No Stationary Points Proof (Calculus Guide)

To prove that a function has no stationary points, state its domain, find its derivative, and show that the derivative is never zero at any point in that domain. Check that the derivative exists where you test it, and reject any solutions outside the domain. A graph or a list of sample values alone cannot prove the claim.

A proof can be short, but each step matters. The most common mistake is solving (f'(x)=0) and forgetting that a solution may not belong to the function’s domain. Another is treating every critical point or endpoint as a stationary point. I use a simple check: define the domain, differentiate, test the whole domain, then state exactly what the result shows.

Problem Setup: Function Domain and Meaning of a Stationary Point

A stationary point is an interior point (c) of a function’s domain where the derivative exists and (f'(c)=0). In geometric terms, the curve has a horizontal tangent at (c). The domain tells you which values of (x) are allowed, so it must be clear before you solve for possible points.

For example, (f(x)=x^2) has domain (\mathbb R), and its derivative is (2x). Since (f'(0)=0), the point ((0,0)) is stationary. By contrast, (f(x)=x) has derivative (1), so it has no stationary point on (\mathbb R).

The definition depends on both the derivative and the domain. A value that solves (f'(x)=0) is only a candidate until you confirm that it is in the domain. A domain endpoint is not an interior point, so it is not a stationary point under the standard definition, even if an endpoint is where a maximum or minimum occurs.

Before differentiating, write down:

• The function (f(x)).
• Its domain, including excluded values.
• Whether the domain includes its endpoints.
• Where the derivative exists.

These details set the scope of the proof. A statement such as “there are no stationary points” should always mean “there are no stationary points on this specified domain.”

Differentiation and Domain-Sensitive Requirements

The derivative measures the function’s instantaneous rate of change. To establish that no stationary points exist, you must show that (f'(x)\neq 0) everywhere in the domain where the derivative exists. A sign proof is often enough: if the derivative is always positive or always negative, it cannot equal zero.

Differentiate using the relevant rules, then simplify the result. For a rational derivative, check both its numerator and denominator: the numerator may determine where the derivative is zero, while the denominator affects where the derivative exists.

For instance, consider [ f(x)=\arctan x,\qquad x\in\mathbb R. ] Its derivative is [ f'(x)=\frac{1}{1+x^2}. ] Since (1+x^2>0) for every real (x), the derivative is positive everywhere on the domain. Therefore (f'(x)) is never zero, and (f) has no stationary points on (\mathbb R).

A derivative that is undefined somewhere needs a separate check. Under the standard definition, a point where (f’) is undefined is not stationary, because a stationary point requires (f'(c)=0). Such a point may still be a critical point if it belongs to the domain. It could also be a domain boundary, depending on the function.

Check Every Candidate Against the Domain

When solving (f'(x)=0), keep the domain beside your work. A candidate outside the domain cannot be a stationary point, even if it solves an algebraic equation formed from the derivative.

Take (f(x)=x^2), but restrict its domain to ((2,\infty)). Then (f'(x)=2x), and solving (2x=0) gives (x=0). But (0\notin(2,\infty)), so there is no stationary point on the stated domain. In fact, (f'(x)>0) for every (x>2).

This is not the same as proving that (x^2) has no stationary points on its usual domain. On (\mathbb R), it does have one, at (x=0). Changing the domain changes the answer.

Proof Strategies, Common Errors, and Worked Verification

A proof of no stationary points must rule out (f'(x)=0) across the full domain. You can do this by showing the derivative has a fixed nonzero sign, or by solving the derivative equation and showing that none of its solutions are allowed. A numerical graph can suggest a result, but it cannot establish it for every point.

I usually start with the sign method because it gives a direct conclusion. If the derivative is a fraction, I check where the denominator is nonzero before using the numerator to test for zeros. If the derivative is a product or sum, factoring or comparing terms may reveal whether it can equal zero.

The table highlights an important rule: a derivative’s formula alone is not the whole proof. The domain decides which derivative values count.

Worked Proof: A Derivative That Is Always Positive

Let [ f(x)=x+\arctan x,\qquad x\in\mathbb R. ] Both terms are defined for every real number. Differentiating gives [ f'(x)=1+\frac{1}{1+x^2}. ] Because (1+x^2>0), the fraction is positive. Thus (f'(x)>1) for every real (x), so (f'(x)\neq0) throughout the domain. Therefore (f) has no stationary points on (\mathbb R).

The key is not that the graph appears to rise. The proof is the inequality that holds for every (x) in the domain.

Worked Proof: A Candidate Outside the Domain

Let [ g(x)=x^2,\qquad x\in(2,\infty). ] The derivative is (g'(x)=2x). The equation (2x=0) has the solution (x=0), but (0) is not in the domain. Also, for every (x>2), (2x>0). Therefore (g'(x)) never vanishes on ((2,\infty)), and (g) has no stationary points there.

This proof uses two checks: it rejects the algebraic candidate and confirms the derivative’s sign across the domain.

Common Proof Errors to Avoid

• Checking only part of the domain: A sign argument must cover every allowed (x), not just selected values.
• Ignoring a denominator: If the derivative is rational, identify where the denominator is zero. Do not claim the derivative exists there.
• Keeping an inadmissible candidate: A solution of (f'(x)=0) outside the domain is not a stationary point.
• Using a graph as proof: A plot may miss a point or hide behavior between plotted values.
• Confusing stationary and critical points: A derivative may be undefined at a critical point, but that point is not stationary under the standard definition.
• Forgetting endpoints: An endpoint may be a maximum or minimum without being a stationary point.

A complete argument names the domain, establishes where the derivative exists, and rules out every zero of the derivative within that domain.

Interpretation: Stationary Points, Extrema, and Applications

“No stationary points” is a precise claim about where the derivative is zero. It does not automatically mean “no critical points,” “no local or absolute extrema,” or “no endpoint extrema.” Those claims involve different conditions, so state only what your proof supports.

For example, (h(x)=|x|) on (\mathbb R) has no stationary point at (x=0), because its derivative is undefined there. Yet (x=0) is a critical point, and (h) has a minimum there. This shows why the distinction matters.

On the closed interval ([0,1]), the function (p(x)=x) has derivative (p'(x)=1) at every interior point, so it has no stationary points. Still, it has an absolute minimum at (0) and an absolute maximum at (1). Both extrema occur at endpoints, which are not interior stationary points.

For a function with (f'(x)>0) throughout an interval, the function is increasing there; if (f'(x)<0), it is decreasing there. These sign results help describe behavior, but the no-stationary-points conclusion follows specifically because the derivative never equals zero.

A Reliable Proof Checklist

Before submitting your work, confirm that you have:

• Stated the function’s full domain.
• Differentiated correctly and simplified the result.
• Identified any points where the derivative is undefined.
• Shown that (f'(x)\neq0) for every relevant domain point.
• Rejected any candidate zeros outside the domain.
• Limited your conclusion to stationary points, rather than making an unsupported claim about all critical points or extrema.

The final sentence can be simple: “Since (f'(x)\neq0) for every (x) in the stated domain, (f) has no stationary points on that domain.” The words “on that domain” make the conclusion exact.

Frequently Asked Questions

These questions clarify what a no-stationary-points proof must show and how to handle common edge cases. The short answers use the standard single-variable calculus definition: a stationary point is an interior domain point where the derivative exists and equals zero. Always state the domain, since it can change which candidates are admissible.

What is a stationary point?

A stationary point is an interior point (c) in the domain where (f'(c)=0). The derivative exists there, and the tangent to the graph is horizontal.

How do I prove that a function has no stationary points?

State the domain, find (f'(x)), and prove that it is never zero on that domain. A proof that (f'(x)>0) everywhere or (f'(x)<0) everywhere is sufficient.

Does a derivative that is always positive mean there are no stationary points?

Yes. If (f'(x)>0) at every point in the domain, then (f'(x)) cannot equal zero there. The same reasoning applies if (f'(x)<0) everywhere.

Is a point with an undefined derivative stationary?

No, not under the standard definition. A stationary point requires the derivative to exist and equal zero. A point with an undefined derivative may still be a critical point if it belongs to the domain.

Can a function have an endpoint maximum but no stationary points?

Yes. Endpoints are not interior points, so an endpoint maximum need not be stationary. For example, (f(x)=x) on ([0,1]) has endpoint extrema but no stationary points.

What if solving (f'(x)=0) gives a value outside the domain?

Discard that value as a stationary-point candidate. It may solve the derivative equation algebraically, but it cannot be a stationary point of the function on the stated domain.

Is a graph enough to prove there are no stationary points?

No. A graph can help you guess what happens, but it may miss behavior between plotted points or at a small scale. Use an analytic argument that covers the entire domain.

Does having no stationary points mean a function has no extrema?

No. A function can have extrema at endpoints or at points where its derivative is undefined. “No stationary points” only rules out interior points where the derivative exists and equals zero.

(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)