Why the Derivative’s Sign Matters
A critical number can mark a local high point, a local low point, or neither. Knowing only that the derivative is zero or undefined at that number does not distinguish these possibilities. The key question for the first derivative test is how the derivative behaves immediately to the left and right.
If the function changes from increasing to decreasing, it reaches a local maximum. If it changes from decreasing to increasing, it reaches a local minimum. These conclusions require appropriate conditions on the function near the critical number. In particular, continuity at that number matters.
A critical number is an interior number \(c\) in the domain of \(f\) where \(f'(c)=0\) or \(f'(c)\) does not exist. The first derivative test examines the signs of \(f'\) on intervals next to a critical number; \(f'\) need not exist at \(c\) itself.
The First Derivative Test
Here, “just to the left” and “just to the right” mean throughout some open intervals ending or starting at \(c\). The signs must be established on those intervals, not merely at a few selected points. The theorem also applies when \(f'(c)\) is zero or does not exist, provided the stated continuity and differentiability conditions hold.
The reason the sign pattern works can be shown with the Mean Value Theorem. If \(f'>0\) on the left of \(c\), then for any nearby \(x<c\), continuity on \([x,c]\) and differentiability on \((x,c)\) allow the theorem to be applied. For some \(\xi\in(x,c)\),
Thus nearby values to the left are smaller than \(f(c)\). If \(f'<0\) on the right, apply the theorem on \([c,x]\) for nearby \(x>c\). For some \(\eta\in(c,x)\),
Nearby values to the right are also smaller than \(f(c)\). Together these inequalities prove that \(c\) is a strict local maximum. Reversing the signs reverses both comparisons and proves a strict local minimum. Continuity at \(c\) is part of what lets the Mean Value Theorem apply on intervals that have \(c\) as an endpoint.
Worked Example: A Maximum and a Minimum
Worked Example: A Maximum and a Minimum
Classify the critical numbers of \(f(x)=x^3-3x\) using the first derivative test.
Differentiate and factor:
The derivative is zero when \(x=-1\) or \(x=1\). On the intervals determined by these numbers, the signs are:
So \(f'\) changes from positive to negative at \(-1\), giving a local maximum there. It changes from negative to positive at \(1\), giving a local minimum there. The function is a polynomial, so it is continuous and differentiable at both numbers, as required.
Find the corresponding function values:
The local maximum is at \((-1,2)\), and the local minimum is at \((1,-2)\). The sign changes verify the classifications: the function rises and then falls at \(-1\), and falls and then rises at \(1\).
Worked Example: No Extremum When the Strict Sign Is Unchanged
Worked Example: No Extremum When the Strict Sign Is Unchanged
Determine whether \(g(x)=(x-2)^3\) has a local extremum at its critical number \(x=2\).
Differentiate:
For every \(x\neq 2\), the square \((x-2)^2\) is positive, so \(g'(x)>0\) on both sides of \(2\). Also, \(g'(2)=0\), and \(g\) is continuous at \(2\). The derivative has the same strict positive sign on both neighboring intervals.
To check directly, write a nearby input as \(x=2+h\). Then
If \(h<0\), then \(h^3<0\), so \(g(2+h)<g(2)\). If \(h>0\), then \(h^3>0\), so \(g(2+h)>g(2)\). Nearby values occur both below and above \(g(2)\), so \(x=2\) is neither a local maximum nor a local minimum. The unchanged strict positive sign agrees with that conclusion.
When the Sign Pattern Does Not Settle the Question
“No sign change” needs to be interpreted carefully. If the derivative is strictly positive on both sides, or strictly negative on both sides, and the continuity and differentiability conditions hold, then the point is not a local extremum. But a sign chart may include zeros on one side or may not establish a strict sign there. In that situation, the absence of a switch from positive to negative or negative to positive is not, by itself, enough to rule out an extremum.
Worked Example: A Sign Chart with Zeros on One Side
Consider the function
At \(c=0\), \(q(0)=0\). For \(x<0\), \(q'(x)=0\); for \(x>0\), \(q'(x)=2x>0\). The derivative is also defined at \(0\): the left difference quotient is \(0\), and the right difference quotient is \(h^2/h=h\), which tends to \(0\). Hence \(q'(0)=0\).
The derivative is never negative around \(0\), but its sign is zero on the left and positive on the right; it does not have the strict negative-to-positive pattern in the theorem. Nevertheless, \(q(x)\geq 0=q(0)\) for every \(x\) near \(0\), so \(0\) is a local minimum. It is not strict, because \(q(x)=q(0)\) for all \(x<0\). This example shows why a sign chart with zeros should not be treated as if it had established strict signs.
A different issue arises if continuity at \(c\) fails. Even a strictly positive derivative on both sides does not rule out a local extremum when the function is discontinuous at \(c\).
For a concrete counterexample, define \(r(x)=x\) for \(x\neq 0\), and \(r(0)=1\). For every \(x\neq 0\), \(r'(x)=1>0\), including on both sides of \(0\). But when \(0<|x|<1\), \(r(x)=x<1=r(0)\). Thus \(r\) has a strict local maximum at \(0\). The function is not continuous there, so the first derivative test’s hypotheses do not hold.
The corrected conclusion is precise: if \(f\) is continuous at \(c\), differentiable on intervals immediately to either side, and \(f'\) has the same strict sign on both sides, then \(c\) is not a local extremum. If those conditions or strict signs are not established, do not use that conclusion.
A Reliable Classification Process
When asked to justify a local extremum with the first derivative test, organize the reasoning so that the sign evidence and conclusion are explicit.
State the value \(c\) being tested and why it is a critical number.
Confirm that \(f\) is continuous at \(c\) and differentiable on intervals to either side. The derivative need not exist at \(c\).
Establish the sign of \(f'\) throughout each neighboring interval. Do not rely only on the value of \(f'(c)\).
Positive to negative gives a local maximum; negative to positive gives a local minimum. If strict signs are the same, the point is neither, under the conditions in the theorem.
This conclusion is local: it compares function values near \(c\), not across the entire domain. A local maximum need not be the greatest value of the function overall, and a local minimum need not be the least. The first derivative test classifies behavior near a critical number; it does not by itself determine absolute extrema.
Check Your Understanding
For each question, state the relevant derivative signs and any conditions needed for a valid conclusion.
- If \(f'\) changes from positive to negative at a critical number \(c\), what kind of local extremum does the first derivative test establish?
- What continuity condition is needed to apply the test at \(c\), and why does it matter?
- If \(f'\) is strictly negative on both sides of \(c\), under the theorem’s conditions, what can you conclude?
- Why does the sign pattern \(f'=0\) on one side and \(f'>0\) on the other not automatically rule out a local minimum?
- In the discontinuous example \(r(0)=1\), \(r(x)=x\) for \(x\neq 0\), why is the first derivative test not applicable at \(0\)?
