Reading a Derivative Table as Evidence
A derivative table summarizes information about a function without displaying its graph or formula. Its entries can show where \(f'\) is positive or negative, where \(f''\) is positive or negative, or what these derivatives equal at selected inputs. The first task is to determine exactly what the table tells you: a sign listed for an entire interval is stronger evidence than a few values sampled at points.
When the sign of \(f'\) is known throughout an interval, it tells us whether \(f\) increases or decreases there. Changes in that sign around a point can classify a local extremum. Information about \(f''\) can also help, but only when its sign is established over an interval, not merely at a handful of inputs.
What the Derivative Signs Mean
Suppose \(f\) is differentiable on an interval. If \(f'(x)>0\) throughout that interval, then \(f\) is increasing there. If \(f'(x)<0\) throughout, then \(f\) is decreasing there. These facts let us translate a sign chart or interval-based derivative table into a description of \(f\)'s behavior.
A point \(c\) in the domain is a local maximum if \(f(c)\) is at least as large as the function values at all sufficiently nearby domain points. It is a local minimum if \(f(c)\) is at most as large as those nearby values. If \(f'\) changes from positive to negative at \(c\), the function changes from increasing to decreasing, supporting a local maximum. A change from negative to positive supports a local minimum.
Derivative tables may include \(f'(c)=0\), which makes \(c\) a critical point when \(c\) is in the domain. But that entry alone does not classify the point. The signs on either side are what show whether the function changes direction.
Why the Sign Rule Works
The reasoning follows from the Mean Value Theorem. If \(f'\) is positive throughout an interval, then for any two points \(a<b\) in it, the theorem gives a point \(d\) between them such that \(f(b)-f(a)=f'(d)(b-a)\). Both factors on the right are positive, so \(f(b)>f(a)\). Thus \(f\) is increasing. If \(f'\) is negative, the same equation gives \(f(b)<f(a)\), so \(f\) is decreasing.
For the local minimum case, suppose \(f'<0\) just left of \(c\) and \(f'>0\) just right of \(c\). For a nearby \(x<c\), the Mean Value Theorem on \([x,c]\) gives \(f(c)-f(x)=f'(d)(c-x)<0\), since \(d<c\). Therefore \(f(c)<f(x)\). For a nearby \(x>c\), applying the theorem on \([c,x]\) gives \(f(x)-f(c)=f'(d)(x-c)>0\). Therefore \(f(x)>f(c)\). In both directions, nearby values are greater than \(f(c)\), so \(c\) is a local minimum. Reversing the derivative signs proves the local maximum case.
Worked Example: Classifying Extrema from Interval Signs
Worked Example: Classifying Extrema from Interval Signs
Suppose \(f(x)=\frac{x^3}{3}-\frac{x^2}{2}-2x\) on \((-3,4)\). Use its derivative signs to find the intervals of increase and decrease and classify its critical points.
First differentiate:
The critical points satisfy \(f'(x)=0\), so \(x=-1\) and \(x=2\). The factors change sign at these points. Testing one input in each interval gives:
| Interval | Test input | \(f'(x)\) at test input | Behavior of \(f\) |
|---|---|---|---|
| \((-3,-1)\) | \(-2\) | \(({-2}+1)({-2}-2)=(-1)(-4)=4>0\) | Increasing |
| \((-1,2)\) | \(0\) | \((0+1)(0-2)=(1)(-2)=-2<0\) | Decreasing |
| \((2,4)\) | \(3\) | \((3+1)(3-2)=(4)(1)=4>0\) | Increasing |
At \(x=-1\), the sign changes from positive to negative, so \(f\) has a local maximum. At \(x=2\), it changes from negative to positive, so \(f\) has a local minimum.
As a check, differentiate the original function again: \(f''(x)=2x-1\). Then \(f''(-1)=-3<0\), consistent with a local maximum, and \(f''(2)=3>0\), consistent with a local minimum. The first-derivative sign changes establish the classifications; the second-derivative values provide a consistent check.
Using Second-Derivative Information in a Table
A table may give \(f''\) rather than directly list the signs of \(f'\) on either side of a critical point. If \(f''>0\) throughout an interval, then \(f'\) is increasing there; if \(f''<0\) throughout, then \(f'\) is decreasing there. This follows by applying the Mean Value Theorem to \(f'\). In particular, if \(f'(c)=0\) and \(f''>0\) on a neighborhood of \(c\), then \(f'\) is negative just to the left of \(c\) and positive just to the right. Those signs imply a local minimum of \(f\). If \(f''<0\) throughout the neighborhood, the signs are positive then negative, implying a local maximum.
The interval condition matters. A value \(f''(c)>0\) alone does not say that \(f''\) is positive throughout a neighborhood. The usual second-derivative test has its own conditions: \(f'(c)=0\), \(f''\) exists near \(c\), and \(f''(c)>0\) gives a local minimum, while \(f''(c)<0\) gives a local maximum. A table that only supplies scattered values should not be treated as proving an interval-wide sign.
Worked Example: Using a Positive Second Derivative on an Interval
Suppose a table establishes that \(f'(1)=0\) and \(f''(x)=2\) throughout \((0.5,1.5)\). Determine the behavior of \(f\) near \(x=1\).
Since \(f''(x)=2>0\) throughout the interval, \(f'\) is increasing there. For any \(x\) in \((0.5,1)\), the Mean Value Theorem applied to \(f'\) on \([x,1]\) gives a \(d\) between \(x\) and \(1\) such that
Because \(f'(1)=0\), this means \(f'(x)<0\). For \(x\) in \((1,1.5)\), apply the theorem on \([1,x]\):
so \(f'(x)>0\). Therefore \(f\) decreases just to the left of \(1\) and increases just to the right. It has a local minimum at \(x=1\).
For a numerical check consistent with these data, \(f(x)=(x-1)^2\) has \(f'(x)=2(x-1)\) and \(f''(x)=2\). Thus \(f'(0.5)=-1\), \(f'(1)=0\), \(f'(1.5)=1\), while \(f''\) is \(2\) at each input and throughout the interval. The derivative signs and the minimum at \(1\) agree with the conclusion.
When Point Samples Do Not Settle the Question
It is tempting to read a few entries in a table as though they describe every input between the listed values. That inference is not justified unless the table states a sign on the whole interval or other information supplies the missing guarantee. Even matching values of both derivatives at several sample inputs may leave the behavior near a critical point undecided.
Worked Example: Identical Sampled Derivatives, Different Extrema
Consider two functions, \(F(x)=\frac{x^4}{4}\) and \(G(x)=-\frac{x^4}{4}+\frac{2x^6}{3}-\frac{x^8}{4}\). Compare the derivative information at \(x=-1,0,1\), then decide whether those samples alone classify the extremum at \(0\).
For the first function,
Thus \(F'(-1)=-1\), \(F'(0)=0\), \(F'(1)=1\), and the corresponding second-derivative values are \(F''(-1)=3\), \(F''(0)=0\), \(F''(1)=3\). Since \(x^3\) is negative for \(x<0\) and positive for \(x>0\), \(F\) has a local minimum at \(0\).
For the second function, differentiating gives
and
At \(x=0\), \(G'(0)=0\) and \(G''(0)=0\). At \(x=1\), \(G'(1)=-1+4-2=1\) and \(G''(1)=-3+20-14=3\). At \(x=-1\), \(G'(-1)=1-4+2=-1\) and \(G''(-1)=-3+20-14=3\). So the sampled values of both derivatives match those for \(F\).
Yet near \(0\), the bracket \(1-2(1-x^2)^2\) is negative: it equals \(-1\) at \(0\) and is continuous, so it remains negative sufficiently close to \(0\). Consequently \(G'\) is positive just left of \(0\) and negative just right of \(0\). The function \(G\) has a local maximum at \(0\), opposite to \(F\). The same finite derivative samples therefore support two different classifications; they do not establish the signs between sample points.
A Reliable Table-Reading Process
Before drawing a conclusion, identify whether each table entry is a point value or an interval-wide sign. Then use the first derivative to determine increase and decrease, and inspect the signs on either side of each candidate point. Use second-derivative information only to the extent that its stated conditions allow.
Look for points in the domain where \(f'=0\) or \(f'\) is undefined, as well as any specified point whose behavior is being studied.
Check whether a sign applies throughout an interval or only to listed sample inputs.
Where \(f'>0\) throughout an interval, \(f\) increases; where \(f'<0\) throughout, \(f\) decreases.
Positive-to-negative supports a local maximum; negative-to-positive supports a local minimum. No sign change means this rule gives no extremum.
If signs between samples are not established, say that the table is insufficient rather than assuming a pattern continues.
A Common Mistake: Treating Samples as Interval Signs
Suppose a table shows \(f'(-1)<0\), \(f'(0)=0\), and \(f'(1)>0\). It is not enough to say that these three entries prove \(f\) decreases on \((-1,0)\) and increases on \((0,1)\). They report values at three inputs; they do not rule out extra zeros or sign changes between them. The second worked example demonstrates that even matching \(f''\) values at those same sample points may not settle the behavior.
The correction is to look for an explicit statement that \(f'\) has the given sign throughout each interval, or a valid theorem-based argument that establishes it. If neither is available, the warranted conclusion is that the table alone does not determine the intervals of increase and decrease or the local classification.
Check Your Understanding
Use only the information stated in each question, and distinguish interval-wide signs from point samples.
- If \(f'\) is negative throughout \((a,c)\) and positive throughout \((c,b)\), what does this establish about \(f\) at \(c\), assuming the required continuity and differentiability?
- A table gives \(f'(c)=0\) and \(f''(c)>0\), but no neighborhood information. What conclusion can you make from those entries alone?
- If \(f''>0\) throughout a neighborhood of \(c\) and \(f'(c)=0\), what are the signs of \(f'\) just to the left and right of \(c\)?
- Why can three sampled values of \(f'\) fail to establish that \(f\) is increasing or decreasing between the inputs?
- If \(f'\) is positive on both sides of a critical point, does the first-derivative sign rule classify a local extremum there?
