Why Critical Numbers of Polynomials Are Especially Direct
A polynomial is defined and differentiable at every real number. That makes finding its critical numbers relatively systematic: differentiate, solve an equation, and check the domain. The final check matters because a real solution of the derivative equation is a critical number of a restricted function only if it belongs to that function’s domain.
There is also a domain detail to keep clear. A polynomial may be considered on all of \(\mathbb{R}\), or it may be restricted to an interval. Under the two-sided derivative convention used here, the endpoints of a closed interval are points where the derivative of the restricted function does not exist. They therefore count as critical numbers by the definition below, even when the derivative of the polynomial formula at those endpoints is nonzero.
Definitions and the Endpoint Convention
A stationary number is always a critical number, but a critical number need not be stationary: the derivative might fail to exist there instead. For a polynomial defined on all of \(\mathbb{R}\), the derivative exists everywhere, so its critical numbers are precisely the real solutions of \(f'(x)=0\).
When a polynomial is restricted to an interval, we use the derivative of the restricted function and the usual two-sided derivative. At an included endpoint of a closed interval, there are not domain points on both sides, so this derivative does not exist. Such an endpoint is critical by the definition, whether or not the derivative of the polynomial formula—viewed as a function on all of \(\mathbb{R}\)—is zero there. An endpoint not included in the domain is not a critical number because it is not in the domain.
Some treatments list endpoints separately as candidates for extrema rather than calling them critical numbers. That is a difference in terminology. Here, the critical-number definition includes points in the domain where the two-sided derivative does not exist. Keeping the convention explicit prevents endpoint calculations from being mistaken for derivative-zero solutions.
The Polynomial Rule
Write a polynomial as \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\), where the coefficients are real numbers. The power rule and linearity of differentiation give
The result is another polynomial, and hence is defined for every real \(x\). To find its zeros, simplify or factor \(p'(x)\), solve \(p'(x)=0\) over the real numbers, and then retain only the solutions that belong to the function’s domain. For a restricted closed interval, also include its endpoints as critical numbers under the stated two-sided convention.
Why the Rule Works
The justification follows directly from the definition of a critical number and the differentiability of polynomials. For a polynomial on \(\mathbb{R}\), the derivative exists at every real number. Thus the “derivative does not exist” part of the definition never applies, leaving exactly the condition \(p'(c)=0\). On the interior of a restricted interval, there is a neighborhood on both sides within the domain, and the derivative of the restriction agrees with the derivative of the polynomial formula. At an included endpoint, however, the two-sided difference quotient cannot be formed using domain points on both sides; the derivative does not exist there. This proves the stated rule.
In more detail, at an interior point \(c\), sufficiently small positive and negative values of \(h\) keep \(c+h\) inside the interval. The difference quotient for the restricted function is then the same quotient as for the polynomial on \(\mathbb{R}\), so its limit is \(p'(c)\). At the left endpoint of a closed interval, negative \(h\) moves outside the domain; at the right endpoint, positive \(h\) does. Consequently, neither endpoint has a two-sided derivative as a point of the restricted domain. Each included endpoint meets the “derivative does not exist” condition in the definition.
A Reliable Procedure
Decide whether the polynomial is defined on all real numbers or restricted to an interval. Do not silently replace a restricted function by its polynomial extension.
Apply the power rule to each term and simplify the resulting polynomial.
Find all real solutions of \(p'(x)=0\). A complex solution is not a real critical number.
Keep derivative-zero solutions in the domain. For a closed interval, include its endpoints as critical numbers under the two-sided convention used here.
Substitute each proposed stationary number into the derivative, and confirm that each listed critical number lies in the domain.
Worked Examples
Worked Example: A Polynomial on All Real Numbers
Find the critical numbers of \(f(x)=2x^3-9x^2+12x-1\), with domain \(\mathbb{R}\).
Differentiate each term using the power rule:
Factor out \(6\), then factor the quadratic:
Set the derivative equal to zero. Since \(6\ne0\), the product is zero when \(x-1=0\) or \(x-2=0\), giving \(x=1\) or \(x=2\). Both real numbers are in the domain \(\mathbb{R}\). Check by substitution:
Therefore the critical numbers are \(1\) and \(2\). As a further check on the corresponding points of the graph, \(f(1)=2-9+12-1=4\) and \(f(2)=16-36+24-1=3\). The critical points are \((1,4)\) and \((2,3)\).
Worked Example: A Polynomial Restricted to a Closed Interval
Let \(q(x)=x^4-4x^2+1\) with domain \([-1,2]\). Find its critical numbers using the two-sided derivative convention.
First differentiate:
Factor the derivative:
The real solutions of \(q'(x)=0\) are \(x=0\), \(x=\sqrt{2}\), and \(x=-\sqrt{2}\). Since \(1<\sqrt{2}<2\), both \(0\) and \(\sqrt{2}\) lie in \([-1,2]\), but \(-\sqrt{2}<-1\) does not. The derivative-zero solutions in the domain are therefore \(0\) and \(\sqrt{2}\). Substitution verifies them:
The included endpoints \(-1\) and \(2\) must also be considered. The two-sided derivative of the restricted function does not exist at either endpoint, so both are critical numbers. In contrast, the polynomial extension has \(q'(-1)=4(-1)(1-2)=4\) and \(q'(2)=4(2)(4-2)=16\); these nonzero extension values do not change the endpoint conclusion for the restricted function. Thus the critical numbers of \(q\) on \([-1,2]\) are \(-1,0,\sqrt{2},2\).
As a domain check, \(-1\), \(0\), \(\sqrt{2}\), and \(2\) all belong to \([-1,2]\), while the excluded derivative-zero solution \(-\sqrt{2}\) does not. The endpoint values are \(q(-1)=1-4+1=-2\) and \(q(2)=16-16+1=1\), though evaluating the function is not needed to determine the critical numbers.
Worked Example: A Repeated Derivative Zero
Find the critical numbers of \(r(x)=x^3-3x^2+3x\) on \(\mathbb{R}\).
Differentiate term by term:
Factor the quadratic:
Solving \(3(x-1)^2=0\) gives \(x=1\). The factor is squared, but it still gives only one distinct critical number. Check it directly:
The domain is all of \(\mathbb{R}\), so \(1\) belongs to the domain. Hence the only critical number is \(1\). The corresponding function value is \(r(1)=1-3+3=1\), so the associated critical point is \((1,1)\).
Worked Example: No Real Derivative Zeros in an Open Interval
Let \(s(x)=x^3+3x\) with domain \((0,2)\). Find its critical numbers.
Differentiate:
For every real \(x\), \(x^2\geq 0\), so \(x^2+1\geq1\), and therefore \(s'(x)=3(x^2+1)\geq3>0\). Thus the derivative has no real zeros. The interval \((0,2)\) has no included endpoints, and the polynomial is differentiable at every point in the interval. Consequently there are no critical numbers. This conclusion has been checked both against the derivative condition and against the domain: neither endpoint belongs to the domain, and no interior point makes the derivative zero.
Common Mistakes and How to Avoid Them
A frequent error is to solve \(p'(x)=0\) and report every real solution without checking the domain. In the closed-interval example, \(-\sqrt{2}\) solves \(q'(x)=0\), but it is not in \([-1,2]\). Calling it a critical number of the restricted function would be incorrect. The repair is to compare each solution with the domain before listing it.
A second error is to confuse a zero of the function with a zero of its derivative. Critical numbers are found by solving \(p'(x)=0\), not \(p(x)=0\). For instance, in the first example, \(f(1)=4\), while \(f'(1)=0\). The fact that \(f(1)\ne0\) does not prevent \(1\) from being critical.
Finally, do not mix the derivative of a polynomial extension with the derivative of its restriction at an included endpoint. For \(q\) on \([-1,2]\), the extension has nonzero derivatives at both endpoints, but the restricted function has no two-sided derivative there. Under the definition adopted in this tutorial, the endpoints are critical numbers because they belong to the domain and their derivatives do not exist. When a problem instead asks for derivative-zero points, report only the solutions of \(p'(x)=0\) that lie in the domain; do not silently include endpoints in that different list.
Why Domain Checks Matter
Critical numbers are used when studying local behavior and locating possible extrema, but this tutorial’s calculation only identifies the numbers satisfying the critical-number definition. A derivative equation gives algebraic candidates; the domain determines which candidates belong to the function being studied. On a restricted interval, the endpoints also require attention because the function’s domain affects whether its derivative exists there.
Keep the two parts of the calculation separate: first find the real derivative zeros; then apply the stated domain and endpoint convention. This makes it clear whether a listed number is stationary because its derivative is zero, or critical for the distinct reason that the derivative does not exist.
Check Your Understanding
For each question, distinguish derivative-zero solutions from any endpoint critical numbers required by the stated convention.
- Find the critical numbers of \(u(x)=x^3-6x^2+9x\) on \(\mathbb{R}\).
- For \(v(x)=x^3-3x\) on \([-1,2]\), solve \(v'(x)=0\), check which solutions lie in the domain, and state all critical numbers.
- Why is a real solution of \(p'(x)=0\) outside the domain not a critical number of the restricted function?
- Under the two-sided derivative convention, why is an included endpoint of a closed interval a critical number even if the derivative of the polynomial extension there is nonzero?
- Can a polynomial on \(\mathbb{R}\) have a critical number where its derivative does not exist? Explain.