From a Derivative Sign to a Justified Conclusion
A derivative sign is evidence about how a function behaves: where \(f'(x)>0\), the function is increasing, and where \(f'(x)<0\), it is decreasing. To justify a local-extremum conclusion, identify the relevant critical number, state the signs on the intervals immediately to its left and right, and explain what the change in behavior means.
This is more than labeling a point from a sign chart. A complete justification makes clear which point is being classified, what the derivative does on both sides, and why those signs support a local maximum, a local minimum, or neither conclusion.
The local-extremum test used here requires continuity at the critical number. A local maximum at \(c\) means that \(f(c)\geq f(x)\) for all domain points \(x\) sufficiently close to \(c\); a local minimum means \(f(c)\leq f(x)\) for all such \(x\). Strict inequalities for nearby \(x\ne c\) give a strict local maximum or minimum.
The First Derivative Test
Suppose \(f\) is continuous at an interior critical number \(c\), and differentiable on the intervals immediately to either side of \(c\). If \(f'\) has a consistent, nonzero sign on each of those intervals, the sign change classifies \(c\).
| Sign to the left | Sign to the right | Behavior near \(c\) |
|---|---|---|
| Positive | Negative | Increasing, then decreasing: local maximum |
| Negative | Positive | Decreasing, then increasing: local minimum |
| Positive | Positive | Increasing on both sides: no local extremum |
| Negative | Negative | Decreasing on both sides: no local extremum |
The signs must describe intervals, not just a pair of isolated test points. For example, checking \(f'(a)>0\) at one point to the left of \(c\) does not by itself establish that \(f'\) stays positive throughout the interval up to \(c\). Factor the derivative, use a sign chart, or otherwise justify the sign across each interval.
Why the Sign Change Works
The reasoning comes from the Mean Value Theorem. In the form needed here, if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then for some \(z\in(a,b)\), \(f(b)-f(a)=f'(z)(b-a)\). Thus, if \(f'>0\) throughout an interval, then \(f(b)-f(a)>0\) whenever \(a<b\), so \(f\) is increasing there. If \(f'<0\), the same equation shows that \(f\) is decreasing.
For the maximum case, take a point \(x<c\) close to \(c\). Since \(f'>0\) on the left, the function increases toward \(c\), so \(f(x)<f(c)\). On the right, \(f'<0\), so the function decreases away from \(c\), giving \(f(x)<f(c)\) for nearby \(x>c\) as well. Continuity at \(c\) ensures that the function’s value at the joining point agrees with these one-sided behaviors. Therefore \(f(c)\) is a local maximum. Reversing the signs gives a local minimum. With the same sign on both sides, the function continues in the same direction through \(c\), so it does not turn there.
A Reliable Structure for a Justification
A concise written argument can follow four steps. It does not need to reproduce the full theorem proof each time, but it should include the facts that make the classification valid.
State where \(f'(c)=0\) or where \(f'(c)\) is undefined, and confirm that \(c\) is in the domain.
Specify intervals on either side of \(c\) that contain no other relevant break in the derivative’s sign.
State the sign of \(f'\) on each interval, with enough work to support those signs.
Describe the change in behavior and conclude whether \(f(c)\) is a local maximum or minimum, or whether the sign test gives no extremum.
Worked Examples
Worked Example: A Local Maximum and a Local Minimum
Classify the critical numbers of \(f(x)=x^3-3x\) using derivative signs.
Differentiate and factor:
Set \(f'(x)=0\). Since \(3(x-1)(x+1)=0\), the critical numbers are \(x=-1\) and \(x=1\). The factors change sign at these numbers, so examine the intervals \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\).
For \(x<-1\), both \(x-1\) and \(x+1\) are negative, so their product is positive and \(f'(x)>0\). For \(-1<x<1\), \(x-1<0\) and \(x+1>0\), so \(f'(x)<0\). For \(x>1\), both factors are positive, so \(f'(x)>0\).
At \(x=-1\), the derivative changes from positive to negative, so \(f\) changes from increasing to decreasing. Since \(f\) is a polynomial and hence continuous, \(x=-1\) is a local maximum. Its value is \(f(-1)=(-1)^3-3(-1)=-1+3=2\).
At \(x=1\), the derivative changes from negative to positive, so \(f\) changes from decreasing to increasing. Therefore \(x=1\) is a local minimum, with value \(f(1)=1^3-3(1)=1-3=-2\).
As a check, the derivative factorization gives \(f'(-1)=3(-2)(0)=0\) and \(f'(1)=3(0)(2)=0\), as required for these critical numbers. The conclusions are about the points \((-1,2)\) and \((1,-2)\) on the graph.
Worked Example: A Critical Number Where the Derivative Is Undefined
Classify \(c=0\) for \(f(x)=|x|\).
Write the function piecewise:
At \(x=0\), the left-hand derivative is \(-1\) and the right-hand derivative is \(1\). Since these one-sided derivatives are not equal, \(f'(0)\) does not exist. The function is defined at \(0\), so \(0\) is a critical number. Also, \(f\) is continuous there because both one-sided function values approach \(0=f(0)\).
For \(x<0\), \(f'(x)=-1<0\). For \(x>0\), \(f'(x)=1>0\). Thus the derivative changes from negative to positive, and \(f\) changes from decreasing to increasing. The first derivative test shows that \(x=0\) is a local minimum. Its value is \(f(0)=|0|=0\).
Check the conclusion directly: for every \(x\), \(|x|\geq 0=f(0)\), with strict inequality when \(x\ne0\). Thus the point \((0,0)\) is in fact a strict local minimum (and also an absolute minimum).
Worked Example: A Critical Number Without a Local Extremum
Use derivative signs to classify \(c=0\) for \(f(x)=x^3\).
Differentiate: \(f'(x)=3x^2\). The derivative is zero at \(x=0\), so \(0\) is a critical number. On each adjacent interval, \(x^2>0\), giving \(f'(x)=3x^2>0\) for \(x<0\) and for \(x>0\).
The derivative does not change sign: \(f\) is increasing on both sides. Therefore \(0\) is not a local maximum or a local minimum. Directly, \(f(0)=0\), while \(f(x)<0\) for \(x<0\) and \(f(x)>0\) for \(x>0\), so values near \(0\) occur both below and above \(f(0)\). This confirms that there is no local extremum at \((0,0)\).
As a calculation check, \(f'(0)=3(0)^2=0\), so the point is indeed critical even though it is not an extremum.
Writing the Conclusion Clearly
A good justification is compact but explicit. For example: “The critical number is \(c=1\), where \(f'(1)=0\). Since \(f'<0\) on \((0,1)\) and \(f'>0\) on \((1,2)\), \(f\) decreases before \(1\) and increases after \(1\). Because \(f\) is continuous at \(1\), the first derivative test shows that \(f(1)\) is a local minimum.” This connects the candidate, the signs, the behavior, and the conclusion.
If a problem asks for the location and value of an extremum, report both. The critical number is the input \(c\); the corresponding point on the graph is \((c,f(c))\), and the extremum value is \(f(c)\). These are related but are not interchangeable.
Common Pitfall: Stopping at the Critical Number
Finding \(f'(c)=0\) identifies a candidate, not a classification. For instance, the incorrect calculation “\(f'(x)=3x^2\), so \(f'(0)=0\); therefore \(x=0\) is a local minimum” skips the sign analysis and reaches a false conclusion for \(f(x)=x^3\). The correction is to check both sides: \(f'(x)>0\) for \(x<0\) and for \(x>0\). Since there is no sign change, the function increases on both sides and has no local extremum at \(0\).
Also check the conditions of the test. A sign argument based on intervals on either side applies to an interior point, and continuity at the critical number is needed for the stated first derivative test. If the function is not defined at the point, it is not a critical number under the definition above. Domain endpoints require separate consideration.
Check Your Understanding
For each question, focus on stating the critical number, the signs on both sides, and the conclusion those signs justify.
- If \(f'\) changes from positive to negative at an interior critical number and \(f\) is continuous there, what type of local extremum occurs?
- Why does \(f'(c)=0\) by itself fail to prove that \(c\) is a local extremum?
- For \(f(x)=x^4\), find the critical number at \(x=0\) and report the sign of \(f'\) on each adjacent interval. What does the sign test conclude?
- What condition on \(f\) at \(c\) is needed, in addition to the derivative signs, for the first derivative test as stated here?
- In a written justification, why should the derivative sign be reported on intervals rather than only at one test point on each side?
