First derivative test is conclusive for function with algebraic derivative

This article describes a situation, or broad range of situations, where a particular test or criterion is conclusive, i.e., it works as intended to help us determine what we would like to determine.
The test is first derivative test. See more conclusive cases for first derivative test | inconclusive cases for first derivative test

Statement

Single definition case

For any function of the following type, the first derivative test is always conclusive, i.e., it can be used to definitively determine whether the function has a local extreme value at a given critical point, and if so, what the nature of the local extreme value is:

• Nonconstant polynomial function on the real line, or on an interval or union of finitely many intervals in the real line
• Nonconstant rational function on the real line (whatever subset it's defined on), or on an interval or union of finitely many intervals in its maximum possible domain
• Nonconstant function defined on the real line, or on an interval or union of finitely many intervals in the real line, such that the derivative of the function is a rational function on its domain

Note that we omit constant functions from consideration because the derivative is identically zero and the analysis of local extreme values does not require the use of derivatives.

Piecewise definition case

• Function that has a piecewise definition by interval in terms of polynomials and rational functions, and is continuous at all transition points between intervals. (Note that if it is not continuous at a transition point, we'll need to use the variation of first derivative test for discontinuous function with one-sided limits)

Related facts

• First derivative test is conclusive for locally analytic function
• Higher derivative test is conclusive for function with algebraic derivative

Facts used

1. Nonzero polynomial function has finitely many zeros
2. Any finite set is a discrete set, and all points in that set are isolated
3. First derivative test is conclusive for differentiable function at isolated critical point

Proof

The proofs in all cases are essentially the same, but we present them separately to make them accessible to people who are not familiar with the more advanced cases.

Case of polynomial function on the real line

Given: A nonconstant polynomial function $f$, a critical point $c$ for $f$

To prove: The first derivative test is conclusive for $f$ at $c$. In other words, $f'$ has constant sign on the immediate left of $c$, and it has constant sign on the immediate right of $c$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The derivative $f'$ is a nonzero polynomial function.	(rules for polynomial differentiation)	$f$ is a nonconstant polynomial function		Direct. Note that the fact that $f$ is nonconstant forces $f'$ to not be identically zero.
2	$f'$ is defined everywhere and has finitely many zeros, so $f$ has finitely many critical points.	Fact (1)		Step (1)	Step-fact combination direct
3	All the critical points of $f$ are isolated critical points. In particular, $c$ is an isolated critical point for $f$.	Fact (2)		Step (2)	Step-fact combination direct
4	The first derivative test is conclusive for $f$ at $c$.	Fact (3)		Step (3)	Step-fact combination direct

Case of rational function on its maximum possible domain

Given: A (reduced form, i.e., no common factors between numerator and denominator) nonconstant rational function $f$, a critical point $c$ for $f$

To prove: The first derivative test is conclusive for $f$ at $c$. In other words, $f'$ has constant sign on the immediate left of $c$, and it has constant sign on the immediate right of $c$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The derivative $f'$ is a nonzero rational function.	(rules for polynomial differentiation, plus quotient rule for differentiation)	$f$ is a rational function		Direct. Note that the derivative is nonzero because the function is not constant.
2	$f'$ is defined everywhere on the domain of $f$ and has finitely many zeros, so $f$ has finitely many critical points given by the zeros of $f'$.	Fact (1)		Step (1)	[SHOW MORE] Note that, when we apply the quotient rule for differentiation on $f$, we don't obtain any new factors in the denominator (the general formula for the denominator is the square of the denominator of the original expression, though we might simplify and cancel a bit). In particular, this means that $f'$ is defined everywhere that $f$ is defined. Further, the zeros of $f'$ are precisely the zeros of its numerator (assuming we have the function in reduced form). The numerator is a polynomial function, so Fact (1) tells us that there are only finitely many zeros.
3	All the critical points of $f$ are isolated critical points. In particular, $c$ is an isolated critical point for $f$.	Fact (2)		Step (2)	Step-fact combination direct
4	The first derivative test is conclusive for $f$ at $c$.	Fact (3)		Step (3)	Step-fact combination direct

Case of function whose derivative is a rational function on its domain

The proof is similar to the rational function case, except that we don't have to do Step (1) of the proof above and can proceed directly from Step (2) onward.