Reading the First Derivative as Evidence
A graph or sign chart of \(f'\) describes how the function \(f\) changes. Where \(f'\) is positive, \(f\) increases; where \(f'\) is negative, \(f\) decreases. A change in the sign of \(f'\) can also provide evidence of a local maximum or minimum—but only when the conditions for the conclusion are met.
In a free-response answer, it is not enough to name an interval or label a point. State the derivative sign on the relevant interval, then connect that sign to the behavior of \(f\). For an extremum, identify the signs on both sides of the point and state that \(f\) is continuous there.
Increasing and Decreasing Intervals
Suppose \(f\) is differentiable on an interval. If \(f'(x)>0\) throughout that interval, then \(f\) is increasing there: as \(x\) increases, \(f(x)\) increases. If \(f'(x)<0\) throughout the interval, then \(f\) is decreasing there. A sign chart often splits the domain at zeros of \(f'\) or at points where \(f'\) is undefined. Read the sign separately on each resulting interval.
The sign at a single point is not enough to describe behavior on an interval. For example, \(f'(c)=0\) at a point \(c\) does not tell us whether \(f\) is increasing on either side. The signs just to the left and right are the relevant evidence for classifying an interior point.
Why the Sign of the Derivative Determines Behavior
The connection between derivative sign and increasing or decreasing behavior follows from the Mean Value Theorem. In the form needed here, if \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there is some \(c\) in \((a,b)\) such that
If \(f'(x)>0\) throughout an interval, choose any \(a<b\) in that interval. The theorem applies, \(f'(c)>0\), and \(b-a>0\). Therefore \(f(b)-f(a)>0\), so \(f(b)>f(a)\). This is exactly the definition of increasing behavior. If \(f'(x)<0\), the same equation gives \(f(b)-f(a)<0\), so \(f(b)<f(a)\), which is decreasing behavior.
This argument also explains what sign-chart evidence should look like in a written response: the sign of \(f'\) must hold on the interval being described, not merely at one test point. When a chart has a zero or an undefined value, use its sign regions on either side to determine the behavior.
Classifying Local Extrema from a Sign Change
An interior point \(c\) is a local maximum if \(f(c)\) is at least as large as the nearby values of \(f\); it is a local minimum if \(f(c)\) is at most as large as those nearby values. For a strict local maximum or minimum, the comparison is strict for nearby \(x\ne c\).
When \(f'\) is positive just to the left of \(c\) and negative just to the right, \(f\) increases toward \(c\) and decreases after \(c\). This supports a local maximum. When \(f'\) changes from negative to positive, \(f\) decreases toward \(c\) and increases after \(c\), supporting a local minimum. In both cases, continuity at \(c\) matters: without it, the value \(f(c)\) might not match the behavior on either side.
Find an interior point where \(f'(c)=0\) or \(f'(c)\) is undefined, or where the sign chart changes regions.
Record whether \(f'\) is positive or negative immediately to the left and right of \(c\).
If \(f\) is continuous at \(c\), a positive-to-negative change gives a local maximum; a negative-to-positive change gives a local minimum. No sign change means this test does not establish an extremum.
The sign-change test can be justified using the Mean Value Theorem on either side of \(c\). For a positive-to-negative change, take a nearby \(x<c\). Continuity on \([x,c]\) and differentiability on \((x,c)\) give a point \(\xi\) with \(f(c)-f(x)=f'(\xi)(c-x)>0\), so \(f(x)<f(c)\). For a nearby \(x>c\), the theorem on \([c,x]\) gives \(f(x)-f(c)=f'(\xi)(x-c)<0\), so \(f(x)<f(c)\). Thus \(f(c)\) is larger than nearby values on both sides. Reversing the signs proves the local-minimum case.
Worked Example: Reading a Sign Chart with Two Sign Changes
Worked Example: Reading a Sign Chart with Two Sign Changes
Suppose \(f(x)=\frac{x^3}{3}-\frac{x^2}{2}-2x\) on \([-2,3]\). Find the intervals where \(f\) increases or decreases and classify its interior local extrema.
First differentiate:
The critical points are \(x=-1\) and \(x=2\), where \(f'=0\). Test one value in each interval. For \(x=-\frac32\),
For \(x=0\), \(f'(0)=(1)(-2)=-2<0\). For \(x=\frac52\),
Thus \(f\) increases on \((-2,-1)\), decreases on \((-1,2)\), and increases on \((2,3)\). The sign changes from positive to negative at \(-1\), so \(f\) has a local maximum there. It changes from negative to positive at \(2\), so \(f\) has a local minimum there. The function is a polynomial, hence continuous at both points. As a numerical check, \(f(-1)=\frac76\) and \(f(2)=-\frac{10}{3}\); these values are consistent with the conclusions but the sign changes provide the local evidence.
Finally, differentiating the stated formula for \(f\) gives \(x^2-x-2\), confirming the derivative used in the sign analysis.
Worked Example: A Zero of the Derivative Without an Extremum
Worked Example: A Zero of the Derivative Without an Extremum
Let \(f(x)=\frac{x^4}{4}-\frac{5x^3}{3}+\frac{7x^2}{2}-3x\) on \([0,4]\). Determine the behavior around its critical points.
Differentiating and factoring gives
The derivative is zero at \(x=1\) and \(x=3\). For \(x=0\), \(f'(0)=(-1)^2(-3)=-3<0\). For \(x=2\), \(f'(2)=(1)^2(-1)=-1<0\). For \(x=\frac72\),
Therefore \(f\) decreases on \((0,1)\) and \((1,3)\), then increases on \((3,4)\). At \(x=1\), the derivative is negative on both sides, so there is no sign change and the first-derivative test gives no local extremum. At \(x=3\), the sign changes from negative to positive, and the polynomial is continuous there, so \(f\) has a local minimum. Differentiating the displayed polynomial for \(f\) reproduces the factored derivative, confirming the calculation.
This example shows why \(f'(c)=0\) is not enough to classify \(c\). At \(x=1\), the zero marks a point to inspect, but the sign evidence shows continued decrease rather than a reversal.
Worked Example: Why Continuity Is Required
Worked Example: Why Continuity Is Required
Consider the function
For \(x<0\), \(f'(x)=1>0\); for \(x>0\), \(f'(x)=-1<0\). A careless application of the sign-change test would call \(x=0\) a local maximum because the derivative changes from positive to negative. That conclusion is wrong: \(f\) is not continuous at \(0\), since the values on either side approach \(0\), while \(f(0)=-1\).
Check the actual local comparison. If \(-1<x<0\), then \(f(x)=x>-1=f(0)\). If \(0<x<1\), then \(f(x)=-x>-1=f(0)\). Thus all sufficiently nearby values other than \(f(0)\) are greater than \(f(0)\), so \(0\) is a strict local minimum, not a local maximum. The sign-change test cannot be used here because its continuity condition fails.
The correction is to check continuity at the point before using the sign change to assert a local extremum. Derivative signs still describe behavior on the intervals to either side, but they do not by themselves determine how the function value at a discontinuity compares with nearby values.
Writing the Evidence Clearly
A concise, complete response links each conclusion to its evidence. For intervals, state the sign of \(f'\) and the resulting behavior of \(f\). For a local extremum, report the change in sign on either side, the classification, and continuity at the point when it is part of the given information or can be established.
Avoid using a sign chart to make conclusions beyond its evidence. A zero of \(f'\) does not automatically imply a maximum or minimum. A sign change without continuity does not justify the usual extremum conclusion. Also distinguish a local conclusion, which compares values near one point, from a claim about the largest or smallest value on an entire domain.
Check Your Understanding
For each question, cite the derivative signs that support your conclusion, and check any condition needed to classify an extremum.
- If \(f'\) is positive on \((-2,1)\) and negative on \((1,4)\), what behavior and local-extremum conclusion are supported at \(x=1\), assuming \(f\) is continuous there?
- If \(f'\) is negative on both sides of \(x=3\) and \(f'(3)=0\), what can you conclude about a local extremum at \(x=3\) from the first-derivative sign test?
- If \(f'\) changes from positive to negative at \(x=c\), what additional condition on \(f\) is needed to conclude that \(c\) is a local maximum?
- What does \(f'(x)<0\) throughout an interval imply about \(f\) on that interval?
- Why can a positive-to-negative derivative sign change fail to indicate a local maximum when \(f\) is discontinuous at the point?
