Why Absolute-Value Functions Need a Careful Candidate Test
For an ordinary differentiable function, an interior absolute maximum or minimum must occur at a point where the derivative is zero. An absolute-value function can have a sharp corner, however, and its derivative may not exist at a point where the expression inside the absolute value is zero.
That means checking only derivative-zero points can miss an extremum. For a function of the form \(h(x)=|g(x)|\), the zeros of \(g\) must be checked separately, along with derivative-based critical points and the endpoints of the interval.
How the Derivative Behaves
Let \(g\) be differentiable and define \(h(x)=|g(x)|\). Where \(g(x)>0\), the absolute value does not change the expression, so \(h(x)=g(x)\). Where \(g(x)<0\), it changes the sign, so \(h(x)=-g(x)\). Thus, at points where \(g(x)\ne 0\),
In either case, \(h'(x)=0\) exactly when \(g'(x)=0\). But this formula does not apply at a zero of \(g\): there, the absolute-value graph may have a corner. Such a zero remains a candidate even if \(h'\) does not exist.
In fact, at an interior point \(a\) where \(g(a)=0\), the function \(|g|\) is differentiable if and only if \(g'(a)=0\). If \(g'(a)\ne0\), its difference quotients from the right and left approach \(|g'(a)|\) and \(-|g'(a)|\), respectively, so the derivative does not exist. If \(g'(a)=0\), then \(g(a+t)/t\to0\), which implies \(|g(a+t)|/|t|\to0\); the derivative of \(|g|\) at \(a\) is therefore zero. Either way, the zero belongs in the candidate test.
Why These Candidates Are Sufficient
The candidate test follows from two standard results. The Extreme Value Theorem says that a continuous function on a closed, bounded interval attains both an absolute maximum and an absolute minimum. Fermat’s theorem says that if a function has a local maximum or minimum at an interior point where it is differentiable, its derivative there is zero.
Here is the justification. Since a differentiable function is continuous, \(g\) is continuous on \([a,b]\). The absolute-value function is continuous, so \(h=|g|\) is also continuous there. By the Extreme Value Theorem, \(h\) attains its absolute maximum and minimum on \([a,b]\).
If an extremum occurs at an endpoint, that endpoint is on the list. If it occurs at an interior zero of \(g\), that zero is on the list. Otherwise \(g(x)\ne0\) at the extremum. The derivative formula above shows that \(h\) is differentiable there. Fermat’s theorem gives \(h'(x)=0\), and the formula then gives \(g'(x)=0\). So every absolute extremum is included among the listed candidates.
This proves that the list is complete; it does not say that every listed point is an extremum. Evaluate \(h\) at the candidates and compare the values to decide which are the absolute maximum and minimum.
A Practical Candidate-Test Process
Include both endpoints in the candidate list.
Solve \(g(x)=0\) on the interval and include each solution.
Solve \(g'(x)=0\), keeping only interior solutions where \(g(x)\ne0\).
Substitute every candidate into \(|g(x)|\). The largest value is the absolute maximum; the smallest is the absolute minimum.
Worked Examples
Worked Example: A Sharp Corner and an Interior Derivative Candidate
Find the absolute maximum and minimum of \(h(x)=|x^2-4|\) on \([-3,3]\).
Set \(g(x)=x^2-4\). The endpoints are \(x=-3\) and \(x=3\). To find zeros, solve
Next, \(g'(x)=2x\), so \(g'(x)=0\) gives \(x=0\). Since \(g(0)=-4\ne0\), this is a derivative-based candidate. The complete candidate list is \(-3,-2,0,2,3\).
Evaluate the function at each candidate:
The largest candidate value is \(5\), attained at both endpoints; the smallest is \(0\), attained at \(x=-2\) and \(x=2\). Thus the absolute maximum is \(5\), and the absolute minimum is \(0\).
At \(x=2\), \(g'(2)=4\ne0\). The expression \(g\) changes sign there, so \(h\) has a corner: its left derivative is \(-4\), and its right derivative is \(4\). This illustrates why zeros must be checked even when the derivative of \(h\) does not exist.
As a direct check, the two endpoint values are \(5\), both zero values are \(0\), and the remaining candidate value is \(4\); these comparisons confirm the stated extrema.
Worked Example: Several Zeros and Tied Absolute Maxima
Find the absolute extrema of \(h(x)=|x^3-3x|\) on \([-2,2]\).
Let \(g(x)=x^3-3x\). Factor to find its zeros:
The endpoints are \(-2\) and \(2\). For derivative-based candidates,
At these two points, \(g(-1)=2\) and \(g(1)=-2\), so neither is a zero. The candidates are \(-2,-\sqrt{3},-1,0,1,\sqrt{3},2\).
Evaluate:
The absolute maximum is \(2\), attained at \(-2,-1,1,\) and \(2\). The absolute minimum is \(0\), attained at \(-\sqrt{3},0,\) and \(\sqrt{3}\). Substitution gives the candidate values shown, and comparing \(0\) with \(2\) verifies both conclusions.
Worked Example: No Zeros on the Interval
Find the absolute extrema of \(h(x)=|x^2+1|\) on \([-1,2]\).
Set \(g(x)=x^2+1\). Since \(x^2+1\ge1\), \(g\) has no zeros. The endpoints are \(-1\) and \(2\). Also \(g'(x)=2x\), so the only derivative-based candidate is \(x=0\), where \(g(0)=1\ne0\). Evaluate:
The absolute minimum is \(1\) at \(x=0\), and the absolute maximum is \(5\) at \(x=2\). The endpoint \(x=-1\) has value \(2\), which lies between them. These evaluations confirm the comparison of all candidates.
Common Mistake: Checking Only Where \(g'(x)=0\)
A tempting but incomplete approach is to solve only \(g'(x)=0\). For \(h(x)=|x^2-4|\) on \([-3,3]\), that approach gives only \(x=0\), with \(h(0)=4\). It misses \(x=-2\) and \(x=2\), where \(h\) equals \(0\). The omitted zeros are where the absolute minimum occurs.
The correction is to distinguish two kinds of interior candidates: zeros of \(g\), and points where \(g\ne0\) and \(g'=0\). Then add the endpoints. Do not assume every candidate is an extremum; compare the function values after the complete list has been formed.
Check Your Understanding
For each question, identify the candidates before deciding what information is needed to determine the extrema.
- Why must the zeros of \(g\) be checked when finding extrema of \(|g|\)?
- At an interior point where \(g(x)\ne0\), what condition on \(g'(x)\) is necessary for an extremum of \(|g(x)|\)?
- For \(h(x)=|x^2-1|\) on \([-2,2]\), list the endpoints, zeros, and derivative-based candidates.
- Does inclusion in the candidate list prove that a point is an absolute extremum? Explain.
- Which theorem ensures that \(|g|\) attains absolute extrema on a closed, bounded interval when \(g\) is differentiable?
