What a Zero Derivative Tells You
A derivative at an input describes the function’s instantaneous rate of change there. Geometrically, it gives the slope of the tangent line to the function’s graph, provided the function is differentiable at that input.
When \(f'(a)=0\), the instantaneous rate is zero and the tangent has slope zero. Such a tangent is horizontal. This is useful information about the function at one input, but it is not enough to determine whether the function is at a local high point or a local low point. A horizontal tangent can occur at a local minimum, a local maximum, or neither.
The word stationary describes the derivative at the point: the instantaneous rate is zero. It does not, by itself, describe what the function does on either side of the point.
Local Maximum and Local Minimum
To say that a function has a local maximum or minimum at \(a\) is to compare its value at \(a\) with nearby values. The comparison is local: values far away from \(a\) do not matter.
A local maximum is not necessarily the largest value the function ever takes, and a local minimum is not necessarily the smallest value it ever takes. The definition only compares values sufficiently close to \(a\).
A Zero Derivative at a Local Minimum
Worked Example: A Horizontal Tangent at a Minimum
Consider \(f(x)=(x-2)^2+1\). First differentiate:
At \(a=2\), the function value and derivative are
The tangent line at \(x=2\) is therefore \(y=f(2)+f'(2)(x-2)=1+0(x-2)=1\), which is horizontal. To check the local behaviour, note that a square is nonnegative:
Thus \(x=2\) is a minimum (in fact, a global minimum). The derivative calculation is consistent with this: the tangent is horizontal at the bottom of the graph.
Check: Substitution gives \(f(2)=1\) and \(f'(2)=0\), as used to obtain the tangent line. The inequality above verifies the minimum directly.
A Zero Derivative Without a Local Extremum
Worked Example: A Horizontal Tangent That Is Neither a Maximum Nor a Minimum
Consider \(g(x)=(x-1)^3+4\). Differentiating gives
At \(a=1\),
So the tangent line there is \(y=4\), which is horizontal. But for \(x<1\), the quantity \(x-1\) is negative, so its cube is negative and \(g(x)<4\). For \(x>1\), its cube is positive and \(g(x)>4\). Values near \(1\) therefore occur both below and above \(g(1)\). The point is neither a local maximum nor a local minimum.
For concrete nearby checks, \(g(0)=(-1)^3+4=3<4\), while \(g(2)=1^3+4=5>4\). These two values illustrate the behaviour; the inequalities for all \(x<1\) and \(x>1\) establish it.
Check: Substituting \(x=1\) into the derivative gives \(g'(1)=0\), and the tangent line has slope zero. The comparison with \(4\) on both sides shows why that zero derivative does not prove an extremum.
A Zero Derivative at a Local Maximum
Worked Example: A Horizontal Tangent at a Maximum
Let \(q(x)=7-(x+1)^2\). Its derivative is
At \(a=-1\),
The tangent line is \(y=7\), so it is horizontal. For every \(x\), the square \((x+1)^2\) is nonnegative. Consequently,
This proves that \(x=-1\) is a maximum, in fact a global maximum. Here the function’s values nearby lie below the value at the stationary point.
Check: The derivative formula gives \(q'(-1)=0\), and substitution into the function gives \(q(-1)=7\). The inequality verifies that no value exceeds \(7\).
Zero as an Instantaneous Rate
The same idea applies when the function represents a changing quantity. If \(s(t)\) is position as a function of time, then \(s'(t)\) is instantaneous velocity. A zero value \(s'(a)=0\) means that the instantaneous velocity at time \(a\) is zero. It does not, on its own, say that the position has a local maximum or minimum at that time.
Worked Example: Zero Instantaneous Velocity Without a Position Extremum
Suppose position is \(s(t)=(t-2)^3+10\) metres, where \(t\) is measured in seconds. Differentiating with respect to time gives the velocity
At \(t=2\) seconds,
To see whether the position has a local extremum there, compare values on either side. At \(t=1\), \(s(1)=(-1)^3+10=9\) metres. At \(t=3\), \(s(3)=1^3+10=11\) metres. More generally, if \(t<2\), then \((t-2)^3<0\) and \(s(t)<10\); if \(t>2\), then \((t-2)^3>0\) and \(s(t)>10\). Thus the position at \(2\) seconds is neither a local maximum nor a local minimum.
Check: Differentiating the stated position gives \(3(t-2)^2\) metres per second, and substituting \(t=2\) gives zero. The units are consistent: differentiating a position measured in metres with respect to time in seconds gives metres per second.
Why a Local Extremum Requires a Zero Derivative
For a differentiable function, the implication in one direction does hold: an interior local maximum or minimum must have derivative zero. The point must be interior to the domain, so that inputs both smaller and larger than \(a\) are available. The function must also be differentiable at \(a\).
Here is the argument for a local maximum. By definition, for all sufficiently small \(h\), \(f(a+h)\leq f(a)\). For positive \(h\), the difference quotient is therefore nonpositive:
For negative \(h\), the numerator is still nonpositive but the denominator is negative, so the quotient is nonnegative:
Differentiability means that the two-sided limit of these difference quotients exists and equals \(f'(a)\). The quotients from the right are nonpositive, so their limit is at most zero; those from the left are nonnegative, so their limit is at least zero. Since both one-sided limits equal the same derivative, that derivative must be zero. For a local minimum, the inequalities reverse: the quotients for \(h>0\) are nonnegative and those for \(h<0\) are nonpositive. The same limit argument again forces \(f'(a)=0\).
The theorem does not say that every point with zero derivative is a local extremum. It says that, under its conditions, an extremum implies a zero derivative. The cubic example shows that the reverse implication fails.
Conditions and a Common Pitfall
The interior-point condition matters. For example, let \(r(x)=x\) on the domain \([0,1]\). The value \(r(0)=0\) is a minimum on this domain, but the derivative is \(r'(0)=1\). The point \(0\) is an endpoint, not an interior point, so inputs from both sides are not available and the theorem does not apply.
A related limitation is that the theorem assumes differentiability at the extremum. A function can have a local extremum where its derivative does not exist. Thus a zero derivative is a necessary condition for an interior local extremum only when differentiability is present; it is not a complete test for finding or classifying extrema.
For instance, in the cubic example the incorrect conclusion would be: \(g'(1)=3(1-1)^2=0\), so \(g\) has a local minimum at \(1\). The correction is to check the function values on both sides: \(g(x)<4=g(1)\) when \(x<1\), while \(g(x)>4=g(1)\) when \(x>1\). Since the nearby values are on both sides of \(g(1)\), there is no local extremum.
Why the Distinction Matters
A derivative value describes a local rate at one input. A local maximum or minimum describes how function values compare across a neighbourhood. These are different kinds of information. The derivative can be zero while values pass smoothly through the point, as in the cubic example; it can also be zero where nearby values lie all on one side, as in the quadratic examples.
When you encounter \(f'(a)=0\), the safe conclusion is that the tangent is horizontal and the instantaneous rate at \(a\) is zero. To decide whether \(a\) is a local extremum, you must examine how \(f(x)\) compares with \(f(a)\) for nearby \(x\). The zero derivative alone does not answer that question.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- If \(f\) is differentiable at \(a\) and \(f'(a)=0\), what is the slope of the tangent line there?
- For \(f(x)=(x-3)^2\), calculate \(f'(3)\) and explain why \(x=3\) is a local minimum.
- For \(h(x)=(x+2)^3\), calculate \(h'(-2)\). Does the zero derivative establish a local extremum? Explain by comparing nearby function values.
- State the assumptions needed for the theorem that a local extremum implies a zero derivative.
- Why does the theorem not rule out a minimum at an endpoint where the derivative is nonzero?
