Why Nondifferentiable Points Need Attention
A derivative-based search for local extrema often begins by solving \(f'(x)=0\). But that equation does not find every possible extremum. A function can have a local minimum or maximum at a point where its derivative does not exist, provided the function itself is defined there.
At such a point, the graph may have a sharp corner or a cusp. The derivative is unavailable at the point, but its signs on either side can still show how the function behaves as it approaches and leaves that point. The first derivative test uses those signs to classify the point.
Local Extrema and Critical Numbers
Let \(f\) be defined on a domain \(D\), and let \(c\) be an interior point of \(D\). The function has a local minimum at \(c\) if there is an interval around \(c\) such that \(f(c)\leq f(x)\) for every \(x\in D\) in that interval. It has a local maximum at \(c\) if \(f(c)\geq f(x)\) for every such \(x\). The inequalities are strict for a strict local extremum when they hold strictly for nearby \(x\neq c\).
A point must belong to the function’s domain to be a local extremum. In particular, a hole in a graph cannot itself be a local maximum or minimum. This tutorial focuses on interior domain points; a domain endpoint has only one side of the domain nearby and requires separate one-sided reasoning.
Fermat’s theorem says that if \(f\) has a local extremum at an interior point \(c\) and is differentiable at \(c\), then \(f'(c)=0\). The differentiability condition matters: when the derivative does not exist, the conclusion \(f'(c)=0\) is not available. Thus, a complete search for interior extrema considers both stationary points and points where the function is defined but not differentiable.
The First Derivative Test at a Nondifferentiable Point
Suppose \(f\) is continuous at \(c\), is differentiable on each side of \(c\) sufficiently nearby, and \(c\) is an interior point of its domain. Determine whether \(f'(x)\) is positive or negative immediately to the left and right of \(c\). A positive derivative indicates that the function is increasing on that interval; a negative derivative indicates that it is decreasing.
| Sign of \(f'\) to the left | Sign of \(f'\) to the right | Conclusion at \(c\) |
|---|---|---|
| Negative | Positive | Local minimum |
| Positive | Negative | Local maximum |
| Positive | Positive | No extremum from this sign pattern |
| Negative | Negative | No extremum from this sign pattern |
The first two rows describe a change in direction: decreasing then increasing produces a valley, while increasing then decreasing produces a peak. If the derivative has the same strict sign on both sides, the function continues in the same direction through \(c\), so there is no local extremum there.
Why the Sign Test Works
The justification comes from the Mean Value Theorem, which says that if \(f\) is continuous on a closed interval and differentiable in its interior, then some point in the interior has derivative equal to the secant slope across the interval. In particular, a strictly negative derivative throughout an interval makes the function strictly decreasing there, and a strictly positive derivative makes it strictly increasing.
For the minimum case, choose a nearby \(x<c\). Since \(f'<0\) between \(x\) and \(c\), the function decreases as the input moves from \(x\) to \(c\), so \(f(x)>f(c)\). For a nearby \(x>c\), \(f'>0\) between \(c\) and \(x\), so \(f(x)>f(c)\) again. Thus \(f(c)\) is less than the values on both sides. The maximum case reverses both inequalities.
Check that \(c\) is in the domain and is an interior point. Verify that the function is defined at \(c\), even if its derivative is not.
Differentiate on each side where the derivative exists, then determine its sign immediately to the left and right of \(c\).
Negative to positive gives a local minimum; positive to negative gives a local maximum. The same strict sign on both sides gives no local extremum.
Evaluate \(f(c)\), then check that the nearby function values lie above or below it as the classification requires.
Worked Examples
Worked Example: A Corner That Is a Local Minimum
Classify the point \(x=2\) for \(f(x)=|x-2|+1\).
The function is defined at \(x=2\), and \(f(2)=|0|+1=1\). For \(x<2\), \(|x-2|=2-x\), so \(f(x)=3-x\) and \(f'(x)=-1\). For \(x>2\), \(|x-2|=x-2\), so \(f(x)=x-1\) and \(f'(x)=1\).
The derivative is negative on the left and positive on the right. The derivative at \(2\) does not exist: the left-hand slope is \(-1\), while the right-hand slope is \(1\). Therefore, the first derivative test gives a strict local minimum at \(x=2\), with value \(f(2)=1\).
Check directly: if \(x\neq 2\), then \(|x-2|>0\), so \(f(x)=|x-2|+1>1=f(2)\). This confirms the minimum.
Worked Example: A Corner That Is a Local Maximum
Classify the point \(x=-1\) for \(g(x)=3-|x+1|\).
At the point, \(g(-1)=3-|0|=3\). For \(x<-1\), \(|x+1|=-(x+1)\), so \(g(x)=3+(x+1)=x+4\), with derivative \(g'(x)=1\). For \(x>-1\), \(|x+1|=x+1\), so \(g(x)=3-(x+1)=2-x\), with derivative \(g'(x)=-1\).
The function is increasing to the left and decreasing to the right, so the sign changes from positive to negative. The derivative does not exist at \(-1\), because the left and right slopes differ. Thus \(g\) has a strict local maximum at \(-1\), with value \(3\).
Check: for \(x\neq -1\), \(|x+1|>0\), and hence \(g(x)=3-|x+1|<3=g(-1)\). The nearby values are indeed lower.
Worked Example: A Cusp That Is a Local Minimum
Classify \(x=1\) for \(h(x)=(x-1)^{2/3}\), using the real-valued interpretation of this power.
The function is defined at \(1\), and \(h(1)=0\). For \(x\neq 1\), the power rule gives \(h'(x)=\frac{2}{3}(x-1)^{-1/3}\). If \(x<1\), then \((x-1)^{1/3}<0\), so \(h'(x)<0\). If \(x>1\), then \((x-1)^{1/3}>0\), so \(h'(x)>0\).
The derivative does not exist as a finite number at \(1\). In the difference quotient, setting the input change to \(t\neq 0\) gives \(\frac{h(1+t)-h(1)}{t}=\frac{t^{2/3}}{t}=t^{-1/3}\), which is unbounded as \(t\) approaches zero. Since the derivative signs change from negative to positive, \(h\) has a strict local minimum at \(1\).
Check: for \(x\neq 1\), \((x-1)^{2/3}=(\sqrt[3]{x-1})^2>0=h(1)\). This confirms the minimum.
Worked Example: A Nondifferentiable Point That Is Not an Extremum
Determine whether \(x=0\) is a local extremum of \(p(x)=\sqrt[3]{x}\).
The function is defined at zero and \(p(0)=0\). For \(x\neq 0\), \(p'(x)=\frac{1}{3x^{2/3}}>0\), on both sides of zero. The derivative at zero does not exist as a finite number: the difference quotient is \(\frac{\sqrt[3]{t}-0}{t}=t^{-2/3}\), which grows without bound as \(t\) approaches zero.
There is no change in derivative sign. More directly, if \(x<0\), then \(\sqrt[3]{x}<0=p(0)\), and if \(x>0\), then \(\sqrt[3]{x}>0=p(0)\). Values occur both below and above \(p(0)\) arbitrarily close to zero, so zero is neither a local minimum nor a local maximum.
A Common Mistake
A common error is to assume that every local extremum must satisfy \(f'(c)=0\). For example, someone might write, “For \(f(x)=|x|\), the minimum occurs at \(x=0\); therefore \(f'(0)=0\).” The first part is correct, but the conclusion is not: \(f'(0)\) does not exist. The left-hand derivative is \(-1\), and the right-hand derivative is \(1\).
Another possible error is to label every nondifferentiable point as an extremum. The cube-root example shows why that fails: the derivative is positive on both sides, and the function keeps increasing through the nondifferentiable point. A point’s nondifferentiability makes it a candidate, not a conclusion.
Check Your Understanding
For each question, use the function’s domain and the behavior on both sides of the indicated point.
- For \(f(x)=|x-4|\), what is \(f(4)\), and is \(x=4\) a local minimum, a local maximum, or neither?
- For \(g(x)=-|x-3|\), determine the derivative signs on either side of \(3\), and classify the point.
- For \(h(x)=(x+2)^{2/3}\), identify the derivative sign on each side of \(-2\) and state the local-extremum conclusion.
- Why does \(p(x)=\sqrt[3]{x}\) have no local extremum at zero even though its derivative does not exist there?
- State the differentiability condition in Fermat’s theorem that prevents it from ruling out extrema at corners or cusps.
