Why Sign Analysis Is Easy to Misread
A derivative sign chart can turn information about \(f'\) into conclusions about where \(f\) increases, decreases, and has local extrema. The calculations are often short, but a small error in interpreting the chart can reverse the conclusion or create an extremum where none exists.
The most reliable approach is to identify the function’s domain, divide it into the relevant intervals, determine the sign of \(f'\) on each interval, and only then interpret the changes in sign. In particular, the value \(f'(c)=0\) is a reason to investigate \(c\), not a conclusion about the graph.
What a Derivative Sign Chart Says
Suppose \(f\) is differentiable on an open interval. If \(f'(x)>0\) throughout that interval, then \(f\) is increasing there. If \(f'(x)<0\) throughout the interval, then \(f\) is decreasing there. A sign chart records these interval-by-interval conclusions.
For an interior point \(c\) in the domain, a local maximum means that \(f(c)\) is at least as large as every function value sufficiently close to \(c\). A local minimum means that \(f(c)\) is at most as large as every sufficiently nearby function value.
The continuity condition matters when \(f'\) does not exist at \(c\). If \(f\) is differentiable at \(c\), continuity there is automatic. The sign test concerns an interior point of the domain; an excluded point cannot be a local extremum of \(f\).
Why the Sign Changes Give the Conclusion
The reasoning comes from the Mean Value Theorem. If \(f'(x)>0\) throughout an interval, then for any two points \(x_1<x_2\) in that interval, the theorem gives a number \(z\) between them such that
Since \(x_2-x_1>0\), it follows that \(f(x_2)-f(x_1)>0\), so \(f(x_2)>f(x_1)\). Thus \(f\) is increasing. If \(f'<0\), the same argument gives \(f(x_2)<f(x_1)\), so \(f\) is decreasing.
For a positive-to-negative change at \(c\), the function increases toward \(c\) from the left and decreases away from \(c\) to the right. Continuity at \(c\) puts the value \(f(c)\) in the appropriate position between nearby values, giving a local maximum. A negative-to-positive change reverses this behavior and gives a local minimum.
Error 1: Checking Only the Critical Point
An interior critical number is a point in the domain where \(f'(c)=0\) or \(f'(c)\) does not exist. It is a candidate for an extremum, not proof that one occurs. To classify it using derivative signs, check the intervals on both sides.
Worked Example: A Zero Derivative Without an Extremum
Consider \(f(x)=x^3\). Its derivative is
Setting \(f'(x)=0\) gives \(3x^2=0\), so \(x=0\) is a critical number. Checking only \(f'(0)=0\) does not determine whether there is an extremum. Instead, use one test point in each interval divided by \(0\):
The factored expression \(3x^2\) is positive for every \(x\ne 0\), so \(f\) increases on both sides of \(0\). There is no change in derivative sign. The nearby function values also show why: \(f(-1)=-1\), \(f(0)=0\), and \(f(1)=1\). Values just to the left are below \(f(0)\), while values just to the right are above it. Thus \(0\) is neither a local maximum nor a local minimum.
Check: Differentiating \(x^3\) gives \(3x^2\), and the derivative is positive on both intervals \((-\infty,0)\) and \((0,\infty)\), as the test values indicate.
Error 2: Reversing the Sign-Change Conclusion
A common chart-reading mistake is to label a positive-to-negative change as a minimum. But positive derivative means increasing, and negative derivative means decreasing. Increasing and then decreasing makes the point a local maximum. Decreasing and then increasing makes it a local minimum.
Worked Example: Reading Both Sign Changes Correctly
Let \(f(x)=x^3-3x\). First calculate and factor the derivative:
The derivative is zero at \(x=-1\) and \(x=1\). These points divide the real line into three intervals. Test \(x=-2\), \(x=0\), and \(x=2\), respectively:
The derivative signs therefore change from positive to negative at \(-1\), and from negative to positive at \(1\). The first change gives a local maximum; the second gives a local minimum. To check the function values:
The value at the local maximum is \(2\), and the value at the local minimum is \(-2\). The incorrect reading would call \(-1\) a minimum because the sign changes there; the correction is to read the direction: \(f\) increases before \(-1\) and decreases after it, so \(-1\) is a maximum.
Check: The signs \(+,-,+\) agree with the increasing, decreasing, increasing behavior, and the function values \(2\) and \(-2\) match the classifications.
Error 3: Ignoring Domain Exclusions
A sign chart must respect the domain of the function. A point excluded from the domain can split the domain into separate intervals, even if algebraic cancellation makes a formula look simpler. An excluded point is not a point at which the function has a local extremum, because the function has no value there.
Worked Example: A Removable Hole Is Still Excluded
Consider
The domain excludes \(x=1\), since the original denominator is zero there. For \(x\ne1\), differentiate using the quotient rule:
The domain consists of the intervals \((-\infty,1)\) and \((1,\infty)\). Test \(x=0\) on the first and \(x=2\) on the second:
The function is increasing on each domain interval. But \(x=1\) is not an interior point of the domain, so it cannot be classified as a local maximum or minimum. Cancelling \(x-1\) gives \(f(x)=x+1\) only for \(x\ne1\); it does not put \(x=1\) back into the original domain. For instance, \(f(0)=1\) and \(f(2)=3\), but \(f(1)\) is undefined.
Check: The derivative is \(1\) at every point where the function is defined, and the original denominator still excludes \(x=1\).
Error 4: Treating One Test Value as the Whole Sign Chart
A test value gives the sign of \(f'\) at that particular value. It establishes the sign on an entire interval only when the derivative’s expression or other reasoning shows that the sign cannot change there. A reliable procedure is to find zeros and points where the derivative is undefined, partition the domain, and then determine the sign on each resulting interval.
For a factored derivative, factor signs can often establish the sign throughout an interval. For example, \(3(x-1)(x+1)\) is positive on \((-\infty,-1)\), negative on \((-1,1)\), and positive on \((1,\infty)\). Testing \(-2\), \(0\), and \(2\) confirms these signs; the factor analysis explains why each sign holds throughout its interval.
A Reliable Sign-Analysis Routine
Before classifying a point, use this sequence to catch the most common errors:
Mark points where \(f\) is undefined. They divide the domain, but an excluded point is not an extremum of \(f\).
Find interior points where \(f'=0\) or \(f'\) does not exist. These are candidates to investigate, not automatic extrema.
Partition at the relevant points and domain breaks. Use algebra and suitable test values to establish each interval’s sign.
Use \(+\to-\) for a local maximum and \(-\to+\) for a local minimum. The same sign on both sides gives no extremum from the sign test.
Confirm that the point belongs to the domain and that the conditions needed for the conclusion hold, including continuity at the point.
Check Your Understanding
For each question, focus on the interval signs and the domain rather than relying on a derivative value alone.
- If \(f'\) changes from negative to positive at an interior point, what type of local extremum does the First Derivative Test indicate?
- Why does \(f'(c)=0\) fail, by itself, to prove that \(f\) has an extremum at \(c\)?
- If \(f'\) is positive on both sides of an interior point and \(f\) is continuous there, what does the sign test conclude?
- Can a point excluded from the domain be a local extremum of the function? Explain briefly.
- What must be established before a test value can represent the sign of \(f'\) on an entire interval?
