Why the Endpoints and Nondifferentiable Points Matter
A derivative can reveal where a differentiable function has a horizontal tangent, but that is not the whole search for absolute extrema. The largest or smallest value on a closed interval may occur at an endpoint, or at an interior point where the derivative does not exist.
For example, a graph may turn sharply at a corner, form a cusp, or have a vertical tangent. Each of these can occur at an absolute maximum or minimum even though the derivative is not zero there. A reliable method therefore builds a candidate list first and compares the function’s values at every point on that list.
The Candidate-Point Rule
Let \(f\) be continuous on the closed interval \([a,b]\). The Extreme Value Theorem guarantees that \(f\) attains an absolute maximum and an absolute minimum on this interval. Any point where either absolute extreme occurs must be an endpoint, or an interior point where \(f'(x)=0\), or an interior point where \(f'(x)\) does not exist.
- the endpoints \(a\) and \(b\);
- interior points \(c\) where \(f'(c)=0\);
- interior points \(c\) where \(f'(c)\) does not exist.
The condition that \(f\) is continuous on \([a,b]\) matters: it ensures that absolute extrema exist. The theorem does not say that every candidate is an extreme point. It says that any absolute extreme must occur somewhere on the list. A candidate may turn out to be neither a maximum nor a minimum.
Endpoints are listed separately from interior critical points. The derivative condition is not required at an endpoint: the endpoint is a candidate because it is the boundary of the interval, whether or not \(f'\) exists there.
Why Interior Candidates Include Nondifferentiable Points
At a differentiable interior maximum or minimum, Fermat’s theorem implies that the derivative is zero. The essential requirements are that the point is interior and the derivative exists. If the derivative does not exist, the theorem does not rule out an extremum.
Here is the reason the derivative must be zero at a differentiable interior extremum. Suppose \(f\) has a local maximum at an interior point \(c\) and is differentiable there. For nearby positive \(h\), the difference quotient is nonpositive; for nearby negative \(h\), it is nonnegative:
Because \(f'(c)\) exists, both one-sided limits of these quotients equal the same number. The right-hand quotients can approach that number only from values at most zero, while the left-hand quotients can approach it only from values at least zero. Thus the common limit is zero, so \(f'(c)=0\). At a local minimum the two inequalities reverse, and the same conclusion follows.
This argument gives a necessary condition for a differentiable interior extremum; it is not a condition that covers every extremum. A corner, cusp, or vertical tangent can prevent the derivative from existing, so such points must be added to the candidate list.
A Practical Search-and-Compare Process
Confirm that the function is defined and continuous throughout the closed interval \([a,b]\). This guarantees that absolute extrema exist.
Include \(a\) and \(b\), even if the derivative exists there or is nonzero there.
Include every interior point where \(f'(x)=0\) and every interior point where \(f'(x)\) does not exist.
Substitute each candidate into \(f\). The largest resulting value is the absolute maximum value; the smallest is the absolute minimum value.
A point belongs on the list because of its location or derivative behavior, not because it is already known to be an extreme. The comparison step decides which candidates actually give the greatest or least values. If two or more candidates tie, the absolute maximum or minimum occurs at each point giving that tied value.
Worked Examples
Worked Example: A Corner Gives the Absolute Minimum
Find the absolute extrema of \(f(x)=|x-1|\) on \([0,3]\).
The function is continuous on the closed interval. The endpoints are \(x=0\) and \(x=3\). For \(x<1\), the expression inside the absolute value is negative, so \(f(x)=1-x\) and \(f'(x)=-1\). For \(x>1\), \(f(x)=x-1\) and \(f'(x)=1\). The derivative changes from \(-1\) to \(1\) at \(x=1\), so the graph has a corner there and the derivative does not exist at that interior point.
The candidate list is \(0,1,3\). Evaluate the function at each point:
The smallest candidate value is \(0\), and the largest is \(2\). Therefore the absolute minimum value is \(0\) at \(x=1\), and the absolute maximum value is \(2\) at \(x=3\). Checking by substitution, \(f(1)=0\) and \(f(3)=2\), as claimed.
Worked Example: A Cusp Is a Candidate but Not an Absolute Maximum
Find the absolute extrema of \(g(x)=|x|^{2/3}\) on \([-1,1]\).
The function is continuous on \([-1,1]\), including at \(x=0\). Its endpoints are \(x=-1\) and \(x=1\). For \(x\ne 0\), differentiating gives
As \(x\) approaches \(0\) from the left, the derivative decreases without bound; from the right, it increases without bound. Thus the graph has a cusp at \(x=0\), and \(g'(0)\) does not exist. Include \(0\) along with both endpoints.
Evaluate all three candidates:
The absolute minimum value is \(0\) at \(x=0\), and the absolute maximum value is \(1\), attained at both \(x=-1\) and \(x=1\). Substitution verifies \(g(0)=0\) and \(g(-1)=g(1)=1\). The cusp is a minimum here, but nondifferentiability alone does not tell us whether a candidate is a maximum or a minimum; we needed to compare values.
Worked Example: A Vertical Tangent Need Not Give an Absolute Extreme
Find the absolute extrema of \(h(x)=\sqrt[3]{x}\) on \([-1,8]\).
The cube-root function is continuous on this closed interval. The endpoints, \(-1\) and \(8\), are candidates. For \(x\ne0\),
At \(x=0\), the derivative would require the limit
This quotient grows without bound as \(t\) approaches zero from either side, so there is no finite derivative at \(0\). The graph has a vertical tangent there, and \(0\) must be included. Since \(h'(x)\) is never zero for \(x\ne0\), the candidate list is \(-1,0,8\).
Evaluate:
The absolute minimum value is \(-1\) at \(x=-1\), and the absolute maximum value is \(2\) at \(x=8\). Substitution confirms \(\sqrt[3]{-1}=-1\) and \(\sqrt[3]{8}=2\). The vertical-tangent point \(x=0\) is a candidate, but its value \(0\) is neither the largest nor the smallest.
Common Mistake: Looking Only for Where \(f'(x)=0\)
A frequent error is to search for derivative-zero points, find none, and conclude that there are no absolute extrema. Consider again \(f(x)=|x-1|\) on \([0,3]\). Away from \(x=1\), the derivative is either \(-1\) or \(1\), so solving \(f'(x)=0\) produces no points. The incorrect conclusion would be: “There are no candidates, so there are no absolute extrema.”
That conclusion misses the endpoints and the corner. The corrected list is \(0,1,3\), with values \(1,0,2\), respectively. Therefore the absolute minimum is \(0\) at \(x=1\), and the absolute maximum is \(2\) at \(x=3\). The derivative-zero test applies only to differentiable interior extrema; it does not replace checking endpoints or nondifferentiable interior points.
Why the Comparison Step Matters
The candidate list narrows the search, but it does not classify each point by itself. A corner can be a minimum, as in the absolute-value example, or a nondifferentiable point can fail to be either absolute extreme, as in the cube-root example. Similarly, an endpoint may or may not give the largest or smallest value.
The final comparison is therefore essential. Use function values, not just derivative information, to decide the absolute extrema. On a closed interval where the function is continuous, evaluating all candidates and choosing the greatest and least values completes the method.
Check Your Understanding
For each question, focus on which points belong in the candidate list and how their function values determine the answer.
- Why must both endpoints be tested when finding absolute extrema on a closed interval?
- What two kinds of interior points belong in the candidate list?
- For \(f(x)=|x-2|\) on \([0,4]\), list the candidates and evaluate \(f\) at each one.
- Does a point where the derivative does not exist have to be an absolute extremum? Explain.
- What role does continuity on the closed interval play in the candidate method?
