x)=x^4-4x^2\) on \([-3,3]\).
- For \(g(x)=\sqrt{x-2}\) on \([1,5]\), what domain restriction must be considered before differentiating?
- For \(r(x)=\frac{x^2-9}{x-3}\), why can \(x=3\) not be a critical number of the original function?
- Can a point where \(f'(x)\) does not exist be a critical number? State the additional condition it must meet.
- When searching for interior critical numbers on \([a,b]\), why are \(a\) and \(b\) handled separately?
