Why Sign-Chart Errors Matter
A derivative sign chart is useful only when it represents the whole domain accurately. Missing one critical number can hide an extremum. Treating a point outside the domain as a candidate can invalidate a comparison. Assuming that a zero of the derivative must be a maximum or minimum can misclassify a stationary point. And even a correct sign chart does not replace comparing candidate values when the question asks for absolute extrema.
This tutorial focuses on checking those steps. The setting is an interval \([a,b]\), where \(a<b\), and a function \(f\) defined there. For an absolute maximum, one seeks a point \(x_{\max}\) such that \(f(x)\leq f(x_{\max})\) for every \(x\) in the domain. An absolute minimum is defined by reversing the inequality.
On a closed interval, the usual candidate list consists of the two endpoints and all interior critical numbers. This list is justified when \(f\) is continuous on \([a,b]\): the Extreme Value Theorem guarantees that absolute extrema are attained, and an interior extremum where \(f\) is differentiable must have derivative zero. Interior points where the derivative does not exist must also be checked.
Audit the Domain Before Making a Sign Chart
Start by identifying where the function is defined. A denominator that vanishes, a logarithm whose input is not positive, or another restriction may split the domain into separate intervals. Mark every such break before interpreting derivative signs. A point excluded from the domain may separate two sign-chart intervals, but it cannot be evaluated as a candidate value.
Within the domain, find every interior point where the derivative is zero or does not exist. Factor the derivative where possible, and solve the resulting equations completely. A sign chart should be divided at every such point and at every domain break. On each resulting interval, determine the sign of the derivative by testing a point or by using a justified factor-sign analysis.
Worked Example: A Critical Number That Changes the Absolute Maximum
Worked Example: A Critical Number That Changes the Absolute Maximum
Find the absolute extrema of \(f(x)=x^3-3x\) on \([-\frac{3}{2},\frac{3}{2}]\), and identify what goes wrong if one critical number is omitted.
Differentiate and factor:
Thus \(f'(x)=0\) at \(x=-1\) and \(x=1\). Both are interior points of the interval. The derivative is positive on \((-\frac{3}{2},-1)\), negative on \((-1,1)\), and positive on \((1,\frac{3}{2})\): for instance, its signs at \(-\frac{3}{2},0,\frac{3}{2}\) are respectively positive, negative, and positive.
Evaluate the function at both endpoints and both critical numbers:
Therefore, the absolute maximum is \(2\) at \(x=-1\), and the absolute minimum is \(-2\) at \(x=1\). As a check, these are the largest and smallest entries in the complete list \(\{\frac{9}{8},2,-2,-\frac{9}{8}\}\).
A faulty solution might solve \(x^2-1=0\) incompletely and record only \(x=1\). Its candidate values would then be \(\frac{9}{8},-2,-\frac{9}{8}\), leading to the incorrect claim that the maximum is \(\frac{9}{8}\). The correction is to solve for every root: \(x^2=1\) gives both \(x=-1\) and \(x=1\).
Why a Sign Change Matters
A critical number is a place to investigate, not automatic proof of an extremum. The signs of the derivative immediately to its left and right indicate the local direction of change. A change from positive to negative indicates a local maximum; a change from negative to positive indicates a local minimum. If the signs are the same, the function does not switch between increasing and decreasing there.
This interpretation follows from the Mean Value Theorem. If \(f\) is continuous on a closed subinterval and differentiable in its interior, then for its endpoints \(u<v\) there is a point \(r\in(u,v)\) such that \(f(v)-f(u)=f'(r)(v-u)\). Since \(v-u>0\), a positive derivative throughout an interval makes \(f(v)>f(u)\) for any two points \(u<v\) there; a negative derivative makes \(f(v)<f(u)\). Thus the sign on each side describes the direction of change there.
Worked Example: A Stationary Point with No Sign Change
Find the absolute extrema of \(f(x)=x^3\) on \([-1,2]\). Check whether its interior critical number is an extremum.
The derivative is \(f'(x)=3x^2\), which is zero at \(x=0\) and positive for \(x<0\) and \(x>0\). The sign chart must still be split at \(0\), even though the signs on the two sides agree. The function increases on both intervals, so the derivative sign does not indicate a local maximum or minimum at \(0\).
Evaluate all candidates, including the endpoints and the critical number:
The absolute minimum is \(-1\) at \(x=-1\), and the absolute maximum is \(8\) at \(x=2\). Substitution into \(f(x)=x^3\) confirms each listed value. In particular, including \(0\) as a candidate does not make it an extremum; comparison shows that its value is neither the greatest nor the least.
A common error is to declare \(x=0\) a local extremum solely because \(f'(0)=0\). That conclusion does not follow: the derivative is positive on both sides. The correction is to check both adjacent signs and, for absolute extrema, compare the values of all candidates.
Worked Example: A Domain Break Is Not a Candidate
Worked Example: A Domain Break Is Not a Candidate
Consider \(f(x)=\frac{1}{x-1}\) on the domain \([0,2]\setminus\{1\}\). Decide whether it has an absolute maximum or minimum, and explain how the domain break affects the sign chart.
The point \(x=1\) is excluded from the domain. For every other point in the domain,
The sign chart has two intervals, \([0,1)\) and \((1,2]\). The derivative is negative on each, so the function decreases on each interval. But \(x=1\) is not a critical number: \(f(1)\) is undefined, and the point is not in the domain. It cannot be entered in the candidate list or assigned a function value.
At the included endpoints, \(f(0)=-1\) and \(f(2)=1\). Those two values alone do not give absolute extrema. As \(x\) approaches \(1\) from the left, \(x-1\) is negative and approaches zero, so \(f(x)\) decreases without bound. From the right, \(f(x)\) grows without bound. Consequently, the function has neither an absolute minimum nor an absolute maximum on this domain.
The Extreme Value Theorem does not guarantee extrema here: the function is not continuous on the whole interval \([0,2]\), and indeed is not defined at \(1\). The correction to an endpoint-only calculation is to identify the domain break and examine what happens near it. The derivative sign chart describes monotonicity on each separate interval; it does not fill in the missing point.
Worked Example: Include a Nondifferentiable Interior Candidate
Worked Example: Include a Nondifferentiable Interior Candidate
Find the absolute extrema of \(f(x)=|x-1|\) on \([0,3]\). Show why checking only points where \(f'(x)=0\) would be inadequate.
For \(x<1\), \(f(x)=1-x\), so \(f'(x)=-1\). For \(x>1\), \(f(x)=x-1\), so \(f'(x)=1\). At \(x=1\), the left and right slopes differ, so \(f'(1)\) does not exist. Since \(1\) lies in the domain and is interior to the interval, it is a critical number and must be checked.
Evaluate both endpoints and the interior critical number:
The absolute minimum is \(0\) at \(x=1\), and the absolute maximum is \(2\) at \(x=3\). Substitution verifies all three values. The derivative changes from negative to positive at \(1\), matching the minimum there. If one checked only solutions of \(f'(x)=0\), the point \(1\) would be missed, because the derivative does not exist there.
Why the Closed-Interval Candidate List Is Complete
For a function continuous on \([a,b]\), the Extreme Value Theorem guarantees that it attains both an absolute maximum and an absolute minimum. Each attained extremum is either at an endpoint or at an interior point. If an interior extremum \(c\) is a point where \(f\) is differentiable, Fermat’s theorem gives \(f'(c)=0\). If \(f\) is not differentiable there, \(c\) is an interior critical number by definition.
The differentiable case in Fermat’s theorem can be seen directly. Suppose \(c\) is an interior local maximum and \(f'(c)\) exists. For small positive \(h\), \(f(c+h)-f(c)\leq 0\), so the right difference quotient is at most zero. For small negative \(h\), \(f(c+h)-f(c)\leq 0\), and dividing by \(h<0\) makes the left difference quotient at least zero. If their common limit \(f'(c)\) exists, it must be both nonpositive and nonnegative, hence \(f'(c)=0\). At an interior local minimum, the inequalities reverse, and the same conclusion follows. Therefore, every absolute extremum on the closed interval is among the endpoints and interior critical numbers.
This argument explains both why omitted candidates are dangerous and why domain assumptions matter. Continuity gives attainment on the whole closed interval; it cannot be invoked when the function is undefined or discontinuous there.
A Reliable Error-Checking Routine
Mark endpoints and every excluded point before solving derivative equations.
Find every in-domain point where the derivative is zero or does not exist. Keep excluded points marked as breaks, not candidates.
Determine the derivative sign on both sides of every critical number and on every separate domain interval. Do not infer a sign change merely from \(f'(c)=0\).
For a continuous function on a closed interval, evaluate both endpoints and every interior critical number. For a domain with breaks, investigate behavior near the breaks as well.
The key distinction is between identifying candidates and deciding which candidates actually give absolute extrema. A sign chart helps describe local direction; a complete value comparison determines the largest and smallest attained values. Keeping those tasks separate prevents many errors.
Check Your Understanding
For each question, explain the check that prevents the indicated error.
- For \(f'(x)=2(x+3)(x-1)\) on \([-4,2]\), which interior numbers must split the sign chart?
- If \(f'(c)=0\) and the derivative is positive on both sides of \(c\), what can you conclude about a sign change there?
- Why can a point excluded from the domain split a sign chart but not be included as a function-value candidate?
- For a continuous function on \([a,b]\), which points must be checked when finding absolute extrema?
- Why is comparing endpoint values alone not enough for a continuous function on a closed interval?
